Exercise 21.11

Question

Prove the following standard facts about infinite series:
  \begin{enumerate}[label=(\alph*)]
  \item Show that if $(s_n)$ is a bounded sequence of real numbers and $s_n \leq s_{n+1}$ for each $n$, then $(s_n)$ converges.
  \item Let $(a_n)$ be a sequence of real numbers; define
    \begin{gather*}
      s_n = \sum_{i=1}^n a_i \,.
    \end{gather*}
    If $s_n 	o s$, we say that the 	extbf{	extit{infinite series}}
    \begin{gather*}
      \sum_{i=1}^\infty a_i
    \end{gather*}
    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 $cs+t$.
  \item Prove the 	extbf{	extit{comparison test}} for infinite series: If $\left|a_iight| \leq b_i$ for each $i$, and if the series $\sum b_i$ converges, then the series $\sum a_i$ converges.
    [Hint: Show that the series $\sum \left|a_iight|$ and $\sum c_i$ converge, where $c_i = \left|a_iight| + a_i$.]
  \item Given a sequence of functions $f_n: X 	o {\mathbb{R}}$, let
    \begin{gather*}
      s_n(x) = \sum_{i=1}^n f_i(x) \,.
    \end{gather*}
    Prove the 	extbf{	extit{Weierstrass M-test}} for uniform convergence: If $\left|f_i(x)ight| \leq M_i$ 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 $\left|s_k(x) - s_n(x)ight| \leq r_n$; conclude that $\left|s(x) - s_n(x)ight| \leq r_n$.]
  \end{enumerate}

