Trying to find the number of terms in a series so it is accurate to 3 decimal places

1. Feb 17, 2012

skyturnred

1. The problem statement, all variables and given/known data

Find the sum to three decimal places

$\sum^{\infty}_{n=1}$$\frac{1}{n^{3}}$

2. Relevant equations

3. The attempt at a solution

So the following is the method that I learned how to do it.. but I think it is wrong.

$\int^{\infty}_{n}$$\frac{dx}{x^{3}}$

to get

$\frac{1}{2n^{2}}$

I then take that and do

$\frac{1}{2n^{2}}$$\leq$0.0005

solving for n gets n=31.6

so there should be 32 terms for the sum to be accurate to 3 decimal places.

But I think I'm wrong because when I plug 32 in for n of the original function, I get something like 0.0000305. But how would anything like that small affect the third decimal place?

Thanks!

2. Feb 17, 2012

dirk_mec1

Re: Trying to find the number of terms in a series so it is accurate to 3 decimal pla

You can use a Riemann zeta function.

3. Feb 17, 2012

Dick

Re: Trying to find the number of terms in a series so it is accurate to 3 decimal pla

It's because there are lots of terms after the 32nd term that have similar size. The exact value of the sum is the Riemann zeta function evaluated at 3. If you sum the first 32 terms and find the difference with zeta(3) you'll get something pretty close to your estimate.

4. Feb 20, 2012

skyturnred

Re: Trying to find the number of terms in a series so it is accurate to 3 decimal pla

OK thanks I didn't quite understand but now I do!