How do you calculate this limit $\lim_{n\to\infty}\sum_{k=1}^{n} \frac{k}{n^2+k^2}$? How to find the value of $\lim_{n\to\infty}S(n)$, where $S(n)$ is given by $$S(n)=\displaystyle\sum_{k=1}^{n} \dfrac{k}{n^2+k^2}$$
Wolfram alpha is unable to calculate it.
This is a question from a questions booklet, and the options for the answer are--
$\begin{align}
&A) \dfrac{\pi}{2} \\
&B) \log 2 \\
&C) \dfrac{\pi}{4} \\
&D) \dfrac{1}{2} \log 2
\end{align}$
 A: You could get away with a simple estimate of the expected range of the limit:
With $S_1(n)=\sum_{k=1}^{n}\frac{k}{n^2+n^2}$ and $S_2(n)=\sum_{k=1}^{n}\frac{k}{n^2}$, we have
$$S_1(n)<S(n)<S_2(n)$$
The limits for $S_1$ and $S_2$ are
$$\lim_{n\rightarrow\infty}S_1(n)=\lim_{n\rightarrow\infty}\frac{1}{2n^2}\sum_{k=1}^nk=\lim_{n\rightarrow\infty}\frac{1}{2n^2}\frac{n^2-n}{2}=\frac{1}{4}$$
$$\lim_{n\rightarrow\infty}S_2(n)=\lim_{n\rightarrow\infty}\frac{1}{n^2}\sum_{k=1}^nk=\lim_{n\rightarrow\infty}\frac{1}{n^2}\frac{n^2-n}{2}=\frac{1}{2}$$
So we know that 
$$\frac{1}{4}<\lim_{n\rightarrow\infty}S(n)<\frac{1}{2}.$$
Out of the given answers, that leaves only $\frac{1}{2}\log2$.
A: Clearly, 
\begin{align}
\lim_{n\to\infty}\sum_{k=1}^n\frac{k}{n^2+k^2} &=\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^n\frac{\frac{k}{n}}{1+\frac{k^2}{n^2}}
\stackrel{\text{Riemann sum}}\longrightarrow \int_0^1 \frac{x\,dx}{1+x^2}=\left.\frac{1}{2}\log (1+x^2)\right|_0^1\\
&=\frac{1}{2}\log 2.
\end{align}
A: The general term is equivalent to $\frac{1}{k}$, so I don't think this converges at all.
