Prove that $\sum_{n\geqslant 1} x^n/n$ is not uniformly convergent

Question

Prove that $\displaystyle\sum_{n\geqslant 1} \frac{x^n}{n}$ is not uniformly convergent on $(-1,1)$ using the Uniform Cauchy Criterion.

The negation of Uniform Cauchy Criterion is the following: $$\exists \epsilon\, \forall N \,\exists x\in(-1,1)\,\exists n,m>N:\displaystyle\left\vert \sum_{k=m}^n \dfrac{x^k}{k}\right\vert\geqslant\epsilon.$$ I derived that $$\sum_{k=m}^n \frac{x^k}{k}\leqslant \frac{1}{m}\sum_{k=m}^n x^k=\frac{1}{m}\frac{x^{m+1}-x^{n+1}}{1-x}.$$ How should I proceed? I always find it difficult to prove these things, since I need to pick an $\epsilon$ and pick an $x$ et cetera..

Answer 1

Assume that it were, then for some $N$, for all $x\in(-1,1)$ and $m\geq N$, we have \begin{align*} \left|\sum_{k=m}^{n}\dfrac{x^{k}}{k}\right|<\epsilon, \end{align*} taking $x\rightarrow 1^{-}$, we have \begin{align*} \sum_{k=m}^{n}\dfrac{1}{k}\leq\epsilon,~~~~m\geq N, \end{align*} this shows that $\left\{\displaystyle\sum_{k=1}^{n}\dfrac{1}{k}\right\}_{n=1}^{\infty}$ is Cauchy and hence convergent, a contradiction.

Answer 2

First, your last term in the UCC should be $\left\vert \sum_{k=m}^n \dfrac{x^k}{k}\right\vert\geqslant\epsilon$.

Hint 1: Try picking $n=2m$. You can do this because you need to find $m,n,x$.

Your problem boils to $$\left\vert \sum_{k=m}^{2m} \dfrac{x^k}{k}\right\vert\geqslant\epsilon$$

Hint 2 Note that $$\left\vert \sum_{k=m}^{2m} \dfrac{x^k}{k}\right\vert\geqslant \left\vert \sum_{k=m}^{2m} \dfrac{x^{2m}}{2m}\right\vert=\frac{x^{2m}}{2}$$

If you can make this greater than $\epsilon$ you are done.

Hint 3 Use that for a fixed $N$, if you pick any $m>N$, for this fixed $m$ you have $$\lim_{x \to 1} \frac{x^{2m}}{2}=\frac{1}{2}$$

All you need to do is pick the right $\epsilon$, then for all $N$ fix some $m>N$, pick $n=2m$ and find the right $x$.