# $a_n$ converges to $a$, but why $\frac{1}{n+1}(a_0+a_1+\cdots+a_n)$ also?

i have asked this question today already, but i couldnot find the answer to my second question (which is this)which came after the answer to the first question. First of all, sorry duplicate question, and please dont redirect me to another question.

i cannot follow the thought that the sequence $\frac{1}{n+1}(a_0+a_1+\cdots+a_n)$ converges to $a$. my idea is this:

$$\frac{1}{n+1}(a_0+a_1+\cdots+a_n)=\frac{a_0}{n+1}+\frac{a_1}{n+1}+\cdots+\frac{a_n}{n+1}$$ and each term goes to $0$ as $n$ grows, but how can the sum of $0+0+0+0+0$ can converge to $a$ ?

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This is a duplicate not of doniyor's earlier question, but (somewhat) of the following older question: math.stackexchange.com/questions/207910/… (I also linked to this on doniyor's earlier question.) –  Jonas Meyer Dec 28 '12 at 18:01
This phenomenon is called Cesaro summability, and it is strictly weaker than convergence (For instance take the sequence $1,-1,1,-1,1,\ldots \to 0$) –  Brett Frankel Dec 28 '12 at 18:11
@Brett, isnot it alternating and diverging sequence? –  doniyor Dec 28 '12 at 18:13
@doniyor: Yes, that is Brett's point. The sequence $a_n=(-1)^n$ diverges in the normal sense, but $\frac{1}{n+1}\sum_{k=0}^n a_k$ converges to $0$ in that case anyway. –  Jonas Meyer Dec 28 '12 at 18:15

Your point makes some sense but is not true. Just like $$(1+\frac1n)^n\to e\neq 1$$ We have $$1+\frac1n\to 1$$ but this doesn't mean that the sequence $(1+\frac1n)^n$ converges to $1^n=1$. In other words you can't disregard the exponent that gets arbitrarily large. Same here, you can't disregard the fact that you have $n$ terms.

The fact that $$\frac{1}{n+1}(a_0+a_1+\cdots+a_n)\to a$$ when $a_n\to a$ is a theorem by Cauchy (the actual theorem says: $$\frac{1}{n}(a_0+a_1+\cdots+a_n)\to a$$) Its proof is somewhat complicated. I could post it here if you'd like however.

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wow, thats it man, Cauchy.. i didnot know that.. thank you so much –  doniyor Dec 28 '12 at 18:06
@doniyor By "Cauchy" I don't mean a Cauchy sequence, but rather the actual person, Augustin Luis Cauchy. –  Nameless Dec 28 '12 at 18:08
oh, so there isnot Cauchy Theorem regarding this? i thought now that it is reasonable why it is getting hard to understand the sequence, because it comes from cauchy and its proof is complicated.. –  doniyor Dec 28 '12 at 18:10
@doniyor: It is not that complex. Just because a great mathematician's name is mentioned is no reason to assume it is hard to understand. –  Jonas Meyer Dec 28 '12 at 18:11
@doniyor This is one of the many theorems to be proven by Cauchy. The proof is complicated, but not "complex" (in the sense of complex numbers) –  Nameless Dec 28 '12 at 18:12

The intuitive way to think of this (and you can make this into a rigorous proof) is to think of the sum as

$$\frac{a_0+a_1+\dots+a_K}{n+1}+\frac{a_{K+1}+\dots+a_n}{n+1}$$

where $K$ is a fixed integer chosen to make all the terms $a_{K+1}$ onwards very close to the limit $a$, and then look at what happens to the two fractions above as $n$ tends to infinity.

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Nice approach! Isn't it possible to make it a strict proof by setting $K=\frac{n}{2}$ or something and then letting $n\to\infty$?... But the difference between intuition and a proof does not seem very large to me. –  barto Dec 28 '12 at 21:27
Thanks. I'm not convinced that taking $K=n/2$ works, though. I think we need to take $K$ to be an integer such that for $j>K$ we have $|a_j-a|<\epsilon$, so that as we let $n$ tend to infinity, the value of $K$ stays fixed - it is important that the number of terms in the first fraction is fixed. –  Old John Dec 28 '12 at 21:31
Yes, of course, the number of terms in the first sum has to be fixed. I missed that for a moment. Thanks for explaining. –  barto Dec 28 '12 at 21:41
Another try: take a function $f$ such that $\lim_{x\to\infty}f(x)=\infty$ and $\lim_{x\to\infty}\frac{f(x)}{x}=0$, for example $\sqrt{x}$ or $\log x$. If you then let $k=\lfloor f(n)\rfloor$, the first fraction goes to $0$ anyway, so $K$ doesn't have to stay fixed necessarily. –  barto Dec 29 '12 at 11:09
True - but why complicate the proof unnecessarily? –  Old John Dec 29 '12 at 11:10

Each term gets small as $n$ gets large, but the number of terms gets equally large.

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why should number of terms interest me? –  doniyor Dec 28 '12 at 17:58
Because adding $N$ numbers equal to $1/N$ gives you 1, no matter how large $N$ is. –  Ron Gordon Dec 28 '12 at 17:59
you must think like (small+small+small+...+small) and not as 0+0+0...+0. –  user52188 Dec 28 '12 at 18:10
so, (small+small+..+small) goes to "small", doesnot it?, i think this very thing is what i am not getting right now.. the $Number$ of terms... –  doniyor Dec 28 '12 at 18:15
@doniyor: No, (small+small+..+small) = big$\times$small. And big$\times$small can be anything in general. –  Jonas Meyer Dec 28 '12 at 19:37

Hint: try the following strategy. For a very large $n$, $a_n$ is close to $a$. Large $n$ can also control the size of $\frac{1}{n+1}(a_0 + ... + a_{N-1})$. So try finding some fixed constant $N$ so that $|\frac{1}{n+1}\sum_{j=0}^n a_j - a| \le \epsilon +|\frac{(n-N)(a + \epsilon)}{n+1} - a |$.

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Let $\epsilon>0$ and choose $N$ such that $|a-a_k|< \frac{1}{2}\epsilon$ for $k \geq N$. Now suppose $n >N$, then we have \begin{eqnarray} |\sum_{k=0}^n \frac{a_k}{n+1} -a| &=& |\sum_{k=0}^n \frac{a_k-a}{n+1}| \\ &\leq& \sum_{k=0}^n \frac{|a_k-a|}{n+1} \\ &=& \sum_{k=0}^{N-1} \frac{|a_k-a|}{n+1}+\sum_{k=N}^n \frac{|a_k-a|}{n+1} \\ &<& \sum_{k=0}^{N-1} \frac{|a_k-a|}{n+1}+\frac{n-N+1}{n+1} \frac{1}{2}\epsilon \\ &\leq & \sum_{k=0}^{N-1} \frac{|a_k-a|}{n+1}+\frac{1}{2}\epsilon \end{eqnarray} Now choose $N'\geq N$ large enough (remember $N$ is fixed) so that for $n\geq N'$, we have $\sum_{k=0}^{N-1} \frac{|a_k-a|}{n+1} < \frac{1}{2}\epsilon$. Then we have $|\sum_{k=0}^n \frac{a_k}{n+1} -a| < \epsilon$. Hence $\lim_{n \to \infty} \sum_{k=0}^n \frac{a_k}{n+1} =a$.

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