Section 21: Problem 11 Solution

Working problems is a crucial part of learning mathematics. No one can learn topology merely by poring over the definitions, theorems, and examples that are worked out in the text. One must work part of it out for oneself. To provide that opportunity is the purpose of the exercises.

James R. Munkres

Prove the following standard facts about infinite series:

(a) Show that if $(s_{n})$ is a bounded sequence of real numbers and $s_{n}\le s_{n+1}$ for each $n$ , then $(s_{n})$ converges.

(b) Let $(a_{n})$ be a sequence of real numbers; define $s_{n}=\sum_{i=1}^{n}a_{i}.$ If $s_{n}\rightarrow s$ , we say that the infinite series $\sum_{i=1}^{\infty}a_{i}$ converges to $s$ also. Show that if $\sum a_{i}$ converges to $s$ and $\sum b_{i}$ converges to $t$ , then $\sum(ca_{i}+b_{i})$ converges to $cs+t$ .

(c) Prove the comparison test for infinite series: If $|a_{i}|\le b_{i}$ for each $i$ , and if the series $\sum b_{i}$ converges, then the series $\sum a_{i}$ converges. [Hint: Show that the series $\sum|a_{i}|$ and $\sum c_{i}$ converge, where $c_{i}=|a_{i}|+a_{i}$ .]

(d) Given a sequence of functions $f_{n}:X\rightarrow\mathbb{R}$ , let $s_{n}(x)=\sum_{i=1}^{n}f_{i}(x).$ Prove the Weierstrass M-test for uniform convergence: If $|f_{i}(x)|<M_{i}$ , for all $x\in X$ and all $i$ , and if the series $\sum M_{i}$ converges, then the sequence $(s_{n})$ converges uniformly to a function $s$ . [Hint: Let $r_{n}=\sum_{i=n+1}^{\infty}M_{i}$ . Show that if $k>n$ , then $|s_{k}(x)-s_{n}(x)|\le r_{n}$ ; conclude that $|s(x)-s_{n}(x)|\le r_{n}$ .]

(a) It converges to $s=\sup_{n\in\mathbb{Z}_{+}}\{s_{n}\}$ , as for every $\epsilon>0$ , there is $N$ such that $s_{N}\in(s-\epsilon,s]$ , and, hence, for $n\ge N$ , $s_{N}\le s_{n}\le s$ implies $s_{n}\in(s-\epsilon,s]$ .

(b) If $c=0$ then $\sum(ca_{i}+b_{i})=\sum b_{i}$ converges to $cs+t=t$ . If $c\neq0$ , then for every $\epsilon>0$ , there is a sufficiently large $N$ such that for $n\ge N$ , $|\sum_{i=1}^{n}a_{i}-s|<\tfrac{\epsilon}{2|c|}$ and $|\sum_{i=1}^{n}b_{i}-s|<\tfrac{\epsilon}{2}$ , therefore, $|\sum_{i=1}^{n}(ca_{i}+b_{i})-(cs+t)|<\epsilon$ .

(c) Using the hint, $s_{n}=\sum_{i=1}^{n}|a_{i}|$ is a sequence such that $s_{n}\le s_{n+1}$ , and $s_{n}\le\sum_{i=1}^{n}b_{i}\le\sum_{i=1}^{\infty}b_{i}=b$ , hence, by (a), $(s_{n})$ converges. Similarly, $t_{n}=\sum_{i=1}^{n}c_{i}$ form a non-decreasing sequence of real numbers that is bounded by $2b$ . Therefore, by (b), $\sum a_{i}=\sum(-1\cdot|a_{i}|+c_{i})$ converges.

(d) By (c), $s_{n}$ converges point-wise to some function $s$ . Now, using the hint, for $k>n$ , $|s_{k}(x)-s_{n}(x)|\le\sum_{i=n+1,\ldots,k}|f_{i}(x)|\le r_{n}$ . Therefore, $|s(x)-s_{n}(x)|\le r_{n}$ uniformly for all $x$ (if that was not true than for some $x\in X$ , $|s(x)-s_{n}(x)|-r_{n}=\delta>0$ , and for some sufficiently large $k$ such that $|s(x)-s_{k}(x)|<\frac{\delta}{2}$ , $|s_{k}(x)-s_{n}(x)|\ge$ $|s(x)-s_{n}(x)|-|s(x)-s_{k}(x)|>$ $r_{n}+\delta-\tfrac{\delta}{2}>r_{n}$ ). Now, note that for any $\epsilon>0$ we can find $N$ such that for $n\ge N$ , $r_{n}=\sum_{i\in\mathbb{Z}_{+}}M_{i}-\sum_{i=1}^{n}M_{i}<\epsilon$ .

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Section 21: Problem 11 Solution

Section 21: Problem 11 Solution