Real Analysis

1. To check continuity at $x=0$, we need to see if $\lim_{x \to 0} f(x) = f(0)$.
2. Note that $f(0) = 0$ because $0$ is rational.
3. For any $x$, the value of $f(x)$ is either $x$ or $-x$.
4. In both cases, $|f(x) - 0| = |x|$.
5. As $x$ approaches $0$, $|x|$ approaches $0$ regardless of whether $x$ is rational or irrational.
6. Therefore, the limit is $0$, which equals $f(0)$.

Answer: Yes, it is continuous at $x = 0$.

1. Recall the definition of differentiability: the limit of the difference quotient must exist.
2. Consider the function $f(x) = |x - 1/2|$.
3. This function is continuous everywhere on $[0, 1]$ because it is the composition of continuous functions.
4. However, at $x = 1/2$, the function has a sharp corner.
5. The left-hand derivative is $-1$ and the right-hand derivative is $1$.
6. Since the one-sided limits of the difference quotient are not equal, the derivative does not exist.

Answer: No, the function $f(x) = |x - 1/2|$ is a counterexample.

1. Polynomials are smooth and have a finite number of terms.
2. Consider the Taylor series for the exponential function $e^x = \sum_{k=0}^{\infty} \frac{x^k}{k!}$.
3. Let $f_n(x)$ be the $n$-th partial sum of this series, which is a polynomial of degree $n$.
4. As $n \to \infty$, $f_n(x)$ converges to $f(x) = e^x$ for all $x$.
5. While each $f_n$ is a polynomial, the limit function $e^x$ is not a polynomial (it has infinitely many non-zero derivatives).

Answer: No, the limit of polynomials can be a non-polynomial function like $e^x$.

1. We start with an arbitrary $x$.
2. If $x \neq 0$, we can choose $y = 1/x$, then $x(1/x) = 1$.
3. However, if $x = 0$, we need to find $y$ such that $0 \cdot y = 1$.
4. Since $0 \cdot y = 0$ for all $y$, no such $y$ exists for $x = 0$.

Answer: False

1. The phrase 'There is a real number' indicates an existential quantifier: $\exists y \in \mathbb{R}$.
2. The phrase 'greater than every other real number' indicates a universal quantifier for the other numbers: $\forall x \in \mathbb{R}$.
3. The condition is $y > x$ (though strictly 'every other' implies $x \neq y$, in analysis we usually write $y > x$ for all $x$ to check for a maximum).

Answer: $\exists y \in \mathbb{R}, \forall x \in \mathbb{R} : y > x$

1. For (A), given $x$, we can choose $y = 5 - x$. Then $x + (5 - x) = 5$. This works for all $x$, so (A) is true.
2. For (B), we must pick one $y$ first. If we pick $y = 2$, then $x + 2 = 5$ only when $x = 3$. It fails for $x = 4$.
3. No single $y$ can make $x + y = 5$ for every possible $x$.

Answer: (A) is True, (B) is False

1. Flip the first quantifier: $\exists L$ becomes $\forall L$.
2. Flip the second quantifier: $\forall \epsilon > 0$ becomes $\exists \epsilon > 0$.
3. Flip the third quantifier: $\exists \delta > 0$ becomes $\forall \delta > 0$.
4. Flip the fourth quantifier: $\forall x$ becomes $\exists x$.
5. Negate the implication: $|x-a| < \delta \implies |f(x)-L| < \epsilon$ becomes $|x-a| < \delta$ and $|f(x)-L| \ge \epsilon$.

Answer: $\forall L, \exists \epsilon > 0, \forall \delta > 0, \exists x, (|x-a| < \delta \text{ and } |f(x)-L| \ge \epsilon)$

1. The original statement is $\forall x \in \mathbb{R}, \exists y \in \mathbb{R}, x^2 + y^2 = 1$.
2. Flip $\forall x$ to $\exists x$.
3. Flip $\exists y$ to $\forall y$.
4. Negate the equation $x^2 + y^2 = 1$ to $x^2 + y^2 \neq 1$.

Answer: There exists a real number $x$ such that for all real numbers $y$, $x^2 + y^2 \neq 1$.

1. The original statement is $\exists f, \forall x, \forall y, (x < y \implies f(x) < f(y))$.
2. Flip $\exists f$ to $\forall f$.
3. Flip $\forall x$ to $\exists x$.
4. Flip $\forall y$ to $\exists y$.
5. Negate the implication: $x < y \implies f(x) < f(y)$ becomes $x < y$ and $f(x) \ge f(y)$.

Answer: For all functions $f$, there exist $x, y$ in the domain such that $x < y$ and $f(x) \ge f(y)$.

1. Assume $n$ is even, so $n = 2k$ for some integer $k$.
2. Assume $m$ is odd, so $m = 2j + 1$ for some integer $j$.
3. Compute the product: $nm = (2k)(2j + 1)$.
4. Distribute: $nm = 4kj + 2k$.
5. Factor out a 2: $nm = 2(2kj + k)$.
6. Since $2kj + k$ is an integer, $nm$ is even by definition.

Answer: The product is $2(2kj + k)$, which fits the definition of an even number.

1. Assume $n$ is odd, so $n = 2k + 1$ for some integer $k$.
2. Substitute into the expression: $n^2 - 1 = (2k + 1)^2 - 1$.
3. Expand the square: $n^2 - 1 = (4k^2 + 4k + 1) - 1$.
4. Simplify: $n^2 - 1 = 4k^2 + 4k$.
5. Factor out a 2: $n^2 - 1 = 2(2k^2 + 2k)$.
6. Since $2k^2 + 2k$ is an integer, $n^2 - 1$ is even.

Answer: The expression simplifies to $2(2k^2 + 2k)$, which is even.

1. Let the three consecutive integers be $n$, $n+1$, and $n+2$.
2. Sum them: $S = n + (n + 1) + (n + 2)$.
3. Combine like terms: $S = 3n + 3$.
4. Factor out the 3: $S = 3(n + 1)$.
5. Since $n+1$ is an integer, $S$ is a multiple of 3, and thus divisible by 3.

Answer: The sum is $3(n+1)$, which is divisible by 3.

1. State the contrapositive: If $n$ is odd, then $3n + 2$ is odd.
2. Assume $n$ is odd, so $n = 2k + 1$ for some integer $k$.
3. Substitute into the expression: $3(2k + 1) + 2 = 6k + 3 + 2 = 6k + 5$.
4. Rewrite as $2(3k + 2) + 1$.
5. Since $3k + 2$ is an integer, $3n + 2$ is odd.
6. The contrapositive is true, so the original statement is true.

Answer: Proven via contrapositive.

1. State the contrapositive: If $x$ is odd AND $y$ is odd, then $xy$ is odd.
2. Assume $x = 2k + 1$ and $y = 2m + 1$ for integers $k, m$.
3. Compute the product: $xy = (2k + 1)(2m + 1) = 4km + 2k + 2m + 1$.
4. Factor out a two: $xy = 2(2km + k + m) + 1$.
5. Since $2km + k + m$ is an integer, $xy$ is odd.
6. The contrapositive is true, so the original statement is true.

Answer: Proven via contrapositive.

1. State the contrapositive: If $n$ is even, then $n^2 - 2n + 7$ is odd.
2. Assume $n = 2k$ for some integer $k$.
3. Substitute: $(2k)^2 - 2(2k) + 7 = 4k^2 - 4k + 7$.
4. Rewrite as $4k^2 - 4k + 6 + 1 = 2(2k^2 - 2k + 3) + 1$.
5. Since $2k^2 - 2k + 3$ is an integer, the result is odd.
6. The contrapositive is true, so the original statement is true.

Answer: Proven via contrapositive.

1. Assume $n^2$ is even but $n$ is odd.
2. Since $n$ is odd, we can write $n = 2k + 1$ for some integer $k$.
3. Then $n^2 = (2k + 1)^2 = 4k^2 + 4k + 1 = 2(2k^2 + 2k) + 1$.
4. This shows $n^2$ is odd, which contradicts the hypothesis that $n^2$ is even.
5. Therefore, $n$ must be even.

Answer: n \text{ is even}

1. Assume there exists a largest integer $M$.
2. Consider the number $M + 1$.
3. Since $M$ is an integer, $M + 1$ is also an integer.
4. By definition, $M + 1 > M$.
5. This contradicts the assumption that $M$ is the largest integer.
6. Therefore, no largest integer exists.

Answer: \text{No largest integer exists}

1. Assume $\sqrt{3} = p/q$ in simplest form.
2. Then $3 = p^2/q^2$, so $3q^2 = p^2$.
3. This implies $p^2$ is a multiple of 3, so $p$ must be a multiple of 3. Let $p = 3k$.
4. Substitute: $3q^2 = (3k)^2 = 9k^2$, so $q^2 = 3k^2$.
5. This implies $q^2$ is a multiple of 3, so $q$ must be a multiple of 3.
6. Since both $p$ and $q$ are multiples of 3, the fraction $p/q$ was not in simplest form, a contradiction.

