anonymous
  • anonymous
Find the limit: \[\lim_{n \rightarrow \infty} \frac{ 1 }{ n } \sum_{i=1}^{n}\frac{ 1 }{ 1+(\frac{ i}{ n })^2 }\]
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • 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...
anonymous
  • anonymous
Any ideas anyone?
anonymous
  • anonymous
I think so too but why?

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hba
  • hba
Its a trick that when the power is higher in the denominator it approaches to 0
anonymous
  • anonymous
I know that trick but what about the i^2 term?
anonymous
  • 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.
hba
  • hba
Well , if you know that you will know that these terms approaches to 0
anonymous
  • anonymous
Good point, the i's approach 0 too. Thanks guys :) .
anonymous
  • anonymous
I can't give a medal to both of you sadly :( .
hba
  • hba
Your welcome.
hba
  • hba
Your'e*
anonymous
  • anonymous
You're welcome also. :)

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