How Accurate is Summation to Three Decimal Places for the Series 1/n^3?

[skyturnred]

Homework Statement

Find the sum to three decimal places

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

Homework Equations

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!

 

[dirk_mec1]

You can use a Riemann zeta function.

 

[Dick]

Homework Statement

Find the sum to three decimal places

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

Homework Equations

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!

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.

 

[skyturnred]

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