Answer: \sqrt{3} \notin \mathbb{Q}

1. Base case: For $n=1$, $1^2 = 1$ and $\frac{1(2)(3)}{6} = 1$. True.
2. Inductive hypothesis: Assume $1^2 + 2^2 + \dots + k^2 = \frac{k(k+1)(2k+1)}{6}$.
3. Inductive step: Add $(k+1)^2$ to both sides: $\frac{k(k+1)(2k+1)}{6} + (k+1)^2$.
4. Factor out $(k+1)$: $\frac{k+1}{6} [k(2k+1) + 6(k+1)] = \frac{k+1}{6} [2k^2 + 7k + 6]$.
5. Factor the quadratic: $2k^2 + 7k + 6 = (k+2)(2k+3)$.
6. Result: $\frac{(k+1)(k+2)(2k+3)}{6}$, which is the formula for $n=k+1$.

Answer: \frac{n(n+1)(2n+1)}{6}

1. Base case: For $n=1$, $2^1 = 2$ and $2 > 1$. True.
2. Inductive hypothesis: Assume $2^k > k$.
3. Inductive step: We want to show $2^{k+1} > k+1$.
4. Start with $2^{k+1} = 2 \cdot 2^k$.
5. By hypothesis, $2 \cdot 2^k > 2k$.
6. Since $k \ge 1$, we know $2k = k + k \ge k + 1$.
7. Thus $2^{k+1} > k+1$.

Answer: 2^n > n

1. Base case: For $n=1$, $3^1 = 3$ and $3 > 1$. True.
2. Inductive hypothesis: Assume $3^k > k$ for some $k \ge 1$.
3. Inductive step: Consider $3^{k+1} = 3 \cdot 3^k$.
4. By hypothesis, $3 \cdot 3^k > 3k$.
5. We want to show $3k > k+1$.
6. This is equivalent to $2k > 1$, which is true for all $k \ge 1$.
7. Thus $3^{k+1} > k+1$.

Answer: 3^n > n

1. Base case $n=1$: $2^0 + 2^1 = 1 + 2 = 3$, and $2^{1+1} - 1 = 4 - 1 = 3$. The base case holds.
2. Inductive step: Assume $\sum_{i=0}^{k} 2^i = 2^{k+1} - 1$.
3. For $n=k+1$, the sum is $(\sum_{i=0}^{k} 2^i) + 2^{k+1}$.
4. Substitute the hypothesis: $(2^{k+1} - 1) + 2^{k+1} = 2 \cdot 2^{k+1} - 1 = 2^{k+2} - 1$.
5. This matches the formula for $n=k+1$.

Answer: $\sum_{i=0}^{n} 2^i = 2^{n+1} - 1$

1. Base case $n=2$: $2$ is prime, so it is a product of one prime.
2. Inductive step: Assume every integer $m$ such that $2 \le m \le k$ is a product of primes.
3. Consider $n=k+1$. If $k+1$ is prime, we are done.
4. If $k+1$ is composite, $k+1 = a \cdot b$ for some $1 < a, b < k+1$.
5. By the strong inductive hypothesis, both $a$ and $b$ are products of primes.
6. Thus, $k+1$ is a product of the primes that make up $a$ and $b$.

Answer: Every integer $n \ge 2$ is a product of primes.

1. Base case $n=2$: $(1+x)^2 = 1 + 2x + x^2$. Since $x > 0$, $x^2 > 0$, so $1 + 2x + x^2 > 1 + 2x$.
2. Inductive step: Assume $(1+x)^k > 1 + kx$ for some $k \ge 2$.
3. Multiply both sides by $(1+x)$. Since $x > 0$, $(1+x) > 0$.
4. $(1+x)^{k+1} > (1+kx)(1+x) = 1 + x + kx + kx^2 = 1 + (k+1)x + kx^2$.
5. Since $k \ge 2$ and $x^2 > 0$, the term $kx^2$ is strictly positive.
6. Therefore, $1 + (k+1)x + kx^2 > 1 + (k+1)x$.

Answer: $(1+x)^n > 1 + nx$

1. To check injectivity, assume $f(x) = f(y)$, so $x^3 + x = y^3 + y$.
2. Rearrange to $x^3 - y^3 + x - y = 0$, which factors as $(x-y)(x^2 + xy + y^2 + 1) = 0$.
3. The term $x^2 + xy + y^2 + 1$ is always positive because it is $(x + y/2)^2 + 3y^2/4 + 1$.
4. Thus, we must have $x - y = 0$, so $x = y$. The function is injective.
5. For surjectivity, since $f(x)$ is a polynomial of odd degree, its limits as $x$ goes to plus or minus infinity are plus and minus infinity.
6. By the Intermediate Value Theorem (Calculus, Lesson 20), $f(x)$ must take every real value. Thus, it is surjective.
7. Since it is both injective and surjective, it is a bijection.

Answer: Yes, it is a bijection.

1. The preimage $f^{-1}(T)$ is the set of all $x$ such that $f(x) \in [1, e^2]$.
2. This means $1 \le e^x \le e^2$.
3. Taking the natural logarithm of all parts of the inequality, we get $\ln(1) \le x \le \ln(e^2)$.
4. Since $\ln(1) = 0$ and $\ln(e^2) = 2$, we have $0 \le x \le 2$.

Answer: [0, 2]

1. For injectivity, assume $f(n) = f(m)$, so $2n = 2m$. Dividing by 2 gives $n = m$. Thus, it is injective.
2. For surjectivity, we check if every integer $k$ in the codomain $\mathbb{Z}$ has a preimage $n \in \mathbb{Z}$ such that $2n = k$.
3. If we pick $k = 1$, we need $2n = 1$, which means $n = 1/2$.
4. However, $1/2$ is not an integer, so $1/2 \notin \mathbb{Z}$.
5. Since there is no integer $n$ such that $f(n) = 1$, the function is not surjective.

Answer: Injective, but not surjective.

1. Use the contrapositive: assume it is not the case that ($x = 0$ or $y = 0$).
2. This means $x \neq 0$ and $y \neq 0$.
3. Since both are non-zero, their product $xy$ cannot be zero.
4. Thus, $\neg(x=0 \lor y=0) \implies xy \neq 0$, which proves the original statement.

Answer: Proven via contrapositive.

1. Check injectivity: $x^3 = y^3$. Taking the cube root of both sides gives $x = y$. It is injective.
2. Check surjectivity: For any $y \in \mathbb{R}$, we can choose $x = \sqrt[3]{y}$. Then $f(x) = (\sqrt[3]{y})^3 = y$. It is surjective.
3. Since it is both injective and surjective, it is a bijection.

Answer: Yes, it is a bijection.

1. Base case $n=1$: $1 = \frac{1(2)}{2} = 1$. True.
2. Inductive step: Assume $S_k = \frac{k(k+1)}{2}$.
3. Then $S_{k+1} = S_k + (k+1) = \frac{k(k+1)}{2} + (k+1)$.
4. Factor out $(k+1)$: $(k+1)(\frac{k}{2} + 1) = (k+1)(\frac{k+2}{2}) = \frac{(k+1)(k+2)}{2}$.
5. This matches the formula for $n=k+1$.

Answer: Proven via induction.

1. We know $x_4 = \frac{577}{408}$.
2. Compute $\frac{2}{x_4} = \frac{2}{577/408} = \frac{816}{577}$.
3. Find the average: $x_5 = \frac{1}{2}(\frac{577}{408} + \frac{816}{577})$.
4. Common denominator is $408 \times 577 = 235416$.
5. Numerator is $577^2 + 816 \times 408 = 332929 + 332928 = 665857$.
6. Result is $\frac{665857}{2 \times 235416} = \frac{665857}{470832}$.

Answer: $\frac{665857}{470832}$

1. Each $y_n$ is a terminating decimal, which can be written as a fraction with a power of 10 in the denominator, so $y_n \in \mathbb{Q}$.
2. The sequence converges to $\sqrt{2}$ by construction.
3. Since $\sqrt{2}$ is irrational, the limit is not in $\mathbb{Q}$.

Answer: The terms are terminating decimals (fractions), but the limit is $\sqrt{2}$, which is irrational.

1. No, density does not imply completeness.
2. The rational numbers $\mathbb{Q}$ are dense because between any two rationals there is another rational.
3. However, $\mathbb{Q}$ is not complete because the Babylonian sequence of rationals bunches up but its limit $\sqrt{2}$ is not in $\mathbb{Q}$.

Answer: No; the rationals are dense but incomplete.

1. By the law of trichotomy, either $x > 0$, $x < 0$, or $x = 0$.
2. Case 1: If $x > 0$, then multiplying both sides of $x > 0$ by the positive number $x$ gives $x^2 > 0$.
3. Case 2: If $x < 0$, then $-x > 0$. Multiplying both sides of $-x > 0$ by the positive number $-x$ gives $(-x)^2 > 0$, which simplifies to $x^2 > 0$.
4. Case 3: If $x = 0$, then $x^2 = 0 \cdot 0 = 0$.
5. In all cases, $x^2 \ge 0$.

Answer: x^2 \ge 0

1. We are given $a < b$, which means $b - a > 0$.
2. We are given $c < d$, which means $d - c > 0$.
3. The sum of two positive numbers is positive, so $(b - a) + (d - c) > 0$.
4. Rearranging the terms using commutativity and associativity gives $(b + d) - (a + c) > 0$.
5. By the definition of order, this means $a + c < b + d$.

Answer: a + c < b + d

1. We are given $a < 1$.
2. We are also given $a > 0$.
3. Since $a$ is positive, we can multiply both sides of the inequality $a < 1$ by $a$ without changing the direction of the inequality.
4. Multiplying $a < 1$ by $a$ yields $a \cdot a < 1 \cdot a$.
5. This simplifies to $a^2 < a$.

