$\displaystyle{\sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}}$ .
$\displaystyle{\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{(2n-1)^3} = \frac{\pi^3}{32}}$ .

Proof. Taking $x = 0$ in the given formula we have
$\begin{align*} && \sum_{n=1}^{\infty} \frac{\cos (nx)}{n^2} &= \frac{x^2}{4} - \frac{\pi x}{2} + \frac{\pi^2}{6} \\[9pt] \implies && \sum_{n=1}^{\infty} \frac{\cos (0)}{n^2} &= \frac{\pi^2}{6} \\[9pt] \implies && \sum_{n=1}^{\infty} \frac{1}{n^2} &= \frac{\pi^2}{6}. \qquad \blacksquare \end{align*}$
Proof. Starting with the given equation we have for $0 \leq x \leq 2 \pi$ ,
$\begin{align*} && \sum_{n=1}^{\infty} \frac{\cos (nx)}{n^2} &= \frac{x^2}{4} - \frac{\pi x}{2} + \frac{\pi^2}{6} \\[9pt] \implies && \int \sum_{n=1}^{\infty} \frac{\cos (nx)}{n^2} \, dx &= \int \left( \frac{x^2}{4} - \frac{\pi x}{4} + \frac{\pi^2}{6} \right) \, dx \\[9pt] \implies && \sum_{n=1}^{\infty} \left(\int \frac{ \cos (nx)}{n^2} \, dx \right)&= \frac{x^3}{12} - \frac{\pi x^2}{4} + \frac{\pi^2 x}{6} \\[9pt] \implies && \sum_{n=1}^{\infty} \frac{\sin (nx)}{n^3} &= \frac{x^3}{12} - \frac{\pi x^2}{4} + \frac{\pi^2 x}{6}. \end{align*}$

Take $x = \frac{\pi}{2}$ , then we have

$\begin{align*} && \sum_{n=1}^{\infty} \frac{\sin \left( \frac{\pi}{2} \cdot n \right)}{n^3} &= \frac{ \left(\frac{\pi}{2} \right)^3}{12} - \frac{ \pi \left( \frac{\pi}{2} \right)^2}{4} + \frac{\pi^2 \left( \frac{\pi}{2} \right)}{6} \\[9pt] \implies && \sum_{n=1}^{\infty} \frac{\sin \left( n \pi - \frac{\pi}{2} \right)}{(2n-1)^3} &= \frac{\pi^3}{32} &(\sin (n \pi) = 0 \text{ for even }n)\\[9pt] \implies && \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{(2n-1)^3} &= \frac{\pi^3}{32}. \qquad \blacksquare \end{align*}$

Prove an integral formula for ∑ (sin (nx)) / n²

by RoRi

March 31, 2016

Prove that the series

$\sum_{n=1}^{\infty} \frac{\sin (nx)}{n^2}$

converges for all $x \in \mathbb{R}$ and let $f(x)$ denote the value of this sum for each $x$ . Prove that $f(x)$ is continuous for $x \in [0, \pi]$ and prove that

$\int_0^{\pi} f(x) \, dx = 2 \sum_{n=1}^{\infty} \frac{1}{(2n-1)^3}.$

Proof. First, the series converges for all real $x$ by the comparison test since

$|\sin (nx)| \leq 1 \quad \implies \quad \left|\frac{\sin (nx)}{n^2}\right| \leq \frac{1}{n^2}$

for all $n$ . Therefore, the convergence of $\sum \frac{1}{n^2}$ implies the convergence of $\sum \frac{\sin (nx)}{n^2}$ . Furthermore, this convergence is uniform by the Weierstrass $M$ -test with $M_n$ given by $\frac{1}{n^2}$ , and again $\sum \frac{1}{n^2}$ converges. Thus, by Theorem 11.2 (page 425 of Apostol),

$f(x) = \sum_{n=1}^{\infty} \frac{\sin (nx)}{n^2}$

is continuous on the interval $[0, \pi]$ . Therefore, we may apply Theorem 11.4 (page 426 of Apostol):

