A dubious proof using Weierstrass-M test for $\sum^n_{k=1}\frac{x^k}{k}$ 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?

Weierstrass M-test only gives you uniform convergence on intervals of the form $[-q,q]$, where $0<q<1$. Your proof shows this.
You also get uniform convergence on the interval $[-1,0]$, but to see this you need other methods. For example the standard estimate for the cut-off error of a monotonically decreasing alternating series will work here.
As David Mitra pointed out, the convergence is not uniform on the interval $[0,1)$. 
Elaborating on his argument: No matter how large an $n$ we choose, the divergence of the harmonic series tells us that $\sum_{k=n+1}^{n+p}(1/k)>2$ for $p$ large enough. We can then select a number $a\in(0,1)$ such that the powers $a^k, n<k\le n+p$ are
all larger than $1/2$. Then it follows that for all $x\in(a,1)$ 
$$
\sum_{k=n+1}^{n+p}\frac{x^k}k\ge \sum_{k=n+1}^{n+p}\frac{a^k}k>\frac12\sum_{k=n+1}^{n+p}\frac1k>1.
$$
Thus the Cauchy condition fails on the subinterval $(a,1)$.
