Determine if this specific sequence is a Cauchy sequence

Question

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?

Answer 1

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 $n,m\rightarrow 0$.

Answer 2

Now, use the fact that$$\sum_{k=n+1}^m\frac1{k^2}<\sum_{k=n+1}^m\frac1k-\frac1{k-1}=\frac1n-\frac1m.$$