# Divergent Harmonic Series, Convergent P-Series (Cauchy sequences)

1. Mar 25, 2008

### PingPong

1. The problem statement, all variables and given/known data
(a) Show that $\sum \frac 1n$ is not convergent by showing that the partial sums are not a Cauchy sequence
(b) Show that $\sum \frac 1{n^2}$ is convergent by showing that the partial sums form a Cauchy sequence

2. Relevant equations
Given epsilon>0, a sequence is Cauchy if there exists an N such that $|a_m-a_n|<\epsilon$ for every m,n>N.

3. The attempt at a solution
For part (a), the sequence terms are $a_n=1+1/2+\ldots+1/n$, so assuming that m>n,
$$|a_m-a_n|=1/m+1/(m-1)+\ldots+1/(n+1)<\frac{m-n}{n+1}<\frac{m-n}{n}$$.

Now, if I take epsilon=1/2 and suppose that m>n>N implies that the distance between two elements is less than epsilon. But m=2n>n>N gives the difference to be 1, which is greater than epsilon, so the sequence is not Cauchy. I think I've gotten this half - am I correct??

For part (b), I'm not sure how to do it. The sequence terms are $a_n=1+1/2^2+\ldots+1/n^2$, so again assuming that m>n, we have

$$|a_m-a_n|=1/m^2+1/(m-1)^2+\ldots+1/(n+1)^2<\frac{m-n}{(n+1)^2}<\frac{m-n}{n^2}<\frac{m}{n^2}$$

I want to be able to find N such that m>n>N implies that this difference is less than epsilon, right? To do that, since I've assumed that m>n, if I can eliminate m from the expression, I'm good to go, but I can't figure out how to do it. Help!

Last edited: Mar 25, 2008
2. Mar 26, 2008

### PingPong

I'm grasping at straws here, but I want to show that $m/n^2<\epsilon$. So I can say, since $n<m$, that $n/n^2=1/n<m/n^2<\epsilon$.

I know this probably isn't right, but if anybody could give me a hand it'd be appreciated.