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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?

Daniel Donnelly
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  • 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 08 '13 at 15:16

1 Answers1

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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,

Patrick Da Silva
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    Yeah, that does help. Neat trick! – Daniel Donnelly Oct 04 '13 at 17:52
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    @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 04 '13 at 19:23
  • @PatrickDaSilva if the series diverges it implies that $\sigma_N$ diverges ? – Maman Jun 01 '15 at 01:17
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    @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 01 '15 at 02:15
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    @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 04:32
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    @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. – Patrick Da Silva Feb 15 '16 at 08:33
  • @PatrickDaSilva: I see what you mean. See my comment for my apology in advance. – Procore Feb 15 '16 at 21:08
  • @Procore : Sorry if I sounded rude! This is totally okay. This forum exists for everyone to have doubts and for everyone else to clear these doubts! – Patrick Da Silva Feb 15 '16 at 21:10
  • @PatrickDaSilva: No - no rudeness whatsoever on your part, and no need to apologize. I'm kicking myself for neglecting to hash through the notation. If anything, I knew I had to have missed something somewhere. I appreciate the consideration - overall, your proof is very concise and elegant (concerning the rigorous part); I proved it differently using the $nS$ multiple instead which made things a little more tedious on my end, and, while I was comparing, I noticed the difference up front. No worries, and thank you for setting me straight. – Procore Feb 15 '16 at 21:14
  • @Procore : you didn't prove it "differently", you just added steps that I omitted. Writing $S = \frac{nS}n = \frac 1n \sum_{k=1}^n S$ is the blank you want to fill. – Patrick Da Silva Feb 15 '16 at 21:18
  • @PatrickDaSilva: It's funny you mention that - right after I wrote my comment I was going to add that virtually everything is the same, I just used the word "different" to quickly implicate that I simply worked through some "additional" details that are covered more concisely in both of your estimates (regarding convergence of the given series and the overall Cesaro summability). It was a quick attempt at, "saving face" if you will. I'd post my answer, but it is part of homework that I have to finish (fortunately it is TeX'ed, but I'm not sure I have time - plus it is overkill at this point). – Procore Feb 15 '16 at 21:26
  • @Procore : Just let it go and get your homework done! ;) – Patrick Da Silva Feb 15 '16 at 21:27
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    @PatrickDaSilva: OK. Not sure what I'm letting go. Have a nice life. – Procore Feb 15 '16 at 21:27
  • @Procore : Letting go of posting your answer ;) if we have the same proof don't bother posting it! – Patrick Da Silva Feb 15 '16 at 21:29
  • @PatrickDaSilva Isn't there an implicit assumption in your proof that the chosen $l$ is $<n$? (i.e., what if for the given $n$ you have that the $l$ from which $|S_k-S|<\frac{\epsilon}{2}$ is actually larger than $n$?). Doesn't this actually require using another threshold $N'$ and taking $\max{N,N'}$? Regardless, great proof :) – Anon May 19 '22 at 15:22
  • @Anon Actually the argument is still correct if $\ell > n$ but it makes things awkward and requiring explanation. But I did say "choose $N$ large enough", so I guess that since $\ell$ is fixed, you could take $\max {N,\ell}$, but it wouldn't affect the argument if you just removed terms that would disappear if $\ell > n$. – Patrick Da Silva May 31 '22 at 11:25