# Determine if this specific sequence is a Cauchy sequence

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?

• Note that (as suggested in one of the edits) the last equality in the last line should be an inequality $\leq$. – Yanko Dec 9 '18 at 10:50
• Just saw it, fixed, thanks a lot :) – Tegernako Dec 9 '18 at 10:51

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 $$n,m\rightarrow 0$$.

• Exactly! Thanks. – Tegernako Dec 9 '18 at 10:41

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

• I've edited my answer. Thank you. – José Carlos Santos Dec 9 '18 at 10:39
• Seems like I went into paralysis by analysis trying to over-express the sum I had. Thanks! – Tegernako Dec 9 '18 at 10:40
• 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. – Jonas Lenz Dec 9 '18 at 10:44
• 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. – José Carlos Santos Dec 9 '18 at 10:46
• @JonasLenz I guess that is how one shows that $\sum_{k=1}^\infty \frac{1}{k^2}$ convergent. – Yanko Dec 9 '18 at 10:49