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

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?

marked as duplicate by Yiorgos S. Smyrlis, Hagen von Eitzen, Niels J. Diepeveen, Andrew D. Hwang, Davide GiraudoFeb 16 '14 at 18:12

• As soon as you realize that convergence of a series is the same thing as convergence of the partial sums, it is the same question as asked here. (Several other questions about the same thing are linked there, too.) – Martin Sleziak Oct 8 '13 at 15:16

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,

• Yeah, that does help. Neat trick! – BananaCats Category Theory App Oct 4 '13 at 17:52
• @Enjoys Math : I want to add : understanding the "roughly speaking" part is what's gonna make you a better "idea-producer". Understanding the other part is going to make you a better "proof-maker". You need both to become an amazing mathematician. – Patrick Da Silva Oct 4 '13 at 19:23
• @PatrickDaSilva if the series diverges it implies that $\sigma_N$ diverges ? – Maman Jun 1 '15 at 1:17
• @Maman : No, just consider $\sum_{n \ge 1} (-1)^n$, which is a divergent series since the sequence $S_n$ is equal to $\frac{(-1)^n-1}2$, so $\sigma_N = \frac 1N \sum_{n=1}^n \frac{(-1)^n-1}2 \to \frac 12$. The Césarò mean averages out your sequence, so if your sequence keeps oscillating the mean is going to converge somewhere between the $\liminf$ and the $\limsup$. If $S_n$ grows to infinity however, $\sigma_N$ can diverge. I didn't think of a case where $S_n$ grows to infinity and $\sigma_N$ converges ; this could probably be figured out (if it's true or false) with some thinking. – Patrick Da Silva Jun 1 '15 at 2:15
• @PatrickDaSilva: In the very beginning of the rigorous part of your proof, going from the left-hand side to the right-hand side of the first equality, would it not in fact be $\dfrac{1}{n}\Bigg|{\displaystyle{\sum_{k=1}^{n}(S_{k}-nS)}}\Bigg|$, and shouldn't that which follows contain the multiple $nS$ inside the sum of absolute values? My apologies if I missed something somewhere, if I'm incorrect, etc., but I was curious in this regard. – Procore Feb 15 '16 at 4:32