Dan Whitman · Accepted Answer

\def\a{\alpha} \def\b{\beta} \def\d{\delta} \def\e{\epsilon} \def { ho} (a) \begin{proof} Clearly image of the sequence $S = \left\{s_n \mid n \in {{\mathbb{Z}}_+} ight\}$ is a set of real numbers that is bounded above since the sequence is bounded above. Therefore $s = \sup S$ exists, noting that of course $s_n \leq s$ for any $n \in {{\mathbb{Z}}_+}$ since $s_n \in S$. We claim that $s_n o s$. So consider any $\e > 0$ so that clearly $s-\e < s$, and hence $s-\e$ cannot be an upper bound of $S$ (since $s$ is the least upper bound). Thus there is an $N \in {{\mathbb{Z}}_+}$ where $s - \e < s_N$. Then, for any $n \geq N$ we have that $s-\e < S_N \leq s_n \leq s$ since the sequence is non-decreasing. Thus we have \begin{gather*} s - \e < s_n \ s - s_n < \e \ \left|s - s_n ight| < \e \ d(s_n, s) < \e \end{gather*} since $s \geq s_n$, where $d$ denotes the usual metric on ${\mathbb{R}}$. Hence $s_n \in B_d(s, \e)$, which shows that the sequence converges to $s$ since $n \geq N$ and $\e$ were arbitrary. \end{proof} (b) \begin{proof} Define the partials sums \begin{align*} s_n &= \sum_{i=1}^n a_i & t_n &= \sum_{i=1}^n b_i \end{align*} of $\sum a_i$ and $\sum b_i$ so that $s_n o s$ and $t_n o t$ by the definition of infinite series. Also define the constant sequence $c_n = c$ for real $c$, so that of course $c_n o c$. For any $n \in {{\mathbb{Z}}_+}$, we of course have \begin{gather*} c_n s_n + t_n = c \sum_{i=1}^n a_i + \sum_{i=1}^n b_i = \sum_{i=1}^n c a_i + \sum_{i=1}^n b_i = \sum_{i=1}^n (c a_i + b_i) \end{gather*} since these are just finite sums. It then follows from what was shown in Exercise~21.5 that \begin{gather*} \sum_{i=1}^\infty (c a_i + b_i) = \lim_{n o \infty} \sum_{i=1}^n (c a_i + b_i) = \lim_{n o \infty} (c_n s_n + t_n) = cs + t \end{gather*} as desired. \end{proof} (c) \begin{proof} Denote the partial sums \begin{align*} s_n &= \sum_{i=1}^n \left|a_i ight| & t_n &= \sum_{i=1}^n b_i \,. \end{align*} We know that $(t_n)$ converges by the definition of an infinite series, since $\sum b_i$ converges. So suppose that $t_n o t$. We also clearly have that $0 \leq \left|a_i ight| \leq b_i$ for all $i \in {{\mathbb{Z}}_+}$, so that each term in both sums is always non-negative. It then follows that the sequence of partial sums $(s_n)$ and $(t_n)$ are non-decreasing, and moreover that $t_n \leq t$ for all $n \in {{\mathbb{Z}}_+}$. Lastly, since each $\left|a_i ight| \leq b_i$, it follows from a simple inductive argument that each $s_n \leq t_n \leq t$. Hence $(s_n)$ is a non-decreasing sequence that is also bounded (by $t$), and so it converges by part~(a). Therefore the infinite series $\sum \left|a_i ight|$ converges by definition. Denote its convergence value by $u$. Now, we also have that clearly \begin{gather*} -\left|a_i ight| \leq a_i \leq \left|a_i ight| \ 0 \leq \left|a_i ight| + a_i \leq 2 \left|a_i ight| \end{gather*} for every $i \in {{\mathbb{Z}}_+}$, and that $\sum 2\left|a_i ight|$ converges to $2u$ by what was shown in part~(b). Therefore, by the same argument as above, the sequence of partial sums of $\sum (\left|a_i ight| + a_i)$ is non-decreasing and bounded by $2u$ so that the series converges, say the the value $v$. Then, again by part~(b), we have that the series \begin{gather*} \sum_{i=1}^\infty a_i = \sum_{i=1}^\infty (\left|a_i ight| + a_i - \left|a_i ight|) = \sum_{i=1}^\infty \left[(\left|a_i ight| + a_i) + (-1)\left|a_i ight| ight] \end{gather*} converges to $v - u$ since both $v = \sum (\left|a_i ight| + a_i)$ and $u = \sum \left|a_i ight|$ have been shown to converge. This shows the desired result. \end{proof} (d) \begin{proof} First, since $\left|f_i(x) ight| \leq M_i$ for all $x \in X$ and $\sum M_i$ converges, it follows from part~(c) that the series $\sum f_i(x)$ converges (as does the sequence $(s_n(x))$) for each $x \in X$. So set the function $s : X o {\mathbb{R}}$ to $s(x) = \lim_{n o \infty} s_n(x)$ for each $x \in X$. Thus $(s_n)$ converges pointwise to $s$ by \href{./exercise_21-6}{Definition~21.6.1}, and $s$ can be thought of as the (pointwise) infinite sum of the functions $f_i$. To show that $s_n o s$ uniformly, first define \begin{gather*} r_n = \sum_{i=n+1}^\infty M_i \end{gather*} for $n \in {{\mathbb{Z}}_+}$ as the partial series of $\sum M_i$ as in \href{./exercise_20-8}{Definition~20.8.1}. Now consider any $x \in X$, any $n \in {{\mathbb{Z}}_+}$, and any $k > n$. Then we have that \begin{gather*} \left|s_k(x) - s_n(x) ight| = \left|\sum_{i=1}^k f_i(x) - \sum_{i=1}^n f_i(x) ight| = \left|\sum_{i=n+1}^k f_i(x) ight| \leq \sum_{i=n+1}^k \left|f_i(x) ight| \leq \sum_{i=n+1}^k M_i \leq \sum_{i=n+1}^\infty M_i = r_n \,, \end{gather*} where we note that $\sum_{i=n+1}^k M_i \leq \sum_{i=n+1}^\infty M_i$ since each $M_i$ is non-negative. Since the absolute value function is continuous on ${\mathbb{R}}$, it follows from Theorem~21.3 and \href{./exercise_21-5}{Exercise~21.5} that the sequence $\left|s_k(x) - s_n(x) ight|$ converges as $k o \infty$, and moreover converges to $\left|s(x) - s_n(x) ight|$ since $s_k(x) o s(x)$ as $k o \infty$ as shown above. Then, since the above inequality holds for any $k > n$, it follows from \href{./exercise_20-8}{Lemma~20.8.1} that \begin{gather*} \left|s(x) - s_n(x) ight| = \lim_{k o \infty} \left|s_k(x) - s_n(x) ight| \leq \lim_{k o \infty} r_n = r_n \,. \end{gather*} Now consider any $\e > 0$. We know from \href{./exercise_20-8}{Lemma~20.8.4} that the sequence of partial series $(r_n)$ converges to zero since each $M_i$ is non-negative. Hence there is an $N \in {{\mathbb{Z}}_+}$ where $\left|r_n - 0 ight| < \e$ for all $n \geq N$. Moreover, \href{./exercise_20-8}{Lemma~20.8.4} asserts that each $r_n \geq 0$ so that $\left|r_n - 0 ight| = \left|r_n ight| = r_n < \e$ for all $n \geq N$. Thus, for any $n > N$ and any $x \in X$, we have \begin{gather*} \left|s(x) - s_n(x) ight| \leq r_n < \e \,. \end{gather*} This suffices to show that $(s_n)$ uniformly converges to $s$ since $\e$ was arbitrary. \end{proof}

Exercise 21.11

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Exercise 21.11

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