Answer: a^2 < a

1. Step 1: List some terms: $1/2, 2/3, 3/4, \dots$. All terms are less than 1, so 1 is an upper bound.
2. Step 2: Check if 1 is the least upper bound. For any $\epsilon > 0$, we need $n/(n+1) > 1 - \epsilon$.
3. Step 3: Solve for $n$: $n > (1 - \epsilon)(n+1) \implies n > n + 1 - \epsilon n - \epsilon \implies \epsilon n > 1 - \epsilon \implies n > (1 - \epsilon)/\epsilon$.
4. Step 4: Since such an $n$ always exists in the natural numbers, 1 is the supremum.
5. Step 5: Check if $1 \in S$. $n/(n+1) = 1$ would imply $n = n+1$, which is $0 = 1$, impossible. So no maximum exists.

Answer: \sup S = 1, \text{ no maximum exists}

1. Step 1: The condition $x^2 < 9$ is equivalent to $-3 < x < 3$.
2. Step 2: The set is the open interval $(-3, 3)$.
3. Step 3: Any number $M \ge 3$ is an upper bound. The smallest such number is 3.
4. Step 4: Since $3 \notin S$, there is no maximum.

Answer: \sup S = 3, \text{ no maximum exists}

1. Step 1: By definition of maximum, $m \in S$ and for all $x \in S, x \le m$.
2. Step 2: This means $m$ is an upper bound of $S$.
3. Step 3: To show it is the least upper bound, assume there is an upper bound $L < m$.
4. Step 4: But we know $m \in S$, and since $L$ is an upper bound, $m \le L$.
5. Step 5: This contradicts $L < m$. Therefore, no upper bound can be smaller than $m$.
6. Step 6: Thus, $m$ is the least upper bound, so $\sup S = m$.

Answer: \sup S = m

1. Step 1: Show 1 is an upper bound. Since $n \ge 1$, $\frac{1}{n} > 0$, so $1 - \frac{1}{n} < 1$ for all $n$.
2. Step 2: Use the epsilon characterisation. We want $1 - \frac{1}{n} > 1 - \epsilon$.
3. Step 3: Simplify the inequality: $-\frac{1}{n} > -\epsilon$, which means $\frac{1}{n} < \epsilon$, or $n > \frac{1}{\epsilon}$.
4. Step 4: Since the natural numbers are unbounded, we can always find an $n \in \mathbb{N}$ such that $n > \frac{1}{\epsilon}$. Thus, $\sup S = 1$.

Answer: \sup S = 1

1. Step 1: Guess $\inf S = 1$ because $\frac{n+1}{n} = 1 + \frac{1}{n}$, which approaches 1 as $n$ grows.
2. Step 2: Show 1 is a lower bound. Since $\frac{1}{n} > 0$, $1 + \frac{1}{n} > 1$ for all $n$.
3. Step 3: We want $1 + \frac{1}{n} < 1 + \epsilon$. This simplifies to $\frac{1}{n} < \epsilon$, or $n > \frac{1}{\epsilon}$.
4. Step 4: Since the natural numbers are unbounded, we can pick $n > \frac{1}{\epsilon}$, ensuring an element of $S$ is less than $1 + \epsilon$.

Answer: \inf S = 1

1. Step 1: $x^2 < 9$ implies $-3 < x < 3$. Thus 3 is an upper bound and -3 is a lower bound.
2. Step 2: For $\sup S$, given $\epsilon > 0$, we need $x \in S$ such that $x > 3 - \epsilon$.
3. Step 3: If $\epsilon > 6$, $x=0$ works. If $\epsilon \le 6$, let $x = 3 - \frac{\epsilon}{2}$. Then $x^2 = (3 - \frac{\epsilon}{2})^2 < 3^2 = 9$, so $x \in S$.
4. Step 4: Similarly for $\inf S$, let $x = -3 + \frac{\epsilon}{2}$. Then $x^2 < 9$ and $x < -3 + \epsilon$.

Answer: \sup S = 3, \inf S = -3

1. Since $a > 0$, the number $1/a$ is a well-defined real number.
2. By the Archimedean property, for any real number $x$, there exists $n \in \mathbb{N}$ such that $n > x$.
3. Let $x = 1/a$. Then there exists $n \in \mathbb{N}$ such that $n > 1/a$.
4. Since $a > 0$, we can multiply both sides of the inequality by $a$ without changing the inequality sign.
5. This gives $na > a(1/a)$, which simplifies to $na > 1$.

Answer: The existence of $n$ is guaranteed by the Archimedean property applied to $1/a$.

1. Let $\delta = y - x$. Since $x < y$, $\delta > 0$.
2. By the Archimedean property, there exists $n \in \mathbb{N}$ such that $1/n < \delta$.
3. Consider the set $S = \{k \in \mathbb{Z} : k/n > x\}$. This set is non-empty because for large $k$, $k/n$ exceeds $x$.
4. Let $m$ be the smallest integer in $S$. Then $m/n > x$ and $(m-1)/n \le x$.
5. From $(m-1)/n \le x$, we have $m/n \le x + 1/n$.
6. Since $1/n < y - x$, we have $x + 1/n < x + (y - x) = y$.
7. Thus, $x < m/n < y$, and $q = m/n$ is the required rational.

Answer: The rational $q = m/n$ exists where $n > 1/(y-x)$ and $m = \lceil nx \rceil$ (if $nx$ is not an integer).

1. We are looking for a rational $q$ in the interval $(0, \epsilon)$.
2. By the density of the rationals, for any two real numbers $x < y$, there exists a rational $q$ such that $x < q < y$.
3. Let $x = 0$ and $y = \epsilon$.
4. Since $\epsilon > 0$, the condition $x < y$ is satisfied.
5. Therefore, there exists a rational $q$ such that $0 < q < \epsilon$.

Answer: Such a $q$ exists by the density of the rationals in the interval $(0, \epsilon)$.

1. We need a function that maps $1$ to $1$, $2$ to $3$, $3$ to $5$, and so on.
2. The formula for the $n$-th odd number is $2n - 1$.
3. Check for injection: if $2n - 1 = 2m - 1$, then $2n = 2m$, so $n = m$.
4. Check for surjection: for any odd number $k$, $n = (k+1)/2$ is a natural number such that $f(n) = k$.

Answer: f(n) = 2n - 1

1. We can list all strings of length 1 alphabetically.
2. Then list all strings of length 2 alphabetically.
3. Then length 3, and so on.
4. Since there are only finitely many strings of any fixed length $L$, we will eventually reach any given string in a finite number of steps.

Answer: Yes, it is countable.

1. Let $A = \{a_1, a_2, \dots\}$ and $B = \{b_1, b_2, \dots\}$ be the listings for $A$ and $B$.
2. We can create a new list by interleaving them: $a_1, b_1, a_2, b_2, a_3, b_3, \dots$.
3. If an element appears in both sets, we simply skip it the second time it appears to maintain a bijection.
4. This process creates an exhaustive list of all elements in $A \cup B$.

Answer: The union is countable.

1. Assume $S$ is countable and list the sequences as $s_1, s_2, s_3, \dots$.
2. Construct a new sequence $x = (x_1, x_2, x_3, \dots)$ where $x_n$ is chosen to be different from the $n$-th digit of $s_n$.
3. For example, if the $n$-th digit of $s_n$ is $0$, let $x_n = 1$; otherwise, let $x_n = 0$.
4. Since $x$ differs from every $s_n$ at the $n$-th position, $x$ is not in the list.
5. This contradicts the assumption that the list was exhaustive, so $S$ is uncountable.

Answer: The set $S$ is uncountable by Cantor's diagonal argument.

1. Since $A \subseteq B$, the cardinality of $B$ must be at least as large as the cardinality of $A$.
2. We are given that $A$ is uncountable, meaning $|A| > \aleph_0$.
3. Therefore, $|B| \ge |A| > \aleph_0$.

Answer: $B$ must also be uncountable.

1. Let $I = (0, 1)$ be the set of all real numbers in the interval, and $Q_I$ be the set of rational numbers in that interval.
2. We know from today's lesson that $|I| = c$ (uncountable) and from yesterday that $|Q_I| = \aleph_0$ (countable).
3. The set of irrationals $Irr_I$ is the complement $I \setminus Q_I$.
4. If $Irr_I$ were countable, then $I = Irr_I \cup Q_I$ would be the union of two countable sets, which is countable.
5. But $I$ is uncountable, so $Irr_I$ must be uncountable.

Answer: The set of irrationals is uncountable.

1. Convert $\frac{7}{27}$ to base three.
2. $\frac{7}{27} = \frac{6}{27} + \frac{1}{27} = \frac{2}{9} + \frac{1}{27}$.
3. In base three, this is $0 \cdot 3^{-1} + 2 \cdot 3^{-2} + 1 \cdot 3^{-3}$, which is $(0.021)_3$.
4. Since the expansion contains the digit $1$, the number is removed during the third step of the construction.

Answer: No, $\frac{7}{27} \notin C$.

1. An endpoint is created at some step $n$ and is never part of an open middle third removed in any step $k > n$.
2. For example, $\frac{1}{3}$ is the left endpoint of the first removed interval $(\frac{1}{3}, \frac{2}{3})$.
3. Since we only remove open intervals, the endpoints themselves are never removed.
4. Therefore, all endpoints remain in the intersection of all $C_n$.

