Proving a sequence is a Cauchy Sequence

[TeenieBopper]

Homework Statement

Prove \sum\frac{(-1)^k}{k^2} is a Cauchy sequence.

Homework Equations

Definition of Cauchy sequence: |a_{n} - a_{m}|<ε for all n,m>=N, n>m

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.

 

[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}.