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Let there be $\epsilon>0$.

$\{a_n\}$ is a Cauchy sequence if there is an index $k$, so that for every $n\ge k$ and for every $p \in \mathbb N$, $$|a_{n+p}-a_n|\lt \epsilon$$

And here is an example for a problem:

Prove the following sequence convergence proving it's a Cauchy sequence: $a_n={1 \over 1^2}+{1 \over 2^2}+{1 \over 3^2}+\cdots +{1 \over n^2}$

So I tried proving it the following method: $$|a_{n+p}-a_n|={1 \over (n+p)^2}+{1 \over (n+p-1)^2}+\cdots +{1 \over (n+1)^2} \leq {p \over (n+1)^2} \leq {p \over n^2}$$

Why can't I say ${p \over n^2} \lt \epsilon$? I can always find an n so that it will happen, given $p$ is a constant. My teacher says that's wrong, and I have, in order to prove a sequence is a Cauchy sequence, get an expression without $p$ in it. But I really don't understand why!

Thanks in advance for any help.

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1 Answer 1

up vote 2 down vote accepted

The trouble is that you don't just have to make your bound for $|a_{n+p}-a_n|$ less than $\varepsilon$ for sufficiently large $n$ and a fixed $p$; you have to find an $n$ such that, for any $p>0$, $|a_{n+p}-a_n| <\varepsilon$. In the case of your bound, when we take $p=n^2$, we get $\displaystyle\frac{p}{n^2} = 1$, so this bound doesn't help us for any $\varepsilon < 1$ (in fact it doesn't help for any $\varepsilon$ at all).

A better way to do this is to to use the bound $\displaystyle \frac{1}{n^2} < \frac{1}{n(n-1)} = \frac{1}{n-1}-\frac{1}{n}$. Then $$|a_{n+p}-a_n| = \sum_{k=n+1}^{n+p}\frac{1}{k^2} \leq \sum_{k=n+1}^{n+p} \frac{1}{k-1}-\frac{1}{k} = \frac{1}{n}-\frac{1}{n+p} < \frac{1}{n}.$$ This can be made smaller than $\varepsilon$ for sufficiently large $n$.

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But why can I take $p=n^2$? –  bobby Dec 18 '13 at 10:55
    
Because $p$ can be anything at all -- you don't get to pick it, so your bound needs to work for any $p$. In particular it needs to work when $p$ happens to equal $n^2$. (You can imagine that $p$ is chosen by your worst enemy to mess you up, and they get to pick it after you pick $n$.) –  universalset Dec 18 '13 at 10:57
    
I see. I thought p has to be a constant number... Thanks! –  bobby Dec 18 '13 at 10:58

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