Cesaro Summability implies Abel Summability

 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. Therefore $$f(x)=\sum_{n=0}^{\infty}\frac{\sigma_n}{n+1}x^n$$ converges. Now see $$\sum_{n=0}^{\infty} \sigma_nx^n=\sum_{n=0}^{\infty}x^n\sum_{k=0}^{n}s_k=\sum_{k=0}^{\infty}s_k\sum_{n=k}^{\infty}x^n=\sum_{k=0}^{\infty}s_k\frac{x^k}{1-x}=\frac{1}{1-x}\sum_{n=0}^{\infty}s_nx^n$$ Doing the same process again we get $$\sum_{n=0}^{\infty}s_nx^n=\frac{1}{1-x}\sum_{n=0}^{\infty}a_nx^n$$ Hence $$\sum_{n=0}^{\infty}\sigma_nx^n=\frac{1}{(1-x)^2}\sum_{n=0}^{\infty}a_nx^n$$

Another identity we need is $$\sum_{n=0}^{\infty}(n+1)x^n=\frac{1}{(1-x)^2}$$ In order to show this let $f(x)=\sum\limits_{n=0}^{\infty}x^n=\frac{1}{1-x}$. Then $$\frac{1}{(1-x)^2}=f'(x)=\sum_{n=1}^{\infty}nx^{n-1}=\sum_{n=0}^{\infty}(n+1)x^n$$ Now $$f(x)=\sum_{n=0}^{\infty}\frac{\sigma_n}{n+1}x^n \iff xf(x)=\sum_{n=1}^{\infty}\frac{\sigma_{n-1}}{n}x^n$$ Since $f(x)$ converges, $xf(x)$ converges and therefore $f(x)+xf'(x)$ converges. $$f(x)+xf'(x)=\sum_{n=1}^{\infty}\frac{\sigma_{n-1}}{n}nx^{n-1}=\sum_{n=1}^{\infty}\sigma_{n-1}nx^{n-1}=\sum_{n=0}^{\infty}\sigma_{n}x^{n}=\frac{1}{(1-x)^2}\sum_{n=0}^{\infty}a_nx^n$$ Therefore $$\sum_{n=0}^{\infty}a_nx^n=\frac{f(x)+xf'(x)}{(1-x)^2}$$ Since $|x|<1$ the function $\frac{f(x)+xf'(x)}{(1-x)^2}$ also converges. Hence $\sum\limits_{n=0}^{\infty}a_nx^n$ converges. Therefore Cesaro summability implies Abel summability.

Since $\frac{\sigma}{n+1}$ converges to $s$. For $\epsilon>0$ $\exists\ k\in\mathbb{N}$ such that $$\left|\frac{\sigma_n}{n+1}-s\right|<\epsilon$$ Now $$(1-x)^2\sum_{n=0}^{\infty}\sigma_nx^n = (1-x)^2\sum_{n=0}^{\infty}(n+1)\frac{\sigma_n}{n+1}x^n$$ $$\left| (1-x)^2\sum_{n=0}^{\infty}(n+1)\frac{\sigma_n}{n+1}x^n-L\right|\leq (1-x)^2\sum_{n=0}^{\infty}(n+1)\left|\frac{\sigma_n}{n+1}x^n-L \right|$$ Now breaking the sum up at $k$ we get $$(1-x)^2\sum_{n=0}^{k}(n)\left|\frac{\sigma_n}{n+1}x^n-L \right|+(1-x)^2\sum_{n=k+1}^{\infty}(n+1)\left|\frac{\sigma_n}{n+1}x^n-L \right|$$ As $x\to 1^-$ the first part goes to 0 and the second part of the sum becomes $$(1-x)^2\sum_{n=k+1}^{\infty}(n+1)\left|\frac{\sigma_n}{n+1}x^n-L \right|< (1-x)^2\sum_{n=k+1}^{\infty}(n+1)\epsilon^k=\frac{\epsilon^k}{1-\epsilon}$$ which also goes to 0. Hence $$\lim_{x\to 1^-} (1-x)^2\sum_{n=0}^{\infty}\sigma_nx^n =\lim_{x\to 1^-}\sum_{n=0}^{\infty}a_nx^n=L$$  $\blacksquare$

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