TITLE: Determine if this specific sequence is a Cauchy sequence
QUESTION [3 upvotes]: 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?

REPLY [5 votes]: 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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