TITLE: Proving a limit summation
QUESTION [4 upvotes]: 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?

REPLY [8 votes]: Writing 
$$\sum_{k=0}^{2n} \frac{k}{k^2 + n^2} = \frac{1}{2n}\sum_{k = 1}^{2n} \frac{\frac{k}{2n}}{\left(\frac{k}{2n}\right)^2 + \frac{1}{4}}$$
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 $\dfrac{\log 5}{2}$.
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