# Convergence of Power Series to Its Cesaro Sum

Let $$\{a_n\}_{n\geq 0}$$ be sequence of real numbers, let $$s_n=a_0+\dots+a_n$$ be its partial sums and $$\sigma_n=\dfrac{s_0+\dots+s_n}{n}$$ be partial Cesaro sums. Assume $$\sigma_n\rightarrow \sigma$$. I want to prove if $$f(x)=\sum_{n\geq 0}a_nx^n$$, then $$f$$ is convergennt power series for $$|x|<1$$ and $$f(x)-\sigma)=(1-x)^2\sum_{n\geq 0}(n+1)(\sigma_n-\sigma)x^n$$ so $$\lim_{x\rightarrow 1^-}f(x)=\sigma$$. By theorem 8.2 in baby Rudin, the result is true if $$\sum_{n\geq 0}a_n$$ convergs. But I still have no idea if $$\sum_{n\geq 0}a_n$$ does not converge.

If $$\sum_{n\geq 0}a_n$$ does not converge, $$\lim_{x\to1^-}f(x)$$ may not exist. For example, let $$a_n=\frac{1}{n+1}$$ and then $$f(x)=\sum_{n\geq 0}a_nx^{n}=-\frac{\ln(1-x)}x,|x|<1.$$ So $$\lim_{x\to1^-}f(x)=\infty.$$
• For $|x|<1$, $\sum_{n\geq 0}\frac1{n+1}x^{n}$ converges. Jun 5, 2023 at 14:07
• As I mentioned in the post, it is assumed the sequence is cesaro summable, $\sigma_n$ is a convergent sequence Jun 6, 2023 at 2:09