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Dido525

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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  1. Dido525
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    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
  2. Dido525
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    Any ideas anyone?

    • one year ago
  3. Dido525
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    I think so too but why?

    • one year ago
  4. hba
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    Its a trick that when the power is higher in the denominator it approaches to 0

    • one year ago
  5. Dido525
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    I know that trick but what about the i^2 term?

    • one year ago
  6. Aylin
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    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
  7. hba
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    Well , if you know that you will know that these terms approaches to 0

    • one year ago
  8. Dido525
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    Good point, the i's approach 0 too. Thanks guys :) .

    • one year ago
  9. Dido525
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    I can't give a medal to both of you sadly :( .

    • one year ago
  10. hba
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    Your welcome.

    • one year ago
  11. hba
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    Your'e*

    • one year ago
  12. Aylin
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    You're welcome also. :)

    • one year ago
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