TITLE: limit $\lim_{n\to \infty }\sum_{k=1}^{n}\frac{k}{k^2+n^2}$
QUESTION [5 upvotes]: How do I evaluate this?
$$\lim_{n\to \infty }\sum_{k=1}^{n}\frac{k}{k^2+n^2}$$
I got concerned for that, I've tried make it integral for Riemann but it still undone.

REPLY [3 votes]: Notice that $$\frac{k}{k^2+n^2}$$ is an increasing function on $k \in (0,n)$ as derivative is positive in this range. 
Therefore $$\int_0^{n-1}\frac{k}{k^2+n^2}\leq\sum_1^{n-1}\frac{k}{k^2+n^2}\leq\int_1^{n}\frac{k}{k^2+n^2}$$.
Note that $$\int_0^{n-1}\frac{k}{k^2+n^2}=(1/2) \log \frac{2n^2-2n+1}{n^2}$$ 
So $$\lim_{n \to \infty}\int_0^{n-1}\frac{k}{k^2+n^2}=\lim_{n \to \infty}(1/2) \log \frac{2n^2-2n+1}{n^2}= 1/2 \log2$$.
Also, $$\int_1^{n}\frac{k}{k^2+n^2}=(1/2) \log \frac{2n^2}{n^2+1}$$ 
So $$\lim_{n \to \infty}\int_1^{n}\frac{k}{k^2+n^2}=\lim_{n \to \infty}(1/2) \log \frac{2n^2}{n^2+1}= 1/2 \log2$$.
So $$\lim_{n\to\infty} \sum_1^{n}\frac{k}{k^2+n^2}=\lim_{n\to\infty}\sum_1^{n-1}\frac{k}{k^2+n^2} + \lim_{n\to\infty}\frac{n}{n^2+n^2}=\lim_{n\to\infty}\sum_1^{n-1}\frac{k}{k^2+n^2} + 0$$
But,
$$S=\lim_{n\to\infty}\sum_1^{n-1}\frac{k}{k^2+n^2} = (1/2)\log2$$, as $$(1/2)\log2\leq S\leq(1/2)\log2$$ 
So $$\lim_{n\to\infty} \sum_1^{n}\frac{k}{k^2+n^2}=(1/2)\log2$$