4.13.E: More Problems on Series of Functions
Verify Note 3 and Example \((\mathrm{c})\) in detail.
Show that the so-called hyperharmonic series of order \(p\),
\[
\sum \frac{1}{n^{p}} \quad\left(p \in E^{1}\right) ,
\]
converges iff \(p>1\).
[Hint: If \(p \leq 1\),
\[
\sum_{n=1}^{\infty} \frac{1}{n^{p}} \geq \sum_{n=1}^{\infty} \frac{1}{n}=+\infty \quad(\text { Example }(\mathrm{b})) .
\]
If \(p>1\),
\[
\begin{aligned} \sum_{n=1}^{\infty} \frac{1}{n^{p}} &=1+\left(\frac{1}{2^{p}}+\frac{1}{3^{p}}\right)+\left(\frac{1}{4^{p}}+\cdots+\frac{1}{7^{p}}\right)+\left(\frac{1}{8^{p}}+\cdots+\frac{1}{15^{p}}\right)+\cdots \\ & \leq 1+\left(\frac{1}{2^{p}}+\frac{1}{2^{p}}\right)+\left(\frac{1}{4^{p}}+\cdots+\frac{1}{4^{p}}\right)+\left(\frac{1}{8^{p}}+\cdots+\frac{1}{8^{p}}\right)+\cdots \\ &=\sum_{n=0}^{\infty} \frac{1}{\left(2^{p-1}\right)^{n}} . \end{aligned}
\]
a convergent geometric series. Explain each step.]
\(\Rightarrow 3.\) Prove the refined comparison test:
(i) If two series of constants, \(\sum\left|a_{n}\right|\) and \(\sum\left|b_{n}\right|,\) are such that the sequence \(\left\{\left|a_{n}\right| /\left|b_{n}\right|\right\}\) is bounded in \(E^{1},\) then
\[
\sum_{n=1}^{\infty}\left|b_{n}\right|<+\infty \text { implies } \sum_{n=1}^{\infty}\left|a_{n}\right|<+\infty .
\]
(ii) If
\[
0<\lim _{n \rightarrow \infty} \frac{\left|a_{n}\right|}{\left|b_{n}\right|}<+\infty ,
\]
then \(\sum\left|a_{n}\right|\) converges if and only if \(\sum\left|b_{n}\right|\) does.
What is
\[
\lim _{n \rightarrow \infty} \frac{\left|a_{n}\right|}{\left|b_{n}\right|}=+\infty ?
\]
[Hint: If (\forall n)|a_{n}| / |b_{n}| \leq K\), then \(|a_{n}| \leq K |b_{n}|.\)]
Test \(\sum a_{n}\) for absolute convergence in each of the following. Use Problem 3 or Theorem 2 or the indicated references.
(i) \(a_{n}=\frac{n+1}{\sqrt{n^{4}+1}}\left(\text { take } b_{n}=\frac{1}{n}\right)\);
(ii) \(a_{n}=\frac{\cos n}{\sqrt{n^{3}-1}}\left(\text { take } b_{n}=\frac{1}{\sqrt{n^{3}}} ; \text { use Problem } 2\right)\);
(iii) \(a_{n}=\frac{(-1)^{n}}{n^{p}}(\sqrt{n+1}-\sqrt{n}), p \in E^{1}\);
(iv) \(a_{n}=n^{5} e^{-n}\) (use Problem 18 of Chapter 3, §15);
(v) \(a_{n}=\frac{2^{n}+n}{3^{n}+1}\);
(vi) \(a_{n}=\frac{(-1)^{n}}{(\log n)^{q}} ; n \geq 2\);
(vii) \(a_{n}=\frac{(\log n)^{q}}{n\left(n^{2}+1\right)}, q \in E^{1}\).
[Hint for (vi) and (vii): From Problem 14 in §2, show that
\[
\lim _{y \rightarrow+\infty} \frac{y}{(\log y)^{q}}=+\infty
\]
and hence
\[
\lim _{n \rightarrow \infty} \frac{(\log n)^{q}}{n}=0 .
\]
Then select \(b_{n}.\)
Prove that \(\sum_{n=1}^{\infty} \frac{n^{n}}{n !}=+\infty\).
[Hint: Show that \(n^{n} / n !\) does not tend to 0.]
Prove that \(\lim _{n \rightarrow \infty} \frac{x^{n}}{n !}=0\).