Answer: Endpoints are never removed because only open intervals are deleted.

1. Removing the middle fifth leaves two intervals of length $\frac{2}{5}$ each.
2. The number of intervals at step $n$ would be $2^n$.
3. The points remaining would be those that can be represented in base five using only the digits $0, 1, 3, 4$ (avoiding the middle digit $2$).
4. This still creates a mapping to the set of all infinite sequences of a finite alphabet with at least two symbols.
5. By Cantor's diagonal argument, this set is uncountable.

Answer: Yes, it would still be uncountable.

1. Identify the repeating block as $72$ with a period of $2$.
2. Write the expansion as a series: $\sum_{n=1}^{\infty} \frac{72}{100^n}$.
3. Identify the first term $a = \frac{72}{100}$ and the ratio $r = \frac{1}{100}$.
4. Apply the formula $S = \frac{a}{1-r} = \frac{72/100}{1 - 1/100} = \frac{72/100}{99/100}$.
5. Simplify the fraction $\frac{72}{99}$ by dividing by $9$.

Answer: $\frac{8}{11}$

1. Write $0.4999\dots$ as $0.4 + 0.0999\dots$.
2. Express $0.0999\dots$ as a series: $\sum_{n=2}^{\infty} \frac{9}{10^n}$.
3. The first term is $a = \frac{9}{100}$ and the ratio is $r = \frac{1}{10}$.
4. Calculate the sum: $S = \frac{9/100}{1 - 1/10} = \frac{9/100}{9/10} = \frac{1}{10} = 0.1$.
5. Add the constant: $0.4 + 0.1 = 0.5$.

Answer: $0.5$

1. The repeating block is $123$ with period $3$.
2. The series is $\sum_{n=1}^{\infty} \frac{123}{1000^n}$.
3. Using the formula $S = \frac{123/1000}{1 - 1/1000} = \frac{123}{999}$.
4. Simplify by dividing by $3$: $\frac{41}{333}$.
5. The sequence $S_k$ converges because it is monotonic increasing and bounded above by $1$, and the limit of the geometric series is exactly $\frac{41}{333}$.

Answer: $\frac{41}{333}$

1. Rewrite the expression by multiplying by the conjugate: $\sqrt{n^2+1} - n = \frac{(\sqrt{n^2+1}-n)(\sqrt{n^2+1}+n)}{\sqrt{n^2+1}+n} = \frac{1}{\sqrt{n^2+1}+n}$.
2. For $n=1$, the value is $1/(\sqrt{2}+1) \approx 0.414$. As $n$ increases, the denominator increases, so the values decrease.
3. The maximum value occurs at $n=1$, so $\sup S = \sqrt{2}-1$. Since this value is in the set, it is also the maximum.
4. As $n \to \infty$, the expression $1/(\sqrt{n^2+1}+n)$ approaches $0$. Since the values are always positive, the infimum is $0$.
5. Since $1/(\sqrt{n^2+1}+n)$ is never $0$ for any $n \in \mathbb{N}$, the infimum is not in the set.

Answer: \sup S = \sqrt{2}-1 \text{ (contained)}, \inf S = 0 \text{ (not contained)}

1. Base case: For $n=1$, $2^1 = 2$ and $2 > 1$. The base case holds.
2. Inductive step: Assume $2^k > k$ for some $k \in \mathbb{N}$.
3. We want to show $2^{k+1} > k+1$.
4. Note that $2^{k+1} = 2 \cdot 2^k$. By the inductive hypothesis, $2 \cdot 2^k > 2k$.
5. Since $k \geq 1$, we know $2k = k + k \geq k + 1$.
6. Therefore, $2^{k+1} > k+1$, completing the induction.

Answer: Proven by induction.

1. Let $I = (0, 1)$ be the set of all real numbers in the interval, and let $\mathbb{Q}_I$ be the set of all rational numbers in that interval.
2. We know from the course that $I$ is uncountable ($|I| = c$).
3. We also know that the set of all rationals $\mathbb{Q}$ is countable, so any subset $\mathbb{Q}_I$ must also be countable.
4. The set of irrationals in the interval is $S = I \setminus \mathbb{Q}_I$.
5. If $S$ were countable, then $I = S \cup \mathbb{Q}_I$ would be the union of two countable sets, which must be countable.
6. This contradicts the fact that $I$ is uncountable. Therefore, $S$ must be uncountable.

Answer: The set of irrationals is uncountable.

1. Let $\epsilon > 0$. We want $|\frac{5}{n+2} - 0| < \epsilon$.
2. This simplifies to $\frac{5}{n+2} < \epsilon$.
3. Rearranging gives $n+2 > \frac{5}{\epsilon}$, so $n > \frac{5}{\epsilon} - 2$.
4. Choose $N$ to be the smallest integer greater than $\frac{5}{\epsilon} - 2$.
5. Then for all $n \geq N$, we have $n > \frac{5}{\epsilon} - 2$, which implies $|a_n - 0| < \epsilon$.

Answer: The sequence converges to $0$ with $N = \lceil \frac{5}{\epsilon} - 2 \rceil$ (or any larger integer).

1. Let $\epsilon > 0$. We want $|\frac{n}{n+1} - 1| < \epsilon$.
2. Simplify the expression: $|\frac{n - (n+1)}{n+1}| = |\frac{-1}{n+1}| = \frac{1}{n+1}$.
3. We need $\frac{1}{n+1} < \epsilon$, which means $n+1 > \frac{1}{\epsilon}$, or $n > \frac{1}{\epsilon} - 1$.
4. Choose $N = \lceil \frac{1}{\epsilon} \rceil$.
5. For $n \geq N$, $n+1 > \frac{1}{\epsilon}$, so $|a_n - 1| < \epsilon$.

Answer: The sequence converges to $1$ with $N = \lceil \frac{1}{\epsilon} \rceil$.

1. To prove divergence to $0$, we need to find one $\epsilon > 0$ such that for every $N$, there is some $n \geq N$ where $|a_n - 0| \geq \epsilon$.
2. Let $\epsilon = 0.5$.
3. For any $N$, we can pick $n$ to be any integer greater than or equal to $N$.
4. Since $|a_n| = |(-1)^n| = 1$ for all $n$, we have $|a_n - 0| = 1$.
5. Since $1 \geq 0.5$, the condition $|a_n - 0| \geq \epsilon$ is always satisfied for every $n \geq N$.

Answer: The sequence does not converge to $0$ because for $\epsilon = 0.5$, $|a_n - 0| \geq \epsilon$ for all $n$.

1. Scratch work: $|\frac{3}{n+1} - 0| < \epsilon \implies \frac{3}{n+1} < \epsilon \implies n+1 > \frac{3}{\epsilon} \implies n > \frac{3}{\epsilon} - 1$.
2. Let $\epsilon > 0$. Choose $N = \lceil \frac{3}{\epsilon} - 1 \rceil$ (or $1$ if this is non-positive).
3. Let $n \geq N$. Then $n \geq \frac{3}{\epsilon} - 1$.
4. Adding $1$ gives $n+1 \geq \frac{3}{\epsilon}$, so $\frac{1}{n+1} \leq \frac{\epsilon}{3}$.
5. Multiplying by $3$ gives $\frac{3}{n+1} \leq \epsilon$. Thus $|a_n - 0| \leq \epsilon$.

Answer: The sequence converges to $0$.

1. Scratch work: $|\frac{n}{n+5} - 1| = |\frac{n - (n+5)}{n+5}| = |\frac{-5}{n+5}| = \frac{5}{n+5}$.
2. We want $\frac{5}{n+5} < \epsilon \implies n+5 > \frac{5}{\epsilon} \implies n > \frac{5}{\epsilon} - 5$.
3. Let $\epsilon > 0$. Choose $N = \max(1, \lceil \frac{5}{\epsilon} - 5 \rceil)$.
4. For $n \geq N$, $n+5 \geq \frac{5}{\epsilon}$, so $\frac{5}{n+5} \leq \epsilon$.
5. Therefore $|a_n - 1| \leq \epsilon$.

Answer: The sequence converges to $1$.

1. Scratch work: $|\frac{4n-1}{2n+3} - 2| = |\frac{4n-1 - 2(2n+3)}{2n+3}| = |\frac{4n-1-4n-6}{2n+3}| = \frac{7}{2n+3}$.
2. We want $\frac{7}{2n+3} < \epsilon \implies 2n+3 > \frac{7}{\epsilon} \implies 2n > \frac{7}{\epsilon} - 3 \implies n > \frac{7}{2\epsilon} - 1.5$.
3. Let $\epsilon > 0$. Choose $N = \max(1, \lceil \frac{7}{2\epsilon} - 1.5 \rceil)$.
4. For $n \geq N$, $2n+3 \geq \frac{7}{\epsilon}$, so $\frac{7}{2n+3} \leq \epsilon$.
5. Therefore $|a_n - 2| \leq \epsilon$.

Answer: The sequence converges to $2$.

1. We want to show that for any $\epsilon > 0$, there exists $N$ such that $||a_n| - 0| < \epsilon$ for all $n \geq N$.
2. Note that $||a_n| - 0| = ||a_n|| = |a_n|$.
3. Since $a_n \to 0$, for any $\epsilon > 0$, there exists $N$ such that $|a_n - 0| < \epsilon$ for all $n \geq N$.
4. This simplifies to $|a_n| < \epsilon$ for all $n \geq N$.
5. Thus, $||a_n| - 0| < \epsilon$ for all $n \geq N$, so $|a_n| \to 0$.

