Power series differentiability at endpoints I have the following problem:

Find domain $I$ of the function defined by 
  $f(x)=\sum\limits_{n=1}^{\infty}(3^{\frac{1}{n^2}}-1)x^n.$
  Investigate differentiability of $f(x)$ in the interior of $I$ and at endpoints in case they are in $I$.

Using Cauchy–Hadamard theorem, one easily gets radius of convergence: $R=1$. The series converges at $1,-1$. It is easily proven that $f(x)$ is differentiable in $(-1,1)$. I have problems with endpoints.
Term-by-term differentiating gives series which converges at $-1$ and diverges at $1$. But, as far as I know, this doesn't prove anything about differentiability of the initial series at these points. If I could've proven uniform convergence of differentiated series in $[-1,c]$, where $-1<c<1$, I would get differentiability at $-1$, but that still leaves $1$. 
So how do I investigate differentiability at endpoints? Are there some well-known facts that I'm missing?
Thanks in advance.
 A: Here, the power series
$$f(x) = \sum_{n = 1}^\infty \bigl(3^{\frac{1}{n^2}} - 1\bigr) x^n$$
converges absolutely and uniformly on the compact interval $[-1,1]$, and therefore defines a continuous function there. By the general theory of power series, $f$ is differentiable on the open interval $(-1,1)$, and the derivative is obtained by termwise differentiation,
$$f'(x) = \underbrace{\sum_{n = 1}^\infty n\bigl(3^{\frac{1}{n^2}} - 1\bigr) x^{n-1}}_{s(x)}.$$
The series $s(x)$ converges for $x = -1$ (by the alternating series test), and by Abel's theorem it defines a continuous function on $[-1,1)$. As shown in Uniform convergence in the endpoints of an interval, if a power series converges in an endpoint of its interval of convergence, then the convergence is uniform on the half interval of convergence including that endpoint, so the series $s(x)$ converges uniformly on $[-1,0]$ (in fact, on $[-1,c]$ for every $-1 < c < 1$). The uniform convergence allows the interchange of limit and integration, so the fundamental theorem of calculus yields the differentiability of $f$ at $-1$ and $f'(-1) = s(-1)$.
At the other endpoint of the interval of convergence, the series $s(x)$ diverges.
From that alone, it is not yet obvious that $f$ cannot be differentiable at $1$. But here, all coefficients of $s$ are positive, and therefore $s$ is strictly increasing on the interval $[0,1)$. If we had $L = \lim\limits_{x \to 1^-} s(x) < +\infty$, then it would follow that
$$\sum_{n = 1}^{N} n\bigl(3^{\frac{1}{n^2}} - 1\bigr)\cdot 1^{n-1} = \lim_{x\to 1^-} \sum_{n = 1}^{N} n\bigl(3^{\frac{1}{n^2}} - 1\bigr)\cdot x^{n-1} \leqslant \lim_{x \to 1^-} s(x) = L$$
for all $N \in \mathbb{N}$, and hence the series $s$ would converge at $1$. So it follows that $\lim\limits_{x \to 1^-} s(x) = +\infty$.
Now the mean value theorem shows that
$$\lim_{h \to 0^+} \frac{f(1) - f(1-h)}{h} = \lim_{h\to 0^+} f'(1 - \theta_h\cdot h) = +\infty,$$
so $f$ is not differentiable at $1$.
