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}

• Hint: $S(n) = \frac{1}{n} \sum_{k=1}^{\infty} \frac{(k/n)}{1+(k/n)^2}.$ Think about Riemann sums. – Ragib Zaman Jul 27 '14 at 13:59
• @RagibZaman We haven't yet been taught Riemann sums. – pkwssis Jul 27 '14 at 13:59
• @Pkwssis he means turn it into an integral. – lemon Jul 27 '14 at 14:00
• @ozo We haven't been taught that either, as of yet. – pkwssis Jul 27 '14 at 14:02
• @DanielFischer No..it is k.. – pkwssis Jul 27 '14 at 14:15

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 $\frac{1}{2}\log2$.
The general term is equivalent to $\frac{1}{k}$, so I don't think this converges at all.
• Note: The sum's upper limit has been corrected from $\infty$ to $n$ by the original poster. – mvw Jul 27 '14 at 14:30