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I have just started learning function series and I would like to show you my solutions to two exercises, because I am not really sure that I am doing them right.
Problem 1: Show that the series $\displaystyle \sum_{n=1}^{\infty} x(1-x)^n$ doesn't converge uniformly on $[0,1]$.
Solution: Let $f_n : [0,1] \to \mathbb{R}$, $f_n(x)=x(1-x)^n$.
According to Cauhcy's criterion, our series converges uniformly on $[0,1]$ iff $\displaystyle \lim_{n\to \infty} \sup_{p\in \mathbb{N}}\left|\sum_{k=0}^p f_{n+k}(x)\right|=0$ for all $x\in [0,1]$.
We have that $\displaystyle \sum_{k=0}^p f_{n+k}(x)=(1-x)^n\left[1-(1-x)^{p+1}\right]$ and this is equal to $1$ if $x=0$, so I think that this is enough to reach our conclusion.
Problem 2: Show that the series $\displaystyle \sum_{n=1}^\infty \frac{(-1)^n}{n}e^{-nx}$ converges uniformly on $[0,1]$.
Solution: We have that $\left|\frac{(-1)^n}{n}e^{-nx}\right|\le e^{-nx}, \forall n \in \mathbb{N}, x\in [0,1]$.
If $x\in (0,1]$, then the series $\displaystyle \sum_{n=1}^\infty e^{-nx}$ is convergent and by the Weierstrass M-test our series converges uniformly.
If $x=0$, then the series rewrites as $\displaystyle \sum_{n=1}^\infty \frac{(-1)^n}{n}$ and this is known to converge (by the Leibniz test). I am not sure if breaking it down like this assures me that the series converges uniformly for $x\in [0,1]$.

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    $\begingroup$ Your use of the Weierstrass test for (2) is incorrect since you are comparing to a series with terms that depend on $x$. To show $\sum f_n(x)$ converges uniformly you want $|f_n(x)| \leqslant M_n$ where $\sum M_n$ converges. Furthermore, $\sum e^{-nx}$ itself does not converge uniformly for $x \in [0,1]$ $\endgroup$
    – RRL
    Sep 4, 2020 at 15:09
  • $\begingroup$ @RRL Oh, right, you are correct. How about (1), that one is fine, right? $\endgroup$ Sep 4, 2020 at 15:12
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    $\begingroup$ Also (2) is uniformly convergent by the Dirichlet test or as shown below. $\endgroup$
    – RRL
    Sep 4, 2020 at 15:13
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    $\begingroup$ Your proof of (1) is good. $\endgroup$
    – RRL
    Sep 4, 2020 at 15:16

1 Answer 1

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For (2), since $e^{-kx}/k$ is nonincreasing with respect to $k$, we have for $x \in [0,1]$,

$$\left|\sum_{k=n}^m \frac{(-1)^ke^{-kx}}{k }\right|= \frac{e^{-nx}}{n} - \left(\frac{e^{-(n+1)x}}{n+1}-\frac{e^{-(n+2)x}}{n+2} \right)- \ldots\leqslant \frac{e^{-nx}}{n} \leqslant \frac{1}{n}$$

which implies uniform convergence by the uniform Cauchy criterion.

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