Let $a_n$ be sequence of positive reals with $\lim na_n = a \in (0, \infty)$. Prove that $\lim_{x \to 1^-} f'(x)(1-x) = a$ for $f(x) = \sum a_nx^n$ 
Let $\{a_n\}$ be a sequence of positive numbers such that $\,a=\lim\limits_{n\to\infty}na_n\,$ exists with $a \in (0, \infty)$.
Let $f(x) = \displaystyle\sum_{n=0}^\infty a_nx^n$ for $x \in (-1,1)$.
Prove that $f(x)$ is convergent and continuously differentiable for
all $x \in (-1,1)$. Also, prove that $\lim\limits_{x\to 1^-}f'(x)(1-x) = a$.

Here is my work so far: let $s_N(x) := \displaystyle\sum_{n=0}^N a_nx^n$. Obviously $s_N(0)$ converges to $a_1$. Also, $s_N'(x)$ converges uniformly, since there exists $N^\sim$ such that for all $n \geq N^\sim$ we have, for any arbitrary $r \in (0, 1)$ and for any $x$ such that $x \in [-r,r], |na_nx^n| \leq 2|a||x|^n \leq 2|a|r^n$. Thus, by Weierstrass M-test, $s_N'(x)$ converges uniformly on $[-r,r]$ for all $r \in (0,1)$. Thus we have that, for all $x \in [-r,r]$, $f'(x) = \displaystyle\sum_{n=1}^\infty na_nx^{n-1}$. Furthermore, since $f'(x)$ is a uniform limit of continuous functions, it is itself continuous on $[-r,r]$. But since for any $ \in (-1, 1)$ there exists $r$ such that $|x| < r < 1$, we have that $f'(x)$ is continuously differentiable for all $x \in (-1,1)$.
I am not sure how to prove that $f(x)$ is convergent and, more importantly, how to prove that $\displaystyle\lim_{x \to 1^-} f'(x)(1-x) = a$. I would appreciate any help. Thank you!

If you have proven that the series defining $f'$ converges in $(-1,1)$, then convergence of the series defining $f$ should readily follows. If you want a direct proof, I think you could use the root test:
$$\lim_{n \to \infty} \sqrt[n]{a_n} = \lim_{n \to \infty} \frac{\sqrt[n]{n a_n}}{\sqrt[n]{n}}=1.$$
For the second part of your question, you want to use Abel's theorem. If you know it, it boils down to expressing $f'(x)(1-x)$ for $x\in(-1,1)$ as a series, and see what it gives you if you formally evaluate this series in $x=1$.
