Let \((a_n)\) be a sequence of real numbers. \(\sum\limits_{n=0}^{\infty}a_n\) be a series. Let $$s_n=\sum_{k=0}^{n}a_k$$Then the sequence \((a_n)\) is Cesaro Summable with Cesaro Sum \(s\in \mathbb{R}\) if $$\lim_{n\to\infty}\frac{\sigma_n}{n+1}=\lim_{n\to\infty}\frac{1}{n+1}\sum_{k=0}^n s_n=s$$Let $$f(x)=\sum\limits_{k=0}^{\infty}a_nx^n$$ be power series. Then the sequence is Abel Summable if the power series \(f(x)\) converges with a radius of convergence \(|x|<1\). You can see that if \(s_n\to L\) as \(n\to \infty\) then \(\sigma_n\to s\) as \(n\to \infty\) i.e. convergence of the series implies Cesaro Summability We will prove Abel Summability is much stronger than Cesaro summability i.e. if a series is Cesaro Summable then it is Abel Summable. So assume \(a_n\) is Cesaro summable. Hence $$\lim_{n\to\infty}\frac{\sigma_n}{n+1}=\lim_{n\to\infty}\frac{1}{n+1}\sum_{k=0}^n s_n=L$$Hence the sequence \(\left( \frac{\sigma_n}{n+1}\right)\) is Abel summable...