# Does (sum (1/n)^p) converge?

1. Jul 6, 2011

### ghc

$\sum_{n=2}^{+\infty}(\frac{1}{n})^{2p}$ where $p \in N$.
Does this sum converge? How do we calcultate it?

I am not very familiar with maths, so if there is some theorem that can used please tell it to me.
Thank you.

2. Jul 6, 2011

### D H

Staff Emeritus
We don't do your homework for you at this site, and this looks very much like homework. (And even if it isn't, we treat questions that look very much like homework as if they were homework.)

Try showing some work. What techniques are there to test whether a series converges? Which ones help with this particular series?

3. Jul 6, 2011

Do you know anything about series? It's a series question. and your series shows different behaviors depending what p you use. for example for p=1/2 the series is the harmonic series which is a famous divergent series. for other values you can use theorems or series tests. for example if p<0 then the series is obviously divergent because it fails the necessary condition to converge. I'll give you a hint.
separate values of p into different groups. like 2p<0, 2p=1, 0<2p<1, 2p>1. That'll be the first step to recognize what the problem asks you to do. I gave you the answer for p<0 & p=1. I let you solve the problem for the other 2 cases. you may not even need to consider the case that 0<2p<1.

About calculating its values, you have to know what p is. moreover, It's usually hard and sometimes extremely difficult to calculate the value of a series.

4. Jul 6, 2011

### ghc

It is not a homework and I am not a student (and I never was a math student) but I can't prove it. So I will show what I can do :

$\frac{(n+1)^{-2p}}{n^{-2p}} = (\frac{n}{n+1})^{2p} \rightarrow 1$ if $n\rightarrow +\infty$
We can't conclude.

$\frac{(n+1)^{-2p}}{n^{-2p}} = (\frac{n+1}{n})^{-2p} = (1+\frac{1}{n})^{-2p} = 1-2p\frac{1}{n}+o(\frac{1}{n})$
$2p>1$ the sum converges.

Is this true?

5. Jul 6, 2011

### ghc

p is a positive integer.

6. Jul 6, 2011