Answer: The sequence $|a_n|$ converges to $0$.

1. Consider $a_n = (-1)^n$.
2. The sequence is bounded because $|a_n| = 1$ for all $n$, so we can pick $M=1$.
3. To show it diverges, assume it converges to $L$. Then for $\epsilon = 0.5$, there must be an $N$ such that $|a_n - L| < 0.5$ for all $n \geq N$.
4. For $n=N$ and $n=N+1$, we have $|a_N - L| < 0.5$ and $|a_{N+1} - L| < 0.5$.
5. But $|a_N - a_{N+1}| = |1 - (-1)| = 2$.
6. By triangle inequality, $2 = |a_N - a_{N+1}| \leq |a_N - L| + |L - a_{N+1}| < 0.5 + 0.5 = 1$.
7. Since $2 < 1$ is a contradiction, the sequence diverges.

Answer: $a_n = (-1)^n$ is bounded but diverges due to oscillation.

1. We want to show $|(a_n - b_n) - (L - M)| < \epsilon$.
2. Rewrite the expression as $|(a_n - L) - (b_n - M)|$.
3. Apply the triangle inequality: $|(a_n - L) - (b_n - M)| \leq |a_n - L| + |b_n - M|$.
4. Since $a_n \to L$, choose $N_1$ such that $|a_n - L| < \epsilon/2$ for $n \geq N_1$.
5. Since $b_n \to M$, choose $N_2$ such that $|b_n - M| < \epsilon/2$ for $n \geq N_2$.
6. For $n \geq \max(N_1, N_2)$, the sum is less than $\epsilon/2 + \epsilon/2 = \epsilon$.

Answer: $c_n \to L - M$.

1. Let $\epsilon > 0$.
2. Since $a_n \to L$, there exists $N_1$ such that $|a_n - L| < \epsilon/2$ for $n \geq N_1$.
3. Since $b_n \to M$, there exists $N_2$ such that $|b_n - M| < \epsilon/2$ for $n \geq N_2$.
4. Let $N = \max(N_1, N_2)$. For $n \geq N$, $|(a_n + b_n) - (L + M)| = |(a_n - L) + (b_n - M)| \leq |a_n - L| + |b_n - M| < \epsilon/2 + \epsilon/2 = \epsilon$.

Answer: The sum converges to $L + M$.

1. Identify $a_n = 1 + 1/n$ and $b_n = 2 + 1/n^2$.
2. Note that $a_n \to 1$ and $b_n \to 2$.
3. By the product rule, $\lim s_n = (\lim a_n)(\lim b_n) = 1 \times 2 = 2$.

Answer: 2

1. Let $b_n = a_n$. We are given $a_n \to L$ and $b_n \to L$.
2. The sequence $a_n^2$ is the product $a_n b_n$.
3. By the product rule, $\lim (a_n b_n) = (\lim a_n)(\lim b_n) = L \times L = L^2$.

Answer: The sequence converges to $L^2$.

1. Step 1: Identify the bounds of the cosine function: $-1 \leq \cos(n) \leq 1$.
2. Step 2: Divide the entire inequality by $n^2$, which is always positive: $-\frac{1}{n^2} \leq \frac{\cos(n)}{n^2} \leq \frac{1}{n^2}$.
3. Step 3: Find the limits of the guards: $\lim_{n \to \infty} -\frac{1}{n^2} = 0$ and $\lim_{n \to \infty} \frac{1}{n^2} = 0$.
4. Step 4: Since both guards converge to 0, by the Squeeze Theorem, the middle sequence must also converge to 0.

Answer: 0

1. Step 1: Calculate the limit of the lower guard: $\lim_{n \to \infty} \frac{1}{n+1} = 0$.
2. Step 2: Calculate the limit of the upper guard: $\lim_{n \to \infty} \frac{1}{n-1} = 0$.
3. Step 3: Since $a_n \to 0$ and $c_n \to 0$, and $b_n$ is squeezed between them, $b_n$ must also converge to 0.

Answer: 0

1. Step 1: Note that $(-1)^n$ is always either $-1$ or $1$. Thus, $-1 \leq (-1)^n \leq 1$.
2. Step 2: Divide by $n$: $-\frac{1}{n} \leq \frac{(-1)^n}{n} \leq \frac{1}{n}$.
3. Step 3: The lower guard $a_n = -\frac{1}{n}$ converges to 0.
4. Step 4: The upper guard $c_n = \frac{1}{n}$ converges to 0.
5. Step 5: By the Squeeze Theorem, $b_n \to 0$.

Answer: 0

1. Step 1: Check monotonicity. $a_{n+1} - a_n = \frac{1}{ (n+1)^2 } > 0$, so the sequence is strictly increasing.
2. Step 2: Check boundedness. We know $\frac{1}{k^2} \leq \frac{1}{k(k-1)}$ for $k \geq 2$.
3. Step 3: The sum $\sum_{k=2}^{n} \frac{1}{k(k-1)}$ is a telescoping sum that equals $1 - \frac{1}{n}$.
4. Step 4: Thus $a_n \leq 1 + (1 - \frac{1}{n}) < 2$. The sequence is bounded above by 2.
5. Step 5: Since it is increasing and bounded above, it converges by the MCT.

Answer: The sequence converges because it is monotonically increasing and bounded above by 2.

1. Step 1: Prove $a_n < 2$ by induction. $a_1 = \sqrt{2} < 2$. If $a_n < 2$, then $a_{n+1} = \sqrt{2+a_n} < \sqrt{4} = 2$.
2. Step 2: Prove $a_{n+1} > a_n$. $a_{n+1}^2 - a_n^2 = 2 + a_n - a_n^2 = (2-a_n)(1+a_n)$. Since $a_n < 2$ and $a_n > 0$, this is positive.
3. Step 3: By MCT, the limit $L$ exists. Solve $L = \sqrt{2+L}$.
4. Step 4: $L^2 - L - 2 = 0 \implies (L-2)(L+1) = 0$. Since $a_n > 0$, $L = 2$.

Answer: L = 2

1. Step 1: Monotonicity. $a_n = 1 - \frac{1}{n+1}$. As $n$ increases, $\frac{1}{n+1}$ decreases, so $1 - \frac{1}{n+1}$ increases. It is monotonically increasing.
2. Step 2: Boundedness. Since $n < n+1$, $a_n < 1$ for all $n$. It is bounded above by 1.
3. Step 3: By MCT, the sequence converges to its supremum, which is 1.

Answer: Yes, it converges to 1.

1. The term for $k=2$ is $\frac{1}{2!}(1 - \frac{1}{n})(1 - \frac{2}{n})$
2. Since $n \geq 2$, the factors $(1 - \frac{1}{n})$ and $(1 - \frac{2}{n})$ are both $\leq 1$
3. Therefore, the product is $\leq \frac{1}{2} \cdot 1 \cdot 1 = \frac{1}{2}$

Answer: \frac{1}{2!}(1 - \frac{1}{n})(1 - \frac{2}{n}) \leq \frac{1}{2}

1. Write $a_n$ as a sum of terms $T_{k,n} = \frac{1}{k!} \prod_{j=1}^{k-1} (1 - \frac{j}{n})$
2. Observe that $1 - \frac{j}{n} < 1 - \frac{j}{n+1}$ for all $j \geq 1$
3. Thus $T_{k,n} < T_{k,n+1}$ for all $k \leq n$
4. Since $a_{n+1}$ has all the increased terms of $a_n$ plus an additional positive term $T_{n+1,n+1}$, $a_{n+1} > a_n$

Answer: a_{n+1} > a_n

1. Start with $a_n \leq \sum_{k=0}^{n} \frac{1}{k!}$
2. Separate the first term: $a_n \leq 1 + \sum_{k=1}^{n} \frac{1}{k!}$
3. Substitute the bound $k! \geq 2^{k-1}$: $a_n \leq 1 + \sum_{k=1}^{n} \frac{1}{2^{k-1}}$
4. The sum is a geometric series: $\sum_{i=0}^{n-1} (1/2)^i < \frac{1}{1 - 1/2} = 2$
5. Thus $a_n < 1 + 2 = 3$

Answer: a_n < 3

1. Consider the subsequence of even terms $a_{2k} = 1 + \frac{1}{2k}$. As $k \to \infty$, $a_{2k} \to 1$.
2. Consider the subsequence of odd terms $a_{2k-1} = 2 + \frac{1}{2k-1}$. As $k \to \infty$, $a_{2k-1} \to 2$.
3. Since the two subsequences converge to different limits ($1 \neq 2$), the original sequence diverges.

Answer: Diverges

1. The even subsequence is $a_{2k} = \frac{1}{2k}$, which converges to $0$.
2. The odd subsequence is $a_{2k-1} = \frac{-1}{2k-1}$, which converges to $0$.
3. Since every term of the sequence is either in the even or odd subsequence, and both converge to $0$, the entire sequence $a_n$ converges to $0$.

