Show that

$$\lim_{n \to \infty} \sum_{k=0}^{2n} \frac{k}{k^2+n^2} = \frac{1}{2}\, \log 5$$

How would you prove this? I understand limits, but summations not so much. Would I need to take the derivative of the summation first, then evaluate the limit of that?


marked as duplicate by YuiTo Cheng, José Carlos Santos limits Jun 29 at 17:20

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  • $\begingroup$ This is a Riemann sum. $\endgroup$ – user296602 Oct 30 '17 at 22:07


$$\sum_{k=0}^{2n} \frac{k}{k^2 + n^2} = \frac{1}{2n}\sum_{k = 1}^{2n} \frac{\frac{k}{2n}}{\left(\frac{k}{2n}\right)^2 + \frac{1}{4}}$$

we see that your sum is a Riemann sum for the function $f(x) = \dfrac{x}{x^2 + 1/4}$ over the interval $[0,1]$. So the limit is

$$\int_0^1 \frac{x}{x^2 + 1/4}\, dx$$

Using a $u$-sub $u = x^2 + 1/4$, you can show that integral evaluates to $\dfrac{\log 5}{2}$.

  • $\begingroup$ Could you explain how you found the second summation? I understand everything else except that part, $\endgroup$ – user482578 Oct 30 '17 at 22:56
  • 2
    $\begingroup$ he divided everything the top by $2n$ and the bottom by $4n^2$. That's why the additional $1/2n$ outside balances the top and makes an equality $\endgroup$ – John Lou Oct 30 '17 at 23:10
  • $\begingroup$ @kobe How did you know this was a riemann sum for that function? $\endgroup$ – user482578 Nov 24 '17 at 20:25
  • $\begingroup$ @A.Mercs in the equation above, the right-hand side can be written $\frac{1}{2n}\sum_{k = 1}^n f(\frac{k}{2n})$ for some function $f$, which we see by inspection is $f(x) = \frac{x}{x^2 + 1/4}$. $\endgroup$ – kobe Nov 25 '17 at 15:52