Show that the series $\sum_{k=1}^{\infty}\frac{x^k}{k}$ uniformly converges in $[0,a]$ where $a \in [0,1)$ but not in $[0,1]$.
To show uniformly convergent in $[0,a]$, we have $|\frac{x^k}{k}|\leq|\frac{a^k}{k}|$. How do I show convergence of $\frac{a^k}{k}$ so that I can use Weierstrass and conclude that $\sum_{k=1}^{\infty}\frac{x^k}{k}$ uniformly converges in $[0,a]$ where $a \in [0,1)$?
Next, to show that the series is not uniformly convergent in [0,1], I assume that it is uniformly convergent such that I get a contradiction. But I am stuck here too. I see that $f_n$ is monotonic and continuous but how to exploit these properties to get a contradiction? I note that at $x=1$, $f_n$ blows up for $n \rightarrow \infty$
Root test for $\displaystyle\sum_{k}\dfrac{a^{k}}{k}$: $\lim_{k\rightarrow\infty}\left(\dfrac{a^{k}}{k}\right)^{1/k}=\lim_{k\rightarrow\infty}\dfrac{a}{k^{1/k}}=a<1$, so the convergence follows.
Assume it is convergent on the whole $[0,1]$, then $\displaystyle\sum_{k=1}\dfrac{1^{k}}{k}$ exists, but this is a contradiction. Of course here the uniformity is not involved.