Assignment 6
Table of contents
url: https://ocw.mit.edu/courses/18-100a-real-analysis-fall-2020/resources/mit18_100af20_hw6/
Chapters and exercises given with a numbering are from Basic Analysis: Introduction to Real Analysis (Vol I) by J. Lebl.
Exercise 2.5.3
Decide the convergence or divergence of the following series:
a) $\sum_{n=1}^\infty \frac{3}{9n + 1}$
$\frac{3}{9n + 1} > \frac{3}{9n + n} = \frac{3}{10n} > \frac{1}{4n}$. $\sum_{n=1}^\infty \frac{1}{4n} = \frac{1}{4} \sum_{n=1}^\infty \frac{1}{n}$ is divergent.
By comparison test, $\sum_{n=1}^\infty \frac{3}{9n + 1}$ is also divergent.
b) $\sum_{n=1}^\infty \frac{1}{2n - 1}$
$\frac{1}{2n - 1} > \frac{1}{2n}$. Since $\sum_{n=1}^\infty \frac{1}{2n}$ is divergent, $\sum_{n=1}^\infty \frac{1}{2n - 1}$ is divergent.
c) $\sum_{n=1}^\infty \frac{(-1)^n}{n^2}$
$\left\vert \frac{(-1)^n}{n^2} \right\vert < \frac{1}{n^2}$. By the $p$-test, $\sum_{n=1}^\infty \frac{1}{n^2}$ is convergent.
Then by the comparison test, $\sum_{n=1}^\infty \frac{(-1)^n}{n^2}$ is convergent.
d) $\sum_{n=1}^\infty \frac{1}{n(n + 1)}$
$\frac{1}{n(n + 1)} = \frac{1}{n^2 + n} < \frac{1}{n^2}$. Convergent.
e) $\sum_{n=1}^\infty n e^{-n^2}$
$n e^{-n^2} = \frac{n}{e^{n^2}} < \frac{n}{1 + n^2} < \frac{n^2}{1 + n^2} \to 1$. Convergent.
Exercise 2.5.4
a) Prove that if $\sum_{n=1}^\infty x_n$ converges, then $\sum_{n=1}^\infty (x_{2n} + x_{2n+1})$ also converges.
Let $S_m := \sum_{n=1}^m (x_{2n} + x_{2n+1}) = \sum_{n=2}^{2m+1} x_n$. And $s_m := \sum_{n=1}^m x_n$ is the partial sum of $\sum x_n$.
Since $\sum x_n$ converges, $\lim_{m \to \infty} \sum_{n=1}^m x_n = \lim_{m \to \infty} s_m = x$
Thus, $\lim_{m \to \infty} S_m = \lim_{m \to \infty} s_{2m+1} - x_1 = x - x_1$
b) Find an explicit example where the converse does not hold.
Show that when $\sum_{n=1}^\infty (x_{2n} + x_{2n+1})$ converges, $\sum x_n$ diverges.
Let $x_n := (-1)^n$. Then $x_{2n} + x_{2n+1} = 1 - 1 = 0$. Thus, $\sum_{n=1}^\infty 0 = 0$ converges. But $\sum_{n=1}^\infty x_n = \sum_{n=1}^\infty (-1)^n$ does not converge.
Exercise 2.5.10
Prove the triangle inequality for series: If $\sum_{n=1}^\infty x_n$ converges absolutely, then
\[\left\vert \sum_{n=1}^\infty x_n \right\vert \leq \sum_{n=1}^\infty \vert x_n\vert .\]Claim: For all $m \in \mathbb{N}$, $\left\vert \sum_{n=1}^m x_n \right\vert \leq \sum_{n=1}^m \vert x_n\vert $.
Proof: (i) For $m = 1$, $\vert x_1\vert \leq \vert x_1\vert $ is trivially true.
