Let $S(x) = \sum_{i \geq 0}a_ix^i \in \mathbb{R}[[x]]$.
How do I prove that
$\displaystyle \lim_{n \rightarrow \infty} \frac{\sum_{1 \leq i \leq n} a_i}{n}$ exists iff $\displaystyle \lim_{x \rightarrow 1} S(x)(1-x)$ exists, and when they do, both are equal.
I've also proved that if the first limit exists then the second does, but my proof is too long and messy and maybe wrong and I've no idea how to solve it completely. I have also found this is equivalent to either of these two:
$\lim_{n \rightarrow \infty} \sum_{1 \leq i \leq n} b_i (1 - \frac{i}{n})$ exists iff $\displaystyle \lim_{x \rightarrow 1} \sum_{0 \geq i} x^i b_i$ exists, and when they do, both are equal.
and
Define $ \displaystyle F(x,n):= \frac{(1-x) \sum_{1 \leq i \leq n} x_i a^i}{1-x^n}$. Then $\lim_{x \rightarrow 1} \lim_{n \rightarrow \infty} F(x,n)$ exists iff $\lim_{n \rightarrow \infty} \lim_{x \rightarrow 1} F(x,n)$ and when they do both are same.
Edit: As mentioned in the comments, this is indeed Hardy Littlewood Tauberian theorem and I was unaware that this was a theorem. See Karmata's proof of Hardy-Littlewood Tauberian theorem for a follow up questin.