[Hint: Use Example (d) and Theorem 4.]
Use Theorems \(3,5,6,\) and 7 to show that \(\sum\left|f_{n}\right|\) converges uniformly on \(B,\) provided \(f_{n}(x)\) and \(B\) are as indicated below, with \(0<a<+\infty\) and \(b \in E^{1}.\) For parts \((\text { ix })-(\text { xii }),\) find \(M_{n}=\max _{x \in B}\left|f_{n}(x)\right|\) and use Theorem \(3 .\) (Calculus rules for maxima are assumed known.)
(i) \(\frac{x^{2 n}}{(2 n) !} ;[-a, b]\).
(ii) \((-1)^{n+1} \frac{x^{2 n-1}}{(2 n-1) !} ;[-a, b]\).
(iii) \(\frac{x^{n}}{n^{n}} ;[-a, a]\).
(iv) \(n^{3} x^{n} ;[-a, a](a<1)\).
\(\left.\text { (v) } \frac{\sin n x}{n^{2}} ; B=E^{1} \text { (use Problem } 2\right)\).
(vi) \(e^{-n x} \sin n x ;[a,+\infty)\).
(vii) \(\frac{\cos n x}{\sqrt{n^{3}+1}} ; B=E^{1}\).
(viii) \(a_{n} \cos n x,\) with \(\sum_{n=1}^{\infty}\left|a_{n}\right|<+\infty ; B=E^{1}\).
(ix) \(x^{n} e^{-n x} ;[0,+\infty)\).
(x) \(x^{n} e^{n x} ;\left(-\infty, \frac{1}{2}\right]\).
(xi) \((x \cdot \log x)^{n}, f_{n}(0)=0 ;\left[-\frac{3}{2}, \frac{3}{2}\right]\).
(xii) \(\left(\frac{\log x}{x}\right)^{n} ;[1,+\infty)\).
(xiii) \(\frac{q(q-1) \cdots(q-n+1) x^{n}}{n !}, q \in E^{1} ;\left[-\frac{1}{2}, \frac{1}{2}\right]\).
\(\Rightarrow 8.\) (Summation by parts.) Let \(f_{n}, h_{n},\) and \(g_{n}\) be real or complex functions (or let \(f_{n}\) and \(h_{n}\) be scalar valued and \(g_{n}\) be vector valued). Let \(f_{n}=\) \(h_{n}-h_{n-1}(n \geq 2) .\) Verify that \((\forall m>n>1)\)
\[
\begin{aligned} \sum_{k=n+1}^{m} f_{k} g_{k} &=\sum_{k=n+1}^{m}\left(h_{k}-h_{k-1}\right) g_{k} \\ &=h_{m} g_{m}-h_{n} g_{n+1}-\sum_{k=n+1}^{m-1} h_{k}\left(g_{k+1}-g_{k}\right) . \end{aligned}
\]
[Hint: Rearrange the sum.]
\(\Rightarrow 9.\) (Abel's test.) Let the \(f_{n}, g_{n},\) and \(h_{n}\) be as in Problem \(8,\) with \(h_{n}=\sum_{i=1}^{n} f_{i} .\) Suppose that
(i) the range space of the \(g_{n}\) is complete;
(ii) \(\left|g_{n}\right| \rightarrow 0\) (uniformly) on a set \(B ;\) and
(iii) the partial sums \(h_{n}=\sum_{i=1}^{n} f_{i}\) are uniformly bounded on \(B ;\) i.e.,
\[
\left(\exists K \in E^{1}\right)(\forall n) \quad\left|h_{n}\right|<K \text { on } B .
\]
Then prove that \(\sum f_{k} g_{k}\) converges uniformly on \(B\) if \(\sum\left|g_{n+1}-g_{n}\right|\) does.
\(\text { (This always holds if the } g_{n} \text { are real and } g_{n} \geq g_{n+1} \text { on } B .)\)
[Hint: Let \(\varepsilon>0 .\) Show that
\[
(\exists k)(\forall m>n>k) \quad \sum_{i=n+1}^{m}\left|g_{i+1}-g_{i}\right|<\varepsilon \text { and }\left|g_{n}\right|<\varepsilon \text { on } B .
\]
Then use Problem 8 to show that
\[
\left|\sum_{i=n+1}^{m} f_{i} g_{i}\right|<3 K \varepsilon .
\]
Apply Theorem \(3^{\prime}\) of §12.]
