Convergence of series implies convergence of Cesaro Mean. [duplicate]

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Proof. Let $\sum_{k = 0}^N c_k \rightarrow s$, let $\sigma_N = (S_0 + \dots + S_{N-1})/N$ be the $Nth$ Cesaro sum where $S_K$ is the $Kth$ partial sum of the series. Then $s - \sigma_N \\= s - c_0 - c_1(N-1)/N + c_2(N-2)N +\dots+c_{N-1}/N \\ =c_1/N + c_2 2/N + \dots + c_{N-1}(N-1)/N + c_N + \dots$

Where do I go from here?

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1 Answer

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You don't need the fact that it's a series, which is maybe why this is confusing for you.

Suppose $S_n \to S$ is a converging sequence. Then $\frac 1n \sum_{k=1}^n S_k \to S$ also. Roughly speaking, if you take $n$ large enough, then all the big terms (big index, not big in value) are close to $S$ ; all the small terms (small index) will get killed when $n$ goes to infinity.

Non-roughly speaking, $$ \left| \left( \frac 1n \sum_{k=1}^n S_k \right) - S \right| = \frac 1n \left| \sum_{k=1}^n (S_k - S) \right| \le \frac 1n \sum_{k=1}^n |S_k - S| = \frac {\sum_{k=1}^{\ell} |S_k - S|}{n} + \frac {\sum_{k=\ell+1}^n |S_k - S|}{n} $$ Let $\varepsilon > 0$, and choose $\ell$ such that for all $k > \ell$, $|S_k - S| < \varepsilon/2$ by convergence of $S_k$ to $S$. Now that $\ell$ is fixed, choose $N$ large enough so that for all $n > N$, $$ \frac{\sum_{k=1}^{\ell} |S_k - S|}{n} < \varepsilon / 2. $$ (Note that the numerator does not depend on $n$ so we still have freedom.) It follows that for all $n > N$, $$ \frac {\sum_{k=1}^{\ell} |S_k - S|}{n} + \frac {\sum_{k=\ell+1}^n |S_k - S|}{n} \le \frac {\sum_{k=1}^{\ell} |S_k - S|}{n} + \frac{(n-\ell) (\varepsilon/2)}n \le \varepsilon. $$ For your particular problem, put $S_n = \sum_{k=0}^n c_k$.

Hope that helps,

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