limit $\lim_{n\to \infty }\sum_{k=1}^{n}\frac{k}{k^2+n^2}$ 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.
 A: $$\sum_{k=1}^{n}\frac{k}{k^{2}+n^{2}}=\frac{1}{n}\sum_{k=1}^{n}\frac{k/n}{\frac{k^{2}}{n^{2}}+1}$$
Which is a Riemann sum for the function $f(x)=\frac{x}{1+x^{2}}$ over $[0,1]$. Your sum therefore tends to $$\int_{0}^{1}\frac{x}{1+x^{2}}dx$$
Can you go on from here?
A: Notice, $$\lim_{n\to \infty}\sum_{k=1}^{n}\frac{k}{k^2+n^2}$$
$$=\lim_{n\to \infty}\sum_{k=1}^{n}\frac{\left(\frac{k}{n}\right)\frac{1}{n}}{\left(\frac{k}{n}\right)^2+1}$$ Let $\frac{k}{n}=u\implies \lim_{n\to \infty}\frac{1}{n}=du\to 0$ 
then we have $$\text{upper bound of u}=\lim_{n\to \infty}\frac{k}{n}=\lim_{n\to \infty}\frac{n}{n}=1$$
$$\text{lower bound of u}=\lim_{n\to \infty}\frac{k}{n}=\lim_{n\to \infty}\frac{1}{n}=0$$ Changing summation into integration with proper limits $$\int_{0}^{1}\frac{u\ du}{u^2+1}$$ $$=\frac{1}{2}\int_{0}^{1}\frac{(2u)\ du}{u^2+1}=\frac{1}{2}\int_{0}^{1}\frac{d(u^2)}{u^2+1}$$  $$=\frac{1}{2}[\ln|u^2+1|]_{0}^{1}$$  $$=\frac{1}{2}[\ln|1+1|-\ln|0+1|]$$$$=\color{red}{\frac{1}{2}\ln 2}$$
A: 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 $$\lim_{n\to\infty} \sum_1^{n}\frac{k}{k^2+n^2}=(1/2)\log2$$