(ii) Assume the inequality holds for $m = k$, i.e. $\left\vert \sum_{n=1}^k x_n \right\vert \leq \sum_{n=1}^k \vert x_n\vert $. Then for $m = k + 1$, \(\left\vert \sum_{n=1}^{k+1} x_n \right\vert = \left\vert \sum_{n=1}^k x_n + x_{k+1} \right\vert \leq \left\vert \sum_{n=1}^k x_n \right\vert + \vert x_{k+1}\vert \leq \sum_{n=1}^{k+1} \vert x_n\vert\)
Thus by induction, $\left\vert \sum_{n=1}^m x_n \right\vert \leq \sum_{n=1}^m \vert x_n\vert $ for all $m \in \mathbb{N}$. $\quad\square$
Since $\sum_{n=1}^\infty x_n$ converges absolutely, $\lim_{m \to \infty} \sum_{n=1}^m \vert x_n\vert $ exists. By comparison test, $\left\vert \sum_{n=1}^m x_n \right\vert $ also converges. Taking limits from both sides: $\left\vert \sum_{n=1}^\infty x_n \right\vert \leq \sum_{n=1}^\infty \vert x_n\vert $.
Exercise 2.6.1
Decide the convergence or divergence of the following series:
a) $\sum_{n=1}^\infty \frac{1}{2^{2n+1}}$
Let $a_n := \frac{1}{2^{2n+1}}$. Then $\left\vert \frac{a_{n+1}}{a_n} \right\vert = \left\vert \frac{2^{2n+1}}{2^{2(n+1)+1}} \right\vert = \frac{1}{4} < 1$. By the ratio test, $\sum a_n$ converges absolutely.
b) $\sum_{n=1}^\infty \frac{(-1)^n(n - 1)}{n}$
Let $b_n := \frac{(-1)^n(n - 1)}{n}$. Then $\left\vert \frac{b_{n+1}}{b_n} \right\vert = \left\vert -\frac{n}{n+1} \cdot \frac{n}{n - 1} \right\vert = \frac{n^2}{n^2 - 1} \to 1$. Ratio test does not work.
Notice that $\sum b_n$ converges if and only if $\lim_{n \to \infty} b_n = 0$. But $b_{2n} = \frac{2n - 1}{2n} \to 1$, and $b_{2n - 1} = -\frac{2n - 2}{2n - 1} \to -1$. Thus $b_n$ does not converge to 0. Therefore, $\sum b_n$ is divergent.
c) $\sum_{n=1}^\infty \frac{(-1)^n}{n^{1/10}}$
Let $c_n := \frac{(-1)^n}{n^{1/10}}$. Since $\frac{1}{n^{1/10}} > 0$ and is decreasing, by the alternating series test, $\sum c_n$ converges.
d) $\sum_{n=1}^\infty \frac{n^n}{(n + 1)^{2n}}$
Let $d_n := \frac{n^n}{(n + 1)^{2n}}$. Consider
\[\sup \{ \vert d_k\vert^{1/k} : k \geq n \} = \sup \left\{ \frac{k}{(k + 1)^2} : k \geq n \right\}\]Since $0 < \frac{k}{(k + 1)^2} = \frac{1}{k + 1 + \frac{1}{k}} \leq \frac{1}{4}$, $\frac{k}{(k + 1)^2}$ is bounded. And $\left( \frac{k}{(k + 1)^2} \right)’ = \frac{1 - k}{(1 + k)^3} < 0$ for all $k \in (1, +\infty)$, so $\frac{k}{(k + 1)^2}$ is monotonically decreasing, hence convergent. Therefore,
\[\limsup_{n \to \infty} \vert d_n\vert^{1/n} = \limsup_{n \to \infty} \sup \{ \vert d_k\vert ^{1/k} : k \geq n \} = \lim_{n \to \infty} \vert d_n\vert^{1/n} = \lim_{n \to \infty} \frac{n}{(n + 1)^2} = 0 < 1\]By the root test, $\sum d_n$ is convergent.
Exercise 2.6.13
Find a series $\sum_{n=1}^\infty x_n$ that converges, but $\sum_{n=1}^\infty x_n^2$ diverges.
Let $x_n := \frac{(-1)^n}{\sqrt{n}}$. Since $\lim_{n \to \infty} x_n = 0$, the series $\sum_{n=1}^\infty x_n$ is convergent by the alternating series test.
But $\sum_{n=1}^\infty x_n^2 = \sum_{n=1}^\infty \frac{1}{n}$, which is the harmonic series and diverges.