How to prove $ \lim_{n\to\infty}\sum_{k=1}^{n}\frac{n+k}{n^2+k}=\frac{3}{2}$? How to prove that
$\displaystyle \lim_{n\longrightarrow\infty}\sum_{k=1}^{n}\frac{n+k}{n^2+k}=\frac{3}{2}$?
I suppose some bounds are nedded, but the ones I have found are not sharp enough (changing $k$ for $1$ or $n$ leads to the limit being between 1 and 2).
Any suggestion is welcomed.

Rewrite the expression as $\displaystyle\sum_{k=1}^{n}\dfrac{n+k}{n^2+k} = \dfrac{1}{n}\sum_{k=1}^{n}\dfrac{1+\frac{k}{n}}{1+\frac{k}{n^2}}$. 
For all $1 \le k \le n$ we have $\dfrac{n}{n+1}\left(1 + \dfrac{k}{n}\right) = \dfrac{1 + \frac{k}{n}}{1+\frac{1}{n}} \le \dfrac{1+\frac{k}{n}}{1+\frac{k}{n^2}} \le \dfrac{1 + \frac{k}{n}}{1+0} = 1 + \dfrac{k}{n}$.
Therefore, $\displaystyle\dfrac{n}{n+1} \cdot \dfrac{1}{n}\sum_{k = 1}^{n}\left(1+\dfrac{k}{n}\right) \le \dfrac{1}{n}\sum_{k=1}^{n}\dfrac{1+\frac{k}{n}}{1+\frac{k}{n^2}} \le \dfrac{1}{n}\sum_{k = 1}^{n}\left(1+\dfrac{k}{n}\right)$. 
Now, use the fact that $\displaystyle\lim_{n \to \infty}\dfrac{n}{n+1} = 1$ and that $\displaystyle\dfrac{1}{n}\sum_{k = 1}^{n}\left(1+\dfrac{k}{n}\right)$ is a Riemann sum for $\displaystyle\int_{0}^{1}(1+x)\,dx$ to get the result.  
