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. enter image description here enter image description here


1 Answer 1


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 $$ \lim_{x\to1^-}f(x)=\infty. $$

  • $\begingroup$ How do you know if the sequence is cesaro summable? $\endgroup$ Jun 5, 2023 at 2:10
  • $\begingroup$ For $|x|<1$, $\sum_{n\geq 0}\frac1{n+1}x^{n}$ converges. $\endgroup$
    – xpaul
    Jun 5, 2023 at 14:07
  • $\begingroup$ how does it imply the sequence is cesaro summable? $\endgroup$ Jun 5, 2023 at 14:53
  • $\begingroup$ What is your meaning of cesaro summability? $\endgroup$
    – xpaul
    Jun 5, 2023 at 19:13
  • $\begingroup$ As I mentioned in the post, it is assumed the sequence is cesaro summable, $\sigma_n$ is a convergent sequence $\endgroup$ Jun 6, 2023 at 2:09

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .