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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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