Answer: \lim_{n \to \infty} a_n = 0

1. Pick $n_k = 8k$, then $a_{8k} = \sin(2k\pi) = 0$. This subsequence converges to $0$.
2. Pick $n_k = 8k+2$, then $a_{8k+2} = \sin(2k\pi + \frac{\pi}{2}) = 1$. This subsequence converges to $1$.
3. Because we found two subsequences with different limits ($0$ and $1$), the sequence diverges.

Answer: Diverges

1. Step 1: Check boundedness. Since $|\frac{(-1)^n}{n}| \leq 1$, we have $4 \leq a_n \leq 6$. The sequence is bounded.
2. Step 2: Apply Bolzano-Weierstrass. Since it is bounded, a convergent subsequence must exist.
3. Step 3: Evaluate the limit. As $n \to \infty$, $\frac{(-1)^n}{n} \to 0$. Thus, the original sequence $a_n \to 5$.
4. Step 4: Since the original sequence converges to 5, every subsequence also converges to 5.

Answer: The sequence is bounded and every convergent subsequence converges to $5$.

1. Step 1: List the first few terms: $\sin(\pi/2)=1, \sin(\pi)=0, \sin(3\pi/2)=-1, \sin(2\pi)=0$.
2. Step 2: The sequence repeats the pattern $1, 0, -1, 0$.
3. Step 3: The subsequence $a_{4k+1}$ is constant $1$, so it converges to $1$.
4. Step 4: The subsequence $a_{4k+3}$ is constant $-1$, so it converges to $-1$.
5. Step 5: The subsequence $a_{2k}$ is constant $0$, so it converges to $0$.

Answer: The subsequential limits are $\{-1, 0, 1\}$.

1. Step 1: Split the sequence into even and odd indices.
2. Step 2: For even $n=2k$, $a_{2k} = (-1)^{2k}(1 + \frac{1}{2k}) = 1 + \frac{1}{2k}$. As $k \to \infty$, $a_{2k} \to 1$.
3. Step 3: For odd $n=2k-1$, $a_{2k-1} = (-1)^{2k-1}(1 + \frac{1}{2k-1}) = -(1 + \frac{1}{2k-1})$. As $k \to \infty$, $a_{2k-1} \to -1$.
4. Step 4: Any other subsequence must be a combination of these terms. Since any subsequence of a convergent sequence converges to the same limit, any convergent subsequence must converge to either $1$ or $-1$.

Answer: The subsequential limits are $1$ and $-1$.

1. Step 1: Consider $n > m$. The difference is $|a_n - a_m| = \sum_{k=m+1}^{n} \frac{1}{k(k+1)}$.
2. Step 2: Use partial fractions: $\frac{1}{k(k+1)} = \frac{1}{k} - \frac{1}{k+1}$.
3. Step 3: The sum telescopes: $\sum_{k=m+1}^{n} (\frac{1}{k} - \frac{1}{k+1}) = \frac{1}{m+1} - \frac{1}{n+1}$.
4. Step 4: Bound the result: $|a_n - a_m| < \frac{1}{m+1} < \frac{1}{m}$.
5. Step 5: For any $\epsilon > 0$, choose $N > \frac{1}{\epsilon}$. Then for $n, m \ge N$, $|a_n - a_m| < \frac{1}{N} < \epsilon$.

Answer: The sequence is Cauchy because for any $\epsilon > 0$, choosing $N > 1/\epsilon$ ensures $|a_n - a_m| < \epsilon$ for all $n, m \ge N$.

1. Step 1: To show it is not Cauchy, we must find an $\epsilon > 0$ such that for any $N$, there exist $n, m \ge N$ with $|a_n - a_m| \ge \epsilon$.
2. Step 2: Let $\epsilon = 1$.
3. Step 3: For any $N$, choose $m = N$ and $n$ such that $\sqrt{n} \ge \sqrt{N} + 1$. This is possible by picking $n \ge (\sqrt{N} + 1)^2$.
4. Step 4: Then $|a_n - a_m| = \sqrt{n} - \sqrt{N} \ge 1 = \epsilon$.

Answer: The sequence is not Cauchy because the distance between terms can be made arbitrarily large.

1. Step 1: Assume $n > m$. Use the triangle inequality to write $|a_n - a_m| \le \sum_{k=m}^{n-1} |a_{k+1} - a_k|$.
2. Step 2: Substitute the given bound: $|a_n - a_m| < \sum_{k=m}^{n-1} \frac{1}{2^k}$.
3. Step 3: This is a geometric sum. The sum is $\frac{1}{2^m} (1 + \frac{1}{2} + \dots + \frac{1}{2^{n-m-1}}) < \frac{1}{2^m} \cdot 2 = \frac{1}{2^{m-1}}$.
4. Step 4: For any $\epsilon > 0$, we can choose $N$ such that $\frac{1}{2^{N-1}} < \epsilon$.
5. Step 5: Then for all $n, m \ge N$, $|a_n - a_m| < \frac{1}{2^{m-1}} \le \frac{1}{2^{N-1}} < \epsilon$.

Answer: The sequence is Cauchy because the sum of the gaps between consecutive terms is bounded by a convergent geometric series.

1. Consider $m > n$. The distance $|a_m - a_n| = \sum_{k=n+1}^{m} \frac{1}{k^2}$.
2. Using the hint, $\sum_{k=n+1}^{m} \frac{1}{k^2} < \sum_{k=n+1}^{m} (\frac{1}{k-1} - \frac{1}{k})$.
3. This is a telescoping sum: $(\frac{1}{n} - \frac{1}{n+1}) + (\frac{1}{n+1} - \frac{1}{n+2}) + \dots + (\frac{1}{m-1} - \frac{1}{m}) = \frac{1}{n} - \frac{1}{m}$.
4. Since $\frac{1}{n} - \frac{1}{m} < \frac{1}{n}$, we can make this less than $\epsilon$ by choosing $N > \frac{1}{\epsilon}$.
5. Thus, the sequence is Cauchy and therefore converges by completeness.

Answer: The sequence is Cauchy because $|a_m - a_n| < \frac{1}{n}$, so it converges.

1. To show it is not Cauchy, we need to find $\epsilon > 0$ such that for any $N$, there exist $m, n \ge N$ with $|a_m - a_n| \ge \epsilon$.
2. Let $\epsilon = \ln(2)$.
3. For any $N$, choose $n = N$ and $m = 2N$.
4. Then $|a_m - a_n| = |\ln(2N) - \ln(N)| = |\ln(\frac{2N}{N})| = \ln(2)$.
5. Since we found a constant distance $\ln(2)$ that persists for any $N$, the sequence is not Cauchy.

Answer: The sequence is not Cauchy because $|a_{2n} - a_n| = \ln(2)$ for all $n$.

1. Let $\epsilon > 0$. Since $a_n$ is Cauchy, there exists $N$ such that $|a_n - a_m| < \epsilon/2$ for all $n, m \ge N$.
2. Since $a_{n_k} \to L$, there exists $K$ such that for $k \ge K$, $|a_{n_k} - L| < \epsilon/2$.
3. Pick $k$ large enough such that $k \ge K$ and $n_k \ge N$.
4. For any $n \ge N$, use the triangle inequality: $|a_n - L| \le |a_n - a_{n_k}| + |a_{n_k} - L|$.
5. The first term is $< \epsilon/2$ because $n, n_k \ge N$. The second term is $< \epsilon/2$ because $k \ge K$.
6. Thus $|a_n - L| < \epsilon$ for all $n \ge N$.

Answer: The sequence converges to $L$ by the triangle inequality and the Cauchy property.

1. We know that $-1 \le \sin(n) \le 1$ for all $n$.
2. Adding 1 to all sides gives $0 \le 1 + \sin(n) \le 2$.
3. Dividing by 2 gives $0 \le \frac{1 + \sin(n)}{2} \le 1$.
4. Since $\sin(n)$ gets arbitrarily close to $1$ and $-1$ infinitely often, the supremum of the tail approaches $1$ and the infimum of the tail approaches $0$.

Answer: \limsup a_n = 1, \liminf a_n = 0

1. The term $\frac{(-1)^n}{n}$ converges to $0$ as $n \to \infty$.
2. Therefore, $a_n$ converges to $0 + 2 = 2$.
3. For a convergent sequence, $\limsup a_n = \liminf a_n = L$.
4. Thus, $\limsup a_n = 2$ and $\liminf a_n = 2$.

Answer: \limsup a_n = \liminf a_n = 2, \text{ so it converges to } 2

1. A bounded sequence converges if and only if its $\limsup$ and $\liminf$ are equal.
2. In this case, $\limsup a_n = 5$ and $\liminf a_n = 2$.
3. Since $5 \neq 2$, the condition for convergence is not met.
4. The sequence must oscillate and therefore diverges.

Answer: \text{No, because } \limsup a_n \neq \liminf a_n

1. Step 1: Show $x_n \ge \sqrt{3}$ for $n \ge 2$. Consider $(x_n - \sqrt{3}/x_n)^2 \ge 0$, which implies $x_n^2 + 3/x_n^2 \ge 2\sqrt{3}$. Alternatively, use the fact that $x_{n+1}^2 - 3 = \frac{(x_n^2-3)^2}{4x_n^2} \ge 0$, so $x_n \ge \sqrt{3}$ for $n \ge 2$.
2. Step 2: Show the sequence is decreasing for $n \ge 2$ by calculating $x_{n+1} - x_n = \frac{3 - x_n^2}{2x_n}$. Since $x_n^2 \ge 3$, the difference is $\le 0$.
3. Step 3: By the Monotone Convergence Theorem, the limit $L$ exists.
4. Step 4: Solve $L = \frac{1}{2}(L + 3/L)$, which gives $2L = L + 3/L$, so $L^2 = 3$. Since $x_n > 0$, $L = \sqrt{3}$.

