# 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?

• 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.) Oct 8, 2013 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! Oct 4, 2013 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. Oct 4, 2013 at 19:23
• @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. Jun 1, 2015 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. Feb 15, 2016 at 4:32
• @Procore : there is no mistake in what I wrote. For $n=2$ for instance, $\frac 12 \sum_{k=1}^2 (S_k-S) = \frac 12 ( (S_1 - S) + (S_2 - S) ) = \frac {S_1 + S_2}2 - S. Essentially I wrote$-S = -nS/n and pulled one $-S/n$ in the sum for each $k$ and then factored $1/n$ out. Feb 15, 2016 at 8:33