# Proving a sequence is a Cauchy Sequence

1. Apr 19, 2012

### TeenieBopper

1. The problem statement, all variables and given/known data
Prove $\sum\frac{(-1)^k}{k^2}$ is a Cauchy sequence.

2. Relevant equations
Definition of Cauchy sequence: |a$_{n}$ - a$_{m}$|<ε for all n,m>=N, n>m

3. The attempt at a solution

I thought if I could prove that the above summation was less than the summation of 1/k^2, the problem would be easy. But either that isn't true, or I don't know how to prove it.

So I started some stuff with the triangle inequality. Here's what I have in full so far:

Let ε>0 be given. Choose a positive integer N such that 1/N<ε. Then, if m,n>=N, and n>m, we have |a_n - a_m| <= |$\sum(-1)^n/n^2$| + |-$\sum(-1)^m/m^2$.

But I don't really know where to go from here. I could say that |a_m| = |-a_m|, but I don't think that gets me anywhere. Any hints to get me going again would be greatly appreciated.

2. Apr 19, 2012

### hamsterman

It should be intuitively clear that $|a+b| \leq |a| + |b|$. From this follows that $|\sum f(x)| \leq \sum |f(x)|$, thus $\sum \frac{(-1)^k}{k^2} \leq \sum \frac{|(-1)^k|}{k^2} = \sum \frac{1}{k^2}$.