Tutorial Note 5 (sequence and series of functions; uniform and nonuniform convergence)

Example 1.
(a) If $\sum a_nx_0^n$ converges, then $a_nx_0^n$ is bounded. Let $|a_nx_0^n|\leq M$ for all $n$, then for every $x$ s.t. $|x|<|x_0|$ we have \[
|a_nx^n|=\left|a_nx_0^n\brac{\frac{x}{x_0}}^n\right|\leq M \left| \frac{x}{x_0}\right|^n,
\] therefore since $\sum | \frac{x}{x_0}|^n$ converges, $\sum a_nx^n$ converges by comparison test and absolute convergence test.

Remark. The series $\sum a_nx^n$ converges uniformly on any $[-r,r]$, $r<|x_0|$, by $M$-test. In fact for any $0<r<|x_0|$ it converges uniformly on $[-r,|x_0|]$ by Abel's Limit Theorem in the next tutorial.

(b) Let $|x|>|x_0|$, if $\sum a_nx^n$ converges, so does $\sum a_nx_0^n$ by part (a), a contradiction.

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Example 2. We need to compute \[
\lims \sqrt[n]{\frac{n!}{n^n}}.
\] Actually the limit exists, so $\lims =\lim$. To see this, recall that \[\limi \left|\frac{a_{n+1}}{a_n}\right|\leq \limi \sqrt[n]{|a_n|}\leq \lims \sqrt[n]{|a_n|}\leq \lims \left|\frac{a_{n+1}}{a_n}\right|,\] hence if the limit of the $n$th root of $a_n$ is difficult to compute, let's first  try to compute the limit of its ratio. If $\bm{\lim a_{n+1}/a_n}$ exists, then by the above inequality, so does $\bm{\lim \sqrt[n]{a_n}}$, moreover, $\bm{\lim \sqrt[n]{a_n}=\lim a_{n+1}/a_n}$.

Bearing this in mind, we have \[
\limn \frac{(n+1)!}{(n+1)^{n+1}}\Bigg/\frac{n!}{n^n}=\limn \frac{1}{(1+\frac{1}{n})^n}=\frac{1}{e}
\implies
\limn \sqrt[n]{\frac{n!}{n^n}} = \frac{1}{e},
\] and thus the radius of convergence is $R=e$.

At the point $x=\pm R=\pm e$, note that  \[
\left|\frac{n!}{n^n}(\pm e)^n\right|=\frac{n!e^n}{n^n},
\] by Stirling formula \[
n!\sim  \sqrt{2\pi n}\brac{\frac{n}{e}}^n\implies \frac{n!e^n}{n^n}\sim \sqrt{2\pi n}\to\infty,
\] therefore the power series diverges at $x=\pm e$ by term test.

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Example 3. Easy to see that $f_n\to 0$ pointwise on $(0,\infty)$.

On $(0,1)$, since \[
\|f_n-0\|_{(0,1)}\ge \left|f_n\brac{\frac{1}{n}}\right|=\frac{1}{1+1}=\frac{1}{2},
\] thus $\|f_n\|_{(0,1)}\not{\toto} 0$. Pictorially this is what's happening:

On $(1,\infty)$, since for every $x\in (1,\infty)$, \[
|f_n(x)-0|\leq \frac{nx}{n^2x^2}=\frac{1}{nx}\leq \frac{1}{n},
\] thus $\|f_n-0\|_{(1,\infty)}\leq \frac{1}{n}$, hence $f_n\toto0$ on $(1,\infty)$.

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Example 4. Let $g_n(x)=x^n(1-x)^2$. There are two ways to find a sharp enough upper bound of $|g_n(x)|$.

Method 1.  We consider the derivative \[
g_n'(x)=(1-x)x^{n-1}(n-(n+2)x).
\] If $g_n'(x)=0$, then $x=0,1$ or $n/(n+2)$ ($<1$). Since $g_n(0)=g_n(1)=0$, and since  $g_n(n/(n+2))>0$ is the only local extreme value of $g$ on $(0,1)$, it must be a global maximum.  So for any $x\in [0,1]$, \[|g_n(x)|=g_n(x)\leq g_n\brac{\frac{n}{n+2}} =4\brac{\frac{n}{n+2}}^n \frac{1}{(n+2)^2}\leq \frac{4}{(n+2)^2}.\]

Method 2. We may use AM-GM inequality. For $x\in [0,1]$,
\begin{align*}|g_n(x)|&=x^n(1-x)^2\\
&=4n^n \brac{\frac{x}{n}}^n \brac{\frac{1}{2}-\frac{x}{2}}^2\\
& \leq 4n^n
\Bigg(\frac{\overbrace{\frac{x}{n}+\cdots +\frac{x}{n}}^{\text{$n$ terms}} + (\frac{1}{2}-\frac{x}{2}) + (\frac{1}{2}-\frac{x}{2}) }{n+2}\Bigg)^{n+2}\\
&=\frac{4n^n}{(n+2)^{n+2}}\\
&\leq \frac{4}{(n+2)^2}.
\end{align*}
Therefore, as $\sum_{n=1}^\infty 4/(n+2)^2$ converges, by $M$-test $\sum_{n=1}^\infty g_n(x)$ converges uniformly on $[0,1]$.

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Example 5.
(a) On $(-1,1)$ we have \[
 \left|\sum_{k=0}^n x^k-\frac{1}{1-x}\right| = \left|\sum_{k=n+1}^\infty x^k\right|=\left|\frac{x^{n+1}}{1-x}\right|.
\] Can this quantity be uniformly small for  sufficiently large $n$? No! Observe that for each $n$ \[
\left|\frac{x^{n+1}}{1-x}\right| \to \infty
\] as $x\to 1^-$, in particular, for each $n$, there is an $x_n\in (-1,1)$, close enough to $1$, such that $\left|\frac{x_n^{n+1}}{1-x_n}\right|\ge 1$, thus for each $n$, \[
 \left\|\sum_{k=0}^n x^k-\frac{1}{1-x}\right\|_{(-1,1)}\ge \left|\frac{x_n^{n+1}}{1-x_n}\right|\ge 1,
\] which cannot converges to 0.

(b) The pointwise limit of $f_n$ on $[0,1)$ is $0$. But the quantity \[
|f_n(x)|=x^n
\] cannot be uniformly small for sufficiently large $n$ because $\lim_{x\to 1^-}x^n=1$. Specifically, for each $n$ there is $x_n\in [0,1)$ such that $x_n^n>1/2$, so \[
\|x^n-0\|_{[0,1)}\ge x_n^n>1/2\implies x^n\not{\toto} 0.
\]

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Example 6. Let $f_n= \sum_{k=0}^n x^2/(1+x^2)^k$, we consider the difference \begin{align*}
|f_{2n}(x)-f_{n}(x)|&=\sum_{k=n+1}^{2n} \frac{x^2}{(1+x^2)^k}\\
&\ge \sum_{k=n+1}^{2n} \frac{x^2}{(1+x^2)^{2n}}\\
&=  \frac{nx^2}{(1+x^2)^{2n}}.
\end{align*} For each $n>1$ we let $x=x_n=1/\sqrt{n}\in (0,1)$ such that \[\|f_{2n}-f_{n}\|_{(0,1)}\ge | f_{2n}(x_n)-f_{n}(x_n)|=\frac{1}{(1+\frac{1}{n})^{2n}}\to \frac{1}{e^2},
\] which shows $\{f_n\}$ cannot be uniformly Cauchy on $(0,1)$.