$\begin{align*} && \sum_{k=1}^{\infty} \int_0^{\pi} \frac{\sin (kt)}{k^2} \, dt &= \int_0^{\pi} \sum_{k=1}^{\infty} \frac{ \sin (kt)}{t^2} \, dt \\[9pt] \implies && \sum_{k=1}^{\infty} \left( \frac{-\cos (kt)}{k^3} \Bigr \rvert_0^{\pi} \right) &= \int_0^{\pi} f(x) \, dx \\[9pt] \implies && \int_0^{\pi} f(x) \, dx &= 2 \sum_{n=1}^{\infty} \frac{1}{(2n-1)^3} \end{align*}$

since $\cos 0 - \cos (k \pi) = 0$ if $k = 2n$ and equals 2 if $k = 2n-1. \qquad \blacksquare$

Prove that the improper integral ∫ f(x) dx and ∑ f(n) both converge or both diverge

by RoRi

March 31, 2016

Assume that $f$ is a monotonically decreasing function for all $x \geq 1$ and that
$\lim_{x \to +\infty} f(x) = 0.$

Prove that the improper integral and the series

$\int_1^{\infty} f(x) \, dx \qquad \text{and} \qquad \sum_{n=1}^{\infty} f(n)$

both converge or both diverge.
Give a counterexample to the theorem in part (a) in the case that $f$ is not monotonic, i.e., find a non-monotonic function $f$ such that $\sum f(n)$ converges but $\int_1^{\infty} f(x) \, dx$ diverges.

Incomplete.

Show that the limit of ∑ 1/k = log (p/q) where the sum is from k = qn to pn

by RoRi

March 31, 2016

For given integers $p$ and $q$ with $1 \leq q \leq p$ , prove
$\lim_{n \to \infty} \sum_{k=qn}^{pn} \frac{1}{k} = \log \frac{p}{q}.$
Consider the series
$1 + \frac{1}{3} + \frac{1}{5} - \frac{1}{2} - \frac{1}{4} + \frac{1}{7} + \frac{1}{9} + \frac{1}{11} - \frac{1}{6} - \frac{1}{8} + \ + \ + \ - \ - \ \cdots.$

This is a rearrangement of the alternating harmonic series ( $\sum \frac{(-1)^n}{n}$ ) in which there are three positive terms followed by two negative terms. Prove that the series converges and that the sum is equal to $\log 2 + \frac{1}{2} \log \frac{3}{2}$ .

Incomplete.

Prove that the sum of reciprocals of integers with no zeros in their decimal representation converges

by RoRi

March 31, 2016

Consider the positive integer with no zeros in their decimal representation:

$n_1 = 1, \ n_2 = 2, \ \ldots \ n_9 = 9, \ n_{10} = 11, \ldots, n_{18} = 19, \ n_{19} = 21, \ldots.$

Prove that the series

$\sum_{n=1}^{\infty} \frac{1}{n_k}$

converges. Further, prove that the sum is less than 90.

Incomplete.

Find integers a such that ∑ (n!)³ / (an)! converges

by RoRi

March 29, 2016

Find all positive integers $a$ such that the series

$\sum_{n=1}^{\infty} \frac{(n!)^3}{(an)!}$

converges.

We use the ratio test,

$\begin{align*} \lim_{n \to \infty} \left( \frac{((n+1)!)^3}{(a(n+1))!} \cdot \frac{(an)!}{(n!)^3} \right) = \lim_{n \to \infty} \frac{(n+1)^3}{(an+1)(an+2)\cdots (an+a)}. \end{align*}$

Since the largest power of $n$ in the denominator is $a^n n^a$ , we have that this limit diverges for $a = 1, 2$ . If $a = 3$ then the limit is $\frac{1}{9}$ , and the limit is 0 if $a > 3$ . Thus, the series converges for all $a \geq 3$ .

