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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 phenomenon is called Cesaro summability, and it is strictly weaker than convergence (For instance take the sequence $1,-1,1,-1,1,\ldots \to 0$)2012-12-28
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    @Brett, isnot it alternating and diverging sequence?2012-12-28
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    @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.2012-12-28

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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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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.2012-12-28
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    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.2012-12-28
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    Yes, of course, the number of terms in the first sum has to be fixed. I missed that for a moment. Thanks for explaining.2012-12-28
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    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.2012-12-29
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    True - but why complicate the proof unnecessarily?2012-12-29
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    I understand... But at school we do not learn about the $\epsilon-\delta$ definition of limits (anyway I do know about it) so for me this looks easier.2012-12-29
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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?2012-12-28
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    Because adding $N$ numbers equal to $1/N$ gives you 1, no matter how large $N$ is.2012-12-28
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    you must think like (small+small+small+...+small) and not as 0+0+0...+0.2012-12-28
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    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...2012-12-28
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    @doniyor - such thinking is the basis of integral calculus.2012-12-28
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    @doniyor: No, (small+small+..+small) = big$\times$small. And big$\times$small can be anything in general.2012-12-28
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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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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 much2012-12-28
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    @doniyor By "Cauchy" I don't mean a Cauchy sequence, but rather the actual person, Augustin Luis Cauchy.2012-12-28
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    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..2012-12-28
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    @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.2012-12-28
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    @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)2012-12-28
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    @doniyor Take $a_n=1$. Does the property hold?2012-12-28
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    It's not complex - see below, see link by Jonas Meyer. It's standard $\epsilon - \delta$. Nameless is just being nice/encouraging. Have a look at the hints, and you can do this!2012-12-28
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    @Nameless, yes, it does. because then $\frac{1}{n+1}n \rightarrow 1$, right?2012-12-28
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    @doniyor Exactly right. Do you know understand why $n$ terms matter?2012-12-28
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    @Nameless, wow, :D, now i understand. but wait, i feel still bad. this sequence $\frac{1}{n+1}(a_0+a_1+..+a_n)$ is this $\frac{a_n}{n+1}$, isnot it?2012-12-28
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    @doniyor: No, it is not. How could it be?2012-12-28
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    @JonasMeyer Perhaps my example did more harm than good...2012-12-28
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    then i didnot get again. with $a_n=1$: $\frac{1}{n+1}1+\frac{1}{n+1}1+..+\frac{1}{n+1}1=\frac{1}{n+1}n \rightarrow 1$ and what is the expansion of $\frac{1}{n+1}a_n$ then?2012-12-28
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    @doniyor In the special case where $a_n=1$ it is true. But nit for more general sequences2012-12-28
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    @doniyor: In that example, you would actually have $\frac{1}{n+1}1+\frac{1}{n+1}1+..+\frac{1}{n+1}1=\frac{1}{n+1}(n+1)=1$ for all $n$ (notice there are $n+1$ terms, not $n$, although this does not affect the limit). And $\frac{1}{n+1}a_n$ is $\frac{1}{n+1}$, because $a_n=1$. Those are very different.2012-12-28