calculus - Derivative of power series with nonnegative coefficients

Let $$f(x) = \sum_{k=0}^\infty a_kx^k$$ be a power series mapping reals to reals, with radius of convergence $1$. Suppose $f'(x_0)$ exists in $(-1,1]$ (take the one-sided limit if $x_0 = 1$). Also suppose $a_k \geq 0$ for all $k$. Then is it always true that $$f'(x_0) = \sum_{k=0}^\infty ka_kx^{k-1}?$$

This clearly holds if $x_0 \in (-1,1)$. But what about $x_0 = 1$?

Answer

The answer to the question is yes. This follows from Abel's theorem, since you assume the $a_k$ (and hence $k a_k$) are positive. If you assume that $\sum_k k a_k$ converges, then it is an immedeate consequence of Abel's theorem that $\sum_k k a_kx^{k-1}$ converges for $x\rightarrow 1^-$ to $\sum_k k a_k$.

If, on the other hand, the sum does not converge, it will diverge to $\infty$ (since all terms are nonnegative real numbers), so the generalized Abel theorem (see the link supplied above under remarks) applies again, which states that in this case the function $g(z)=\sum_k k a_kx^{k-1}$ goes off to $\infty$ as well when $x\rightarrow 1$. Since you assume that this is not true you are back in the first case.

What you are requesting here is a special case of what is known as 'Tauberian therom', which you could also refer to. I wonder, though, whether it's possible to give an elementary proof in this case....