Answer: $\sqrt{3}$

1. Step 1: Use induction to show $x_n < 3$. $x_1 = \sqrt{6} < 3$. If $x_n < 3$, then $x_{n+1} = \sqrt{6 + x_n} < \sqrt{6 + 3} = 3$.
2. Step 2: Show the sequence is increasing. $x_{n+1}^2 - x_n^2 = 6 + x_n - x_n^2 = (3 - x_n)(2 + x_n)$. Since $x_n < 3$ and $x_n > 0$, the difference is positive.
3. Step 3: By MCT, the limit $L$ exists.
4. Step 4: Solve $L = \sqrt{6 + L} \implies L^2 - L - 6 = 0 \implies (L-3)(L+2) = 0$. Since $x_n > 0$, $L = 3$.

Answer: 3

1. Step 1: Calculate the first few terms: $x_1 = 1$, $x_2 = 3(1) - 2 = 1$, $x_3 = 3(1) - 2 = 1$.
2. Step 2: Observe that $x_n = 1$ for all $n$.
3. Step 3: A constant sequence is both monotone and bounded.
4. Step 4: The limit is simply the constant value 1.

Answer: Yes, it converges to 1.

1. Use induction to show $x_n < 2$. For $n=1$, $1 < 2$. If $x_n < 2$, then $x_{n+1} = \sqrt{2 + x_n} < \sqrt{2 + 2} = 2$.
2. Show the sequence is increasing. $x_{n+1}^2 - x_n^2 = 2 + x_n - x_n^2 = (2 - x_n)(1 + x_n)$. Since $x_n < 2$ and $x_n > 0$, this is positive, so $x_{n+1} > x_n$.
3. By the Monotone Convergence Theorem, the sequence converges to a limit $L$.
4. Solve $L = \sqrt{2 + L}$, which gives $L^2 - L - 2 = 0$. The roots are $2$ and $-1$. Since $x_n > 0$, $L = 2$.

Answer: $L = 2$

1. The term $\sin(n\pi/2)$ takes values $1, 0, -1, 0$ periodically.
2. The sequence $a_n$ takes values $1 + 1/1, 0 + 1/2, -1 + 1/3, 0 + 1/4, 1 + 1/5, \dots$
3. The subsequence of terms where $\sin(n\pi/2) = 1$ converges to $1$.
4. The subsequence of terms where $\sin(n\pi/2) = -1$ converges to $-1$.
5. The subsequence of terms where $\sin(n\pi/2) = 0$ converges to $0$.
6. The highest limit point is $1$ and the lowest is $-1$.

Answer: $\liminf a_n = -1, \limsup a_n = 1$

1. Consider $m > n$. The difference is $|a_m - a_n| = \sum_{k=n+1}^m \frac{1}{k^2}$.
2. Use the inequality $\frac{1}{k^2} < \frac{1}{k(k-1)} = \frac{1}{k-1} - \frac{1}{k}$.
3. The sum is a telescoping sum: $\sum_{k=n+1}^m (\frac{1}{k-1} - \frac{1}{k}) = \frac{1}{n} - \frac{1}{m}$.
4. Thus $|a_m - a_n| < \frac{1}{n}$.
5. For any $\epsilon > 0$, choose $N > 1/\epsilon$. Then for $m, n \ge N$, $|a_m - a_n| < 1/N < \epsilon$.
6. Therefore, the sequence is Cauchy.

Answer: The sequence is Cauchy because $|a_m - a_n| < 1/n$ for $m > n$.

1. Identify the first term $a$ by plugging in $n=1$: $a = 5(1/4)^0 = 5$.
2. Identify the common ratio $r = 1/4$.
3. Check convergence: $|1/4| < 1$, so the series converges.
4. Apply the formula $S = a / (1 - r) = 5 / (1 - 1/4) = 5 / (3/4)$.
5. Simplify the fraction: $5 \cdot (4/3) = 20/3$.

Answer: 20/3

1. Identify the first term $a = 3(-2/3)^0 = 3$.
2. Identify the common ratio $r = -2/3$.
3. Check convergence: $|-2/3| = 2/3 < 1$, so the series converges.
4. Apply the formula $S = a / (1 - r) = 3 / (1 - (-2/3)) = 3 / (5/3)$.
5. Simplify the fraction: $3 \cdot (3/5) = 9/5$.

Answer: 9/5

1. Identify the first term $a$ by plugging in the starting index $n=2$: $a = (1/3)^2 = 1/9$.
2. Identify the common ratio $r = 1/3$.
3. Check convergence: $|1/3| < 1$, so the series converges.
4. Apply the formula $S = a / (1 - r) = (1/9) / (1 - 1/3) = (1/9) / (2/3)$.
5. Simplify the fraction: $(1/9) \cdot (3/2) = 3/18 = 1/6$.

Answer: 1/6

1. Consider the block sum $\sum_{k=n+1}^{m} \frac{1}{k^2}$.
2. Apply the hint: $\sum_{k=n+1}^{m} \frac{1}{k^2} < \sum_{k=n+1}^{m} (\frac{1}{k-1} - \frac{1}{k})$.
3. This is a telescoping sum: $(\frac{1}{n} - \frac{1}{n+1}) + (\frac{1}{n+1} - \frac{1}{n+2}) + \dots + (\frac{1}{m-1} - \frac{1}{m})$.
4. The sum simplifies to $\frac{1}{n} - \frac{1}{m}$.
5. Since $\frac{1}{n} - \frac{1}{m} < \frac{1}{n}$, we can make this less than $\epsilon$ by choosing $N > \frac{1}{\epsilon}$.
6. Thus, the Cauchy criterion is satisfied.

Answer: The series converges because the block sum is bounded by $\frac{1}{n}$, which can be made arbitrarily small.

1. Since the series converges, it satisfies the Cauchy criterion.
2. For any $\epsilon > 0$, there exists $N$ such that for all $m > n \ge N$, $|\sum_{k=n+1}^{m} a_k| < \epsilon$.
3. Choose $m = n + 1$.
4. The sum becomes $|\sum_{k=n+1}^{n+1} a_k| = |a_{n+1}|$.
5. Therefore, $|a_{n+1}| < \epsilon$ for all $n \ge N$, which means $|a_k| < \epsilon$ for all $k > N$.

Answer: The result follows directly from the Cauchy criterion by setting $m = n + 1$.

1. The series is the alternating harmonic series.
2. We know from the Alternating Series Test (Calculus, Lesson 107) that this series converges.
3. By the theorem proved today, a series converges if and only if it satisfies the Cauchy criterion.
4. Therefore, the alternating harmonic series must satisfy the Cauchy criterion.
5. Specifically, the block sum $|S_m - S_n|$ for an alternating series with decreasing terms is bounded by the magnitude of the first omitted term, $|a_{n+1}| = \frac{1}{n+1}$.
6. Since $\frac{1}{n+1} \to 0$, the Cauchy criterion is satisfied.

Answer: Yes, it satisfies the Cauchy criterion because the series converges.

1. Observe that for $n \ge 1$, $n^2 + n > n^2$.
2. Therefore, $\frac{1}{n^2 + n} < \frac{1}{n^2}$.
3. We know that $\sum \frac{1}{n^2}$ is a convergent p-series with $p=2$.
4. By the Comparison Test, since our series is term-by-term smaller than a convergent series, it must also converge.

Answer: Converges

1. The terms $a_n = \frac{1}{n (\ln n)^3}$ are positive and non-increasing.
2. Apply condensation: $2^k a_{2^k} = 2^k \cdot \frac{1}{2^k (\ln 2^k)^3} = \frac{1}{(k \ln 2)^3}$.
3. This simplifies to $\frac{1}{(\ln 2)^3} \cdot \frac{1}{k^3}$.
4. The series $\sum \frac{1}{k^3}$ is a convergent p-series with $p=3$.
5. Since the condensed series converges, the original series converges.

Answer: Converges

1. Observe that for $n \ge 1$, $n + \sqrt{n} \le n + n = 2n$.
2. Therefore, $\frac{1}{n + \sqrt{n}} \ge \frac{1}{2n}$.
3. The series $\sum \frac{1}{2n} = \frac{1}{2} \sum \frac{1}{n}$ is a multiple of the harmonic series, which diverges.
4. By the Comparison Test, since our series is term-by-term larger than a divergent series, it must also diverge.

Answer: Diverges

1. Identify $a_n = \frac{n!}{n^n}$ and $a_{n+1} = \frac{(n+1)!}{(n+1)^{n+1}}$.
2. Compute the ratio: $\left| \frac{a_{n+1}}{a_n} \right| = \frac{(n+1)!}{(n+1)^{n+1}} \cdot \frac{n^n}{n!} = \frac{(n+1)n!}{(n+1)(n+1)^n} \cdot \frac{n^n}{n!} = \frac{n^n}{(n+1)^n}$.
3. Rewrite the ratio as $\left( \frac{n}{n+1} \right)^n = \frac{1}{(1 + 1/n)^n}$.
4. Take the limit as $n \to \infty$: $\lim_{n \to \infty} \frac{1}{(1 + 1/n)^n} = \frac{1}{e}$.
5. Since $e \approx 2.718$, $1/e < 1$.
6. By the Ratio Test, the series converges absolutely.