\(\Rightarrow \mathbf{9}^{\prime}.\) Prove that if \(\sum a_{n}\) is a convergent series of constants \(a_{n} \in E^{1}\) and if \(\left\{b_{n}\right\}\) is a bounded monotone sequence in \(E^{1},\) then \(\sum a_{n} b_{n}\) converges.
[Hint: Let \(b_{n} \rightarrow b\). Write
\[
a_{n} b_{n}=a_{n}\left(b_{n}-b\right)+a_{n} b
\]
\(\left.\text { and use Problem } 9 \text { with } f_{n}=a_{n} \text { and } g_{n}=b_{n}-b .\right]\)
\(\Rightarrow 10.\) Prove the Leibniz test for alternating series: If \(\left\{b_{n}\right\} \downarrow\) and \(b_{n} \rightarrow 0\) in \(E^{1},\) then \(\sum(-1)^{n} b_{n}\) converges, and the sum \(\sum_{n=1}^{\infty}(-1)^{n} b_{n}\) differs from \(s_{n}=\sum_{k=1}^{n}(-1)^{k} b_{k}\) by \(b_{n+1}\) at most.
\(\Rightarrow 11.\) (Dirichlet test.) Let the \(f_{n}, g_{n},\) and \(h_{n}\) be as in Problem 8 with \(\sum_{n=0}^{\infty} f_{n}\) uniformly convergent on \(B\) to a function \(f,\) and with
\[
h_{n}=-\sum_{i=n+1}^{\infty} f_{i} \text { on } B .
\]
Suppose that
(i) the range space of the \(g_{n}\) is complete; and
(ii) there is \(K \in E^{1}\) such that
\[
\left|g_{0}\right|+\sum_{n=0}^{\infty}\left|g_{n+1}-g_{n}\right|<K \text { on } B .
\]
Show that \(\sum f_{n} g_{n}\) converges uniformly on \(B\).
[Proof outline: We have
\[
\left|g_{n}\right|=\left|g_{0}+\sum_{i=0}^{n-1}\left(g_{i+1}-g_{i}\right)\right| \leq\left|g_{0}\right|+\sum_{i=0}^{n-1}\left|g_{i+1}-g_{i}\right|<K \quad \text { by (ii). }
\]
Also,
\[
\left|h_{n}\right|=\left|\sum_{i=0}^{n} f_{i}-f\right| \rightarrow 0 \text { (uniformly) on } B
\]
by assumption. Hence
\[
(\forall \varepsilon>0)(\exists k)(\forall n>k) \quad\left|h_{n}\right|<\varepsilon \text { on } B .
\]
Using Problem \(8,\) obtain
\[
(\forall m>n>k)\left|\sum_{i=n+i}^{m} f_{i} g_{i}\right|<2 K \varepsilon .]
\]
Prove that if \(0<p \leq 1\), then \(\sum \frac{(-1)^{n}}{n^{p}}\) converges conditionally.
[Hint: Use Problems 11 and 2 .]
\(\Rightarrow 13.\) Continuing Problem 14 in §12, prove that if \(\sum\left|f_{n}\right|\) and \(\sum\left|g_{n}\right|\) converge on \(B\) (pointwise or uniformly), then so do the series
\[
\sum\left|a f_{n}+b g_{n}\right|, \sum\left|f_{n} \pm g_{n}\right|, \text { and } \sum\left|a f_{n}\right| .
\]
\(\left[\text { Hint } :\left|a f_{n}+b g_{n}\right| \leq|a|\left|f_{n}\right|+|b|\left|g_{n}\right| . \text { Use Theorem } 2 .\right]\)
For the rest of the section, we define
\[
x^{+}=\max (x, 0) \text { and } x^{-}=\max (-x, 0) .
\]
\(\Rightarrow 14.\) Given \(\left\{a_{n}\right\} \subset E^{*}\) show the following:
(i) \(\sum a_{n}^{+}+\sum a_{n}^{-}=\sum\left|a_{n}\right|\).
(ii) If \(\sum a_{n}^{+}<+\infty\) or \(\sum a_{n}^{-}<+\infty,\) then \(\sum a_{n}=\sum a_{n}^{+}-\sum a_{n}^{-}\).
(iii) If \(\sum a_{n}\) converges conditionally, then \(\sum a_{n}^{+}=+\infty=\sum a_{n}^{-}\).
