Let
$$f(x)=\sum_{n=1}^{\infty}\frac{x^n}{n^2},\:\:\:x\in(0,2).$$
By Weierstrass M-Test I know, $f(x)$ is a uniformly convergent series on $(0,1]$. Now I want to find all $x\in (0,2)$ at which $f(x)$ is continuously and infinitely many times differentiable.
However for the series $g(x):=\sum_{n=1}^{\infty}\frac{x^{n-1}}{n}$, there is no way to apply Weierstrass M-test. I have no clue how to go further. I appreciate any help.
We can see that $\sum_{n=1}^{\infty} \frac{x^n}{n^2}$ is not convergent for $x > 1$, which is clear considering that $\lim_{n\to \infty} \frac{x^n}{n^2} = \infty$ for $x > 1$ (if this is not obvious, use l'Hôspital's rule twice). Therefore, we consider only $x\in (0, 1]$. Notice that $$\sum_{n=1}^{\infty} \frac{x^{n-1}}{n} = \frac{1}{x}\sum_{n=1}^{\infty} \frac{x^n}{n} = -\frac{\log(1-x)}{x}$$ for $x\in (0, 1)$. Convergence is readily apparent by the ratio test, and then we have uniform convergence on compact sets by Dini's theorem (or by the Weierstrass M-test, but Dini's theorem is easy here). For $x = 1$, we get the harmonic series, which diverges. Note that $g(x) = -\frac{\log(1-x)}{x}$ is $C^{\infty}$-smooth on $(0, 1)$.