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 $x\in [0,1]$.

  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ Also (2) is uniformly convergent by the Dirichlet test or as shown below. $\endgroup$
    – RRL
    Sep 4, 2020 at 15:13
  • 1
    $\begingroup$ Your proof of (1) is good. $\endgroup$
    – RRL
    Sep 4, 2020 at 15:16

1 Answer 1


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.