(iv) If \(\sum\left|a_{n}\right|<+\infty,\) then for any \(\left\{b_{n}\right\} \subset E^{1}\),
\[
\sum\left|a_{n} \pm b_{n}\right|<+\infty \text { iff } \sum\left|b_{n}\right|<\infty ;
\]
\[
\text { moreover, } \sum a_{n} \pm \sum b_{n}=\sum\left(a_{n} \pm b_{n}\right) \text { if } \sum b_{n} \text { exists. }
\]
\(\left.\text { [Hint: Verify that }\left|a_{n}\right|=a_{n}^{+}+a_{n}^{-} \text {and } a_{n}=a_{n}^{+}-a_{n}^{-} . \text {Use the rules of } §4 .\right]\)
\(\Rightarrow 15.\) (Abel's theorem.) Show that if a power series
\[
\sum_{n=0}^{\infty} a_{n}(x-p)^{n} \quad\left(a_{n} \in E, x, p \in E^{1}\right)
\]
converges for some \(x=x_{0} \neq p,\) it converges uniformly on \(\left[p, x_{0}\right]\) (or \(\left.\left[x_{0}, p\right] \text { if } x_{0}<p\right)\).
[Proof outline: First let \(p=0\) and \(x_{0}=1 .\) Use Problem 11 with
\[
f_{n}=a_{n} \text { and } g_{n}(x)=x^{n}=(x-p)^{n} .
\]
As \(f_{n}=a_{n} 1^{n}=a_{n}\left(x_{0}-p\right)^{n}\), the series \(\sum f_{n}\) converges by assumption. The convergence is uniform since the \(f_{n}\) are constant. Verify that if \(x=1\), then
\[
\sum_{k=1}^{\infty}\left|g_{k+1}-g_{k}\right|=0 ,
\]
and if \(0 \leq x<1,\) then
\[
\sum_{k=0}^{\infty}\left|g_{k+1}-g_{k}\right|=\sum_{k=0}^{\infty} x^{k}|x-1|=(1-x) \sum_{k=0}^{\infty} x^{k}=1 \quad \text { (a geometric series). }
\]
Also, \(\left|g_{0}(x)\right|=x^{0}=1 .\) Thus by Problem \(11(\text { with } K=2), \sum f_{n} g_{n}\) converges uniformly on \([0,1],\) proving the theorem for \(p=0\) and \(x_{0}=1 .\) The general case reduces to this case by the substitution \(x-p=\left(x_{0}-p\right) y .\) Verify!]
Prove that if
\[
0<\underline{\lim} a_{n} \leq \overline{\lim } a_{n}<+\infty ,
\]
then the convergence radius of \(\sum a_{n}(x-p)^{n}\) is 1.
Show that a conditionally convergent series \(\sum a_{n}\left(a_{n} \in E^{1}\right)\) can be rearranged so as to diverge, or to converge to any prescribed sum \(s\). [Proof for \(s \in E^{1} :\) Using Problem 14(iii), take the first partial sum
\[
a_{1}^{+}+\cdots+a_{m}^{+}>s .
\]
Then adjoin terms
\[
-a_{1}^{-},-a_{2}^{-}, \ldots,-a_{n}^{-}
\]
until the partial sum becomes less than \(s .\) Then add terms \(a_{k}^{+}\) until it exceeds \(s\). Then adjoin terms \(-a_{k}^{-}\) until it becomes less than \(s,\) and so on.
As \(a_{k}^{+} \rightarrow 0\) and \(a_{k}^{-} \rightarrow 0\) (why?), the rearranged series tends to \(s .\) (Why?)
Give a similar proof for \(s=\pm \infty\). Also, make the series oscillate, with no sum.]
Prove that if a power series \(\sum a_{n}(x-p)^{n}\) converges at some \(x=x_{0} \neq p\), it converges absolutely (pointwise) on \(G_{p}(\delta)\) if \(\delta \leq\left|x_{0}-p\right|\).
[Hint: By Theorem \(6, \delta \leq\left|x_{0}-p\right| \leq r\left(r=\text { convergence radius). Fix any } x \in G_{p}(\delta) .\right.\) Show that the line \(\overrightarrow{p x},\) when extended, contains a point \(x_{1}\) such that \(|x-p|< \left|x_{1}-p\right|<\delta \leq r .\) By Theorem \(6,\) the series converges absolutely at \(x_{1},\) hence at \(x\) as well, by Theorem 7.]