I have been trying to prove the uniform convergence of the series $f_{n}(x)=\sum^n_{k=1}\frac{x^k}{k}$ Obviously, the series converges only for $x\in(-1,1)$. Consequently, I decided to split this into two intervals: $(-1,0]$ and $[0,1)$ and see if it converges on both of them using the Weierstrass M-test.
For $x\in(-1,0]$, let's take $q\in(-1,x)$. We thus have: $\left|\frac{x^k}{k}\right|\leq\left|x^k\right|\leq\left|q^k\right|$ and since $\sum|q^n|$ is convergent, $f_n$ should be uniformly convergent on the given interval. Now let's take $x\in[0,1)$ and $q\in(x,1)$. Now, we have: $\left|\frac{x^k}{k}\right|=\frac{x^k}{k}\leq\ x^k\leq{q^k}$ and once again, we obtain the uniform convergence of $f_n$.
However, not sure of my result, I decided to cross-check it by checking whether $f_n$ is Cauchy. For $x\in(-1,0]$, I believe it was a positive hit, since for $m>n$ we have: $\left|f_{m}-f_{n}\right|=\left|f_{n+1}+f_{n+2}+...f_{m}\right|\leq\left|\frac{x^n}{n}\right|\leq\frac{1}{n}$ which is what we needed. However, I haven't been able to come up with a method to show the same for $x\in[0,1)$. Now, I am not so sure whether $f_n$ is uniformly convergent on $[0,1)$. If it is, then how can we show it otherwise, and if it isn't, then how can we disprove it? Also, what's equally important - what did I do wrong in the Weierstrass-M test?