## anonymous 4 years ago Find the limit: $\lim_{n \rightarrow \infty} \frac{ 1 }{ n } \sum_{i=1}^{n}\frac{ 1 }{ 1+(\frac{ i}{ n })^2 }$

1. anonymous

I tried to simplify it as much as possible but then I end up with: $\lim_{n \rightarrow \infty} \sum_{i=1}^{n} \frac{ n }{ n^2+i^2}$ I am stuck here...

2. anonymous

Any ideas anyone?

3. anonymous

I think so too but why?

4. hba

Its a trick that when the power is higher in the denominator it approaches to 0

5. anonymous

I know that trick but what about the i^2 term?

6. anonymous

Maybe... $\lim_{n \rightarrow \infty}\sum_{i=1}^{n}\frac{ n }{ n^{2}+i^{2} }=\lim_{n \rightarrow \infty}(\frac{ n }{ 1+n^{2} }+\frac{ n }{ 4+n^{2} }+...+\frac{ n }{ 2(n^{2}-2n+2) }+\frac{ n }{ 2n^{2}-2n+1 }+\frac{ 1 }{ 2n })$ All of these terms have a power 1 polynomial over a power 2 polynomial except for the last one (1/2n, which is clearly 0). So as n-> Infinity, all of the terms should go to 0 as well.

7. hba

Well , if you know that you will know that these terms approaches to 0

8. anonymous

Good point, the i's approach 0 too. Thanks guys :) .

9. anonymous

I can't give a medal to both of you sadly :( .

10. hba

11. hba

Your'e*

12. anonymous

You're welcome also. :)