# Using Cauchy sequence , show that this sequence converges

Consider $\left(b_k\right)^{\infty }_{k=0}$ a sequence of integer numbers, and $a_{n\:=\:}\sum _{k=0}^n\:\frac{1}{k^2}\left(-1\right)^{b_k}$.
Using Cauchy theorem, How can i prove that this sequence converges?

So far i said:
Let $\epsilon >0$. we need to find an N such that for any $n,m > N$ , $\left|a_m-a_n\right|<\epsilon$.
So i want to evaluate $\left|a_m-a_n\right|$:
$\left|a_m-a_n\right|$ = $\left|\sum \:_{k=n+1}^m\:\frac{1}{k^2}\left(-1\right)^{b_k}\:\:\right|$ but from here i stuck, because we don't know what is $b_k$. tnx!

• The series is absolutely convergent. Dec 29, 2014 at 23:16
• Next step: Without loss of generality are assuming $n\lt m$. By the Triangle Inequality, $\left|\sum_{n+1}^m \frac{1}{k^2}(-1)^{b_n}\right|\le \sum_{n+1}^m \frac{1}{k^2}$. The step after that is to show that by taking $N$ large enough, we can make $\sum_{n+1}^m \frac{1}{k^2}\lt \epsilon$. Dec 29, 2014 at 23:23

Expanding on André Nicolas's comment, which I upvoted since it should have been an answer,

$$\sum_{n+1}^m \frac{1}{k^2} \le \sum_{n+1}^m \frac{1}{k(k-1)} = \sum_{n+1}^m \left(\frac{1}{k-1}-\frac{1}{k}\right) = \frac1{n}-\frac1{m} < \frac1{n}$$ so the sum is Cauchy since this upper bound goes to zero as $n$ and $m$ go to infinity.

Of course, nothing in this answer is original by me.

• I like that your explanation requires no knowledge of the convergence of $\sum \frac{1}{k^2}$. :-)
– kobe
Dec 30, 2014 at 1:08
• Your comment was so perceptive and insightful that I just had to upvote it:) Dec 30, 2014 at 5:54

First, prove $c_n = \sum_{k = 1}^n \frac{1}{k^2}$ converges.

So ${c_n}$ is Cauchy; i.e. $\exists N$ such that $m \geq n > N$ implies $c_m - c_n = \frac{1}{(n+1)^2} + \dots + \frac{1}{m^2} < \epsilon$ for any given $\epsilon > 0.$ Therefore,

$$\left|\sum_{k = n+1}^m \frac{1}{k^2}(-1)^{b_k}\right| = \left| \frac{(-1)^{b_{n+1}}}{(n+1)^2} + \dots + \frac{(-1)^{b_{m}}}{m^2}\right| \leq \frac{1}{(n+1)^2} + \dots + \frac{1}{m^2} < \epsilon$$ if $m \geq n > N$.