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I have the following sequence: $$a_n =\sum_{k = 1}^n (-1)^{b_k} {1\over k^2} $$ And the hint is that I have to prove that: $$ {1\over k^2} < {1\over k-1} - {1\over k} $$

So assuming $m>n$, I have to prove that: $$\forall \epsilon >0, \exists N \in \mathbb{N},$$ so that $$ \forall m,n > N \Rightarrow \lvert a_m - a_n\rvert < \epsilon $$

What I gathered so far: $ \lvert a_m - a_n\rvert = \lvert \sum_{k = n+1}^m (-1)^{b_k} {1\over k^2}\rvert $

$b_k$ is a sequence of natural numbers ${1,2,3.....}$, so in absolute value, $(-1)^{b_k} $ is $1$. Therefore:

$ \lvert a_m - a_n\rvert = \lvert \sum_{k = n+1}^m (-1)^{b_k} {1\over k^2}\rvert \leq \sum_{k = n+1}^m {1\over k^2}. $

From here on, its not so clear to me as to how to proceed. What should be my next steps?

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    $\begingroup$ Note that (as suggested in one of the edits) the last equality in the last line should be an inequality $\leq$. $\endgroup$ – Yanko Dec 9 '18 at 10:50
  • $\begingroup$ Just saw it, fixed, thanks a lot :) $\endgroup$ – Tegernako Dec 9 '18 at 10:51
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You're almost done. Since $\frac{1}{k^2} \leq \frac{1}{k-1}-\frac{1}{k}$ you have that

$$\sum_{k = n+1}^m {1\over k^2}\leq\sum_{k=n+1}^m \left[\frac{1}{k-1}-\frac{1}{k}\right]$$ This is a telescoping series which is equal to $\frac{1}{n}-\frac{1}{m}$. It converges to zero as $n,m\rightarrow 0$.

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  • $\begingroup$ Exactly! Thanks. $\endgroup$ – Tegernako Dec 9 '18 at 10:41
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Now, use the fact that$$\sum_{k=n+1}^m\frac1{k^2}<\sum_{k=n+1}^m\frac1k-\frac1{k-1}=\frac1n-\frac1m.$$

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    $\begingroup$ I've edited my answer. Thank you. $\endgroup$ – José Carlos Santos Dec 9 '18 at 10:39
  • $\begingroup$ Seems like I went into paralysis by analysis trying to over-express the sum I had. Thanks! $\endgroup$ – Tegernako Dec 9 '18 at 10:40
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    $\begingroup$ Why do we need the hint? Can't we simply say that as $\sum_{k=1}^\infty \frac{1}{k^2}$ is convergent, we know that that $\sum_{k=n}^\infty \frac{1}{k^2}$ converges to $0$ as $n \to \infty$ which shows the claim, too. $\endgroup$ – Jonas Lenz Dec 9 '18 at 10:44
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    $\begingroup$ In order to prove that, for each $\varepsilon>0$, $\sum_{k=n+1}^m\frac1{k^2}<\varepsilon$, if $m$ and $n$ are large enough. $\endgroup$ – José Carlos Santos Dec 9 '18 at 10:46
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    $\begingroup$ @JonasLenz I guess that is how one shows that $\sum_{k=1}^\infty \frac{1}{k^2}$ convergent. $\endgroup$ – Yanko Dec 9 '18 at 10:49

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