Find real c such that ∑ (n!)^c / (3n)! converges

by RoRi

March 29, 2016

Find all $c \in \mathbb{R}$ such that the following series is convergent:

$\sum_{n=1}^{\infty} \frac{(n!)^c}{(3n)!}.$

Consider the ratio test,

$\begin{align*} \lim_{n \to \infty} \frac{a_{n+1}}{a_n} &= \lim_{n \to \infty} \left( \frac{((n+1)!)^c}{(3(n+1))!} \cdot \frac{(3n)!}{(n!)^c} \right) \\[9pt] &= \lim_{n \to \infty} \left( \left( \frac{(n+1)!}{n!}\right)^c \cdot \frac{1}{(3n+1)(3n+2)(3n+3)} \right) \\[9pt] &= \lim_{n \to \infty} \frac{(n+1)^c}{(3n+1)(3n+2)(3n+3)}. \end{align*}$

This limit is 0 if $c < 3$ and is $\frac{1}{9}$ if $c = 3$ . If $c > 3$ then the limit diverges. Hence, by the ratio test we have that the series converges for $c \leq 3$ and diverges for $c > 3$ .

Prove relationships between a given series

by RoRi

March 29, 2016

Let $\{ a_n \}$ be a sequence of positive terms.

Prove or give a counterexample: if $\sum_{n=1}^{\infty} a_n$ diverges then $\sum_{n=1}^{\infty} a_n^2$ also diverges.
Prove or give a counterexample: if $\sum_{n=1}^{\infty} a_n^2$ converges then $\sum_{n=1}^{\infty} \frac{a_n}{n}$ also converges.

Counterexample. Let $a_n = \frac{1}{n}$ . Then
$\sum_{n=1}^{\infty} \frac{1}{n}$

diverges. Furthermore, $a_n^2 = \frac{1}{n^2}$ and

$\sum_{n=1}^{\infty} \frac{1}{n^2}$

converges.
Proof. We know from a previous exercise (Section 10.20, Exercise #51) we know
$\sum a_n \quad \text{converges} \quad \implies \quad \sum \sqrt{a_n}n^{-p} \quad \text{converges}$

for $p > \frac{1}{2}$ . So, for this case define $b_n = a_n^2$ . Then $\sum b_n = \sum a_n^2$ converges by hypothesis. Furthermore, (since $a_n > 0$ so $\sqrt{a_n^2} = a_n$ ),

$\sum \sqrt{b_n} n^{-1} = \sum \frac{a_n}{n}.$

Since this is the case $p = 1$ , we have proved the statement $. \qquad \blacksquare$

Determine the convergence of ∑ ( (n^a + 1)^1/2 – (n^a)^1/2

by RoRi

March 29, 2016

Prove that if $a > 2$ then the series

$\sum_{n=0}^{\infty} \left( \sqrt{n^a + 1} - \sqrt{n^a} \right)$

converges, and that it diverges if $a = 2$ .

Proof. We have

$\begin{align*} \sqrt{n^a + 1} - \sqrt{n^a} &= \frac{\left( \sqrt{n^a+1} - \sqrt{n^a} \right) \left( \sqrt{n^a + 1} + \sqrt{n^a} \right)}{\sqrt{n^a+1} + \sqrt{n^a}} \\[9pt] &= \frac{1}{\sqrt{n^a + 1} + \sqrt{n^a}}. \end{align*}$

Then, we use the limit comparison test with the series $\sum \frac{1}{n^{\frac{a}{2}}}$ .

$\begin{align*} \lim_{n \to \infty} \frac{ \frac{1}{n^{\frac{a}{2}}}}{\frac{1}{\sqrt{n^a+1} + \sqrt{n^a}}} &= \lim_{n \to \infty} \left( \sqrt{ 1 + \frac{1}{n^a}} + \sqrt{1} \right) \\[9pt] &= 2. \end{align*}$

Thus, the given series converges or diverges as $\sum \frac{1}{n^{\frac{a}{2}}}$ does. But, we know $\sum \frac{1}{n^{\frac{a}{2}}}$ converges for $a > 2$ and diverges for $a = 2$ as desired $. \qquad \blacksquare$