Answer: Converges absolutely

1. Identify $a_n = \left( \frac{3n+1}{2n-1} \right)^n$.
2. Apply the n-th root: $\sqrt[n]{|a_n|} = \frac{3n+1}{2n-1}$.
3. Take the limit as $n \to \infty$: $\lim_{n \to \infty} \frac{3n+1}{2n-1} = \frac{3}{2}$.
4. Since $3/2 > 1$, the Root Test implies the series diverges.

Answer: Diverges

1. Identify $a_n = \frac{(-1)^n}{n^2}$.
2. Apply the Ratio Test: $\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \frac{n^2}{(n+1)^2} = 1$.
3. The Ratio Test is inconclusive because $L=1$.
4. Check for absolute convergence by looking at $\sum |a_n| = \sum \frac{1}{n^2}$.
5. This is a p-series with $p=2$, which converges (Lesson 105).
6. Therefore, the series converges absolutely.

Answer: Converges absolutely

1. Check if terms alternate: The factor $(-1)^{n+1}$ ensures the signs flip every term.
2. Check if absolute values $a_n = \frac{1}{\ln(n+1)}$ are non-increasing: Since $\ln(x)$ is an increasing function, $\ln(n+2) > \ln(n+1)$, which means $\frac{1}{\ln(n+2)} < \frac{1}{\ln(n+1)}$.
3. Check the limit: $\lim_{n \to \infty} \frac{1}{\ln(n+1)} = 0$ because the logarithm grows without bound.
4. Since all three conditions are met, the series converges.

Answer: Converges

1. The error bound is $|S - S_n| \le a_{n+1}$.
2. We need $a_{n+1} \le 0.01$, where $a_{n+1} = \frac{1}{(n+1)^2}$.
3. Solve $\frac{1}{(n+1)^2} \le \frac{1}{100}$.
4. This implies $(n+1)^2 \ge 100$, so $n+1 \ge 10$.
5. Therefore, $n \ge 9$.

Answer: n = 9

1. Check the limit of the absolute terms: $a_n = \frac{n+1}{n} = 1 + \frac{1}{n}$.
2. Compute the limit: $\lim_{n \to \infty} (1 + \frac{1}{n}) = 1$.
3. The Alternating Series Theorem requires the limit to be $0$.
4. Since the limit is $1$, the terms of the series do not approach $0$.
5. By the Divergence Test, the series diverges.

Answer: Diverges

1. The series is alternating with $a_n = \frac{1}{n \ln(n)}$. Since $a_n$ decreases to $0$, the series converges by the Alternating Series Theorem.
2. The absolute series is $\sum \frac{1}{n \ln(n)}$. Using the integral test, $\int_2^{\infty} \frac{1}{x \ln(x)} dx = [\ln(\ln(x))]_2^{\infty} = \infty$. Thus, it diverges.
3. Since the series converges but not absolutely, it is conditionally convergent.
4. By today's theorem, $\sum a_n^+ = \infty$ and $\sum a_n^- = -\infty$.

Answer: Conditionally convergent; $\sum a_n^+ = \infty$ and $\sum a_n^- = -\infty$.

1. We know $a_n = a_n^+ + a_n^-$.
2. If $\sum a_n^-$ converges, then $\sum a_n$ is the sum of two convergent series, so it must converge.
3. If $\sum a_n$ converges, then $\sum a_n^- = \sum a_n - \sum a_n^+$.
4. Since the difference of two convergent series is convergent, $\sum a_n^-$ must converge.

Answer: The statement is true by the linearity of convergent series.

1. If $\sum a_n^+$ converges, then for the series to be conditionally convergent, $\sum |a_n|$ must diverge.
2. We know $\sum |a_n| = \sum a_n^+ - \sum a_n^-$.
3. If $\sum a_n^+$ converges and $\sum |a_n|$ diverges, then $\sum a_n^-$ must diverge to $-\infty$.
4. However, if $\sum a_n^+$ converges and $\sum a_n^-$ diverges, then the total sum $\sum a_n = \sum a_n^+ + \sum a_n^-$ must also diverge to $-\infty$.
5. This contradicts the definition of conditional convergence, which requires the total sum to converge.

Answer: No, because if $\sum a_n^+$ converges, the total series can only converge if $\sum a_n^-$ also converges, which would imply absolute convergence.

1. The first term is $1$.
2. Since the current sum is $1$, and our target is $1$, the algorithm immediately switches to negative terms to avoid exceeding the target.
3. We add negative terms $-\frac{1}{2}, -\frac{1}{4}, \dots$ until the sum is less than $1$.
4. Then we add positive terms $\frac{1}{3}, \frac{1}{5}, \dots$ until the sum exceeds $1$ again.

Answer: Start with $1$, then add negative terms until the sum is $< 1$, then positive terms until the sum is $> 1$.

1. No, such a rearrangement does not exist.
2. A fundamental property of absolutely convergent series is that every rearrangement converges to the same sum.
3. Therefore, for any permutation $\sigma$, $\sum a_{\sigma(n)} = S$.
4. Since $S \neq S + 1$, the target sum is impossible.

Answer: No, because absolutely convergent series are invariant under rearrangement.

1. Use the same strategy as the divergence to $+\infty$, but mirrored.
2. Add negative terms until the partial sum is less than $-1$.
3. Add exactly one positive term to ensure all terms are used.
4. Add negative terms until the partial sum is less than $-2$, and so on.
5. Since $\sum a_n^- = -\infty$, we can always push the sum below any negative integer $-k$.
6. Since $a_n \to 0$, the single positive terms cannot stop the divergence.

Answer: By greedily adding negative terms to pass $-1, -2, -3, \dots$ and adding single positive terms to maintain the bijection.

1. The individual sums are geometric series: $\sum (1/3)^n = 1/(1 - 1/3) = 3/2$. The product of the sums is $(3/2) \cdot (3/2) = 9/4$.
2. Compute $c_n = \sum_{k=0}^n (1/3)^k (1/3)^{n-k} = \sum_{k=0}^n (1/3)^n = (n+1)/3^n$.
3. Sum the resulting series: $\sum_{n=0}^{\infty} \frac{n+1}{3^n} = \sum \frac{n}{3^n} + \sum \frac{1}{3^n}$.
4. The geometric part is $3/2$. For the arithmetico-geometric part, we use the known result $\sum_{n=0}^{\infty} x^n = 1/(1-x)$. Differentiating term-by-term (Calculus L114) gives $\sum n x^{n-1} = 1/(1-x)^2$. Multiplying by $x$ gives $\sum n x^n = x/(1-x)^2$. For $x=1/3$, this is $(1/3)/(4/9) = 3/4$.
5. Total sum is $3/4 + 3/2 = 9/4$.

Answer: $\sum c_n = 9/4$

1. The terms are $c_n = \sum_{k=0}^n \frac{1}{k!} \frac{1}{(n-k)!}$.
2. Multiply and divide by $n!$: $c_n = \frac{1}{n!} \sum_{k=0}^n \frac{n!}{k!(n-k)!}$.
3. Recognize the binomial coefficient: $c_n = \frac{1}{n!} \sum_{k=0}^n \binom{n}{k}$.
4. Use the identity $\sum \binom{n}{k} = 2^n$: $c_n = \frac{2^n}{n!}$.
5. The resulting series is $\sum \frac{2^n}{n!}$, which is the Taylor series for $e^2$.

Answer: $c_n = \frac{2^n}{n!}$

1. The series $\sum \frac{(-1)^n}{n+1}$ converges conditionally, not absolutely.
2. Mertens' Theorem requires at least one series to converge absolutely to guarantee the product of sums.
3. As shown in the lesson, the terms $c_n$ of the Cauchy product of the alternating harmonic series behave like $2 \ln n / n$ for large $n$.
4. The series $\sum 2 \ln n / n$ diverges by comparison with the harmonic series $\sum 1/n$.
5. Since the Cauchy product diverges, it cannot converge to the finite value $(\ln 2)^2$.

Answer: The series is conditionally convergent and the resulting Cauchy product diverges.

1. Multiply $\frac{7}{16}$ by 2: $\frac{14}{16} = 0.875$, so $d_1 = 0$.
2. Multiply $0.875$ by 2: $1.75$, so $d_2 = 1$. Remainder is $0.75$.
3. Multiply $0.75$ by 2: $1.5$, so $d_3 = 1$. Remainder is $0.5$.
4. Multiply $0.5$ by 2: $1.0$, so $d_4 = 1$. Remainder is $0$.
5. The expansion is $0.0111_2$.

Answer: 0.0111_2

1. The terminating expansion is $0.1_5$.
2. To find the non-terminating version, subtract 1 from the last non-zero digit: $1-1 = 0$.
3. Append an infinite string of the digit $b-1$, which is $5-1 = 4$.
4. The expansion is $0.0444\dots_5$.

Answer: 0.0444\dots_5

1. Identify the first term $a = \frac{2}{3^1} = \frac{2}{3}$.
2. Identify the common ratio $r = \frac{1}{3}$.
3. Apply the formula $S = \frac{a}{1-r}$.
4. Compute $S = \frac{2/3}{1 - 1/3} = \frac{2/3}{2/3} = 1$.

Answer: 1