# Two questions about function series

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 $$x\in [0,1]$$.

• 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 $x \in [0,1]$
– RRL
Sep 4, 2020 at 15:09
• @RRL Oh, right, you are correct. How about (1), that one is fine, right? Sep 4, 2020 at 15:12
• Also (2) is uniformly convergent by the Dirichlet test or as shown below.
– RRL
Sep 4, 2020 at 15:13
• Your proof of (1) is good.
– RRL
Sep 4, 2020 at 15:16

For (2), since $$e^{-kx}/k$$ is nonincreasing with respect to $$k$$, we have for $$x \in [0,1]$$,
$$\left|\sum_{k=n}^m \frac{(-1)^ke^{-kx}}{k }\right|= \frac{e^{-nx}}{n} - \left(\frac{e^{-(n+1)x}}{n+1}-\frac{e^{-(n+2)x}}{n+2} \right)- \ldots\leqslant \frac{e^{-nx}}{n} \leqslant \frac{1}{n}$$