Question

Find the limit:

\[\lim_{n \rightarrow \infty} \frac{ 1 }{ n } \sum_{i=1}^{n}\frac{ 1 }{ 1+(\frac{ i}{ n })^2 }\]

Answer 1

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...

Answer 2

Any ideas anyone?

Answer 3

I think so too but why?

Answer 4

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

Answer 5

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

Answer 6

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.

Answer 7

Well ,

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

Answer 8

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

Answer 9

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

Answer 10

Your welcome.

Answer 11

Your'e*

Answer 12

You're welcome also. :)