How do I prove that $$\lim_{n \to \infty }n^2\left ( \frac{1}{(n^2+1)^2} +\frac{2}{(n^2+2^2)^2}+...+\frac{n}{(n^2+n^2)^2} \right )=\frac{1}{4}$$ using the definition of the integral?


$$ n^2\sum_{k=1}^n\frac{k}{(n^2+k^2)^2}=n^2\sum_{k=1}^n\frac{k}{n^4(1+(k/n)^2)^2}=\frac{1}{n}\sum_{k=1}^n\frac{k/n}{(1+(k/n)^2)^2}\ , $$ which is a Riemann sum, yielding $$ \int_0^1 dx \frac{x}{(1+x^2)^2}=1/4 $$ in the limit.

  • $\begingroup$ Lightening fast response. I posted the same, but deleted after seeing your post. +1 - Mark $\endgroup$
    – Mark Viola
    Jan 23 '16 at 17:54
  • $\begingroup$ Thanks Mark, no need to delete yours though. The more inputs and collective effort, the better. (Mine had also a small typo I've now fixed, shouldn't rush answers ;-) ) $\endgroup$ Jan 23 '16 at 17:56
  • $\begingroup$ @PierpaoloVivo I was just about to ask you regarding the integral bounds you had typed. +1 for fast response $\endgroup$
    – FreeMind
    Jan 23 '16 at 17:57
  • $\begingroup$ @PierpaoloVivo Thank you, but it is nearly identical to yours and completely redundant. $\endgroup$
    – Mark Viola
    Jan 23 '16 at 18:17

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