Supoose that the power series $f(x)=\displaystyle\sum_{i=0}^\infty a_ix^i$ has radius of convergence $R=1$, and that $f$ converges at $x=1$. I am aware that $f'$ may not exist at $1$, but must the one-sided derivative of from the left of $f$ exist at $1$? That is, must the following limit exist? $$f'_-(1)=\displaystyle\lim_{x\to 1^-}\dfrac{f(x)-f(1)}{x-1}$$
If not, are there easy counter-examples?
Consider the series defined by $$ a_n \stackrel{\rm def}{=} \frac{1}{n(n+1)}$$ for $n\geq 1$.
Then the corresponding power series $f(x)=\sum_{n=1}^\infty a_n x^n$ has radius of convergence $1$, converges at $R=1$ (with $f(1) = \sum_{n=1}^\infty a_n = 1$), but $$ \lim_{x\to 1^-} \frac{f(x)-1}{1-x} = -\infty\,. $$