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Dido525
Group Title
Find the limit:
\[\lim_{n \rightarrow \infty} \frac{ 1 }{ n } \sum_{i=1}^{n}\frac{ 1 }{ 1+(\frac{ i}{ n })^2 }\]
 one year ago
 one year ago
Dido525 Group Title
Find the limit: \[\lim_{n \rightarrow \infty} \frac{ 1 }{ n } \sum_{i=1}^{n}\frac{ 1 }{ 1+(\frac{ i}{ n })^2 }\]
 one year ago
 one year ago

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Dido525 Group TitleBest ResponseYou've already chosen the best response.0
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...
 one year ago

Dido525 Group TitleBest ResponseYou've already chosen the best response.0
Any ideas anyone?
 one year ago

Dido525 Group TitleBest ResponseYou've already chosen the best response.0
I think so too but why?
 one year ago

hba Group TitleBest ResponseYou've already chosen the best response.2
Its a trick that when the power is higher in the denominator it approaches to 0
 one year ago

Dido525 Group TitleBest ResponseYou've already chosen the best response.0
I know that trick but what about the i^2 term?
 one year ago

Aylin Group TitleBest ResponseYou've already chosen the best response.1
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.
 one year ago

hba Group TitleBest ResponseYou've already chosen the best response.2
Well , if you know that you will know that these terms approaches to 0
 one year ago

Dido525 Group TitleBest ResponseYou've already chosen the best response.0
Good point, the i's approach 0 too. Thanks guys :) .
 one year ago

Dido525 Group TitleBest ResponseYou've already chosen the best response.0
I can't give a medal to both of you sadly :( .
 one year ago

Aylin Group TitleBest ResponseYou've already chosen the best response.1
You're welcome also. :)
 one year ago
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