Infinite series (i think it's riemann)

1. Jul 4, 2011

flyerpower

1. The problem statement, all variables and given/known data
$\sum_{1}^{inf}$ k^2/(n^3+k^2)

3. The attempt at a solution
I think it's Riemann but i cannot find a suitable function to integrate.

2. Jul 4, 2011

Susanne217

Isn't it suppose to be

$$\sum_{n = 1}^{\infty} \frac{k^2}{n^3 + k^2}$$

3. Jul 4, 2011

Gib Z

Please take more care to expressing to others (and yourself) what it is you want to find. As written the sum doesn't make total sense. It could be what Susanne217 said above, or Riemann sum comment makes me think you could have also meant $\displaystyle\lim_{n\to\infty} \sum_{k=1}^n \frac{k^2}{n^3 + k^2}$.

4. Jul 4, 2011

Susanne217

I agree...

I have been away for a while but I think it could be that the young man is suppose re-write the sum to an improper integral and thereby finding the sum of the series??

5. Jul 4, 2011

Gib Z

In the case that it is what I thought, then it's not as simple as recognizing it as a pre-prepared Riemann sum. With some careful estimates to bound the sum, you should get the result to be $1/3$.

6. Jul 4, 2011

Susanne217

Isn't it true that you need to rewrite this series into more easy to work with??

7. Jul 4, 2011

flyerpower

First of all sorry for misspelling.

I used the bounds k^2/(n^3+n^2) <= k^2/(n^3+k^2) <= k^2/(n^3+1) and i worked it out to 1/3.

8. Jul 4, 2011

Susanne217

Then I have made a mistake cause it get to be 0.373551 and not a 1/3 :(

9. Jul 4, 2011

flyerpower

I may have been wrong but wolframalpha says it's 1/3 so it must be correct :).
Use those bounds and take the limits, you'll see they both converge to 1/3.