The limit of $S_n=\sum_{k=1}^{n} \frac{kn}{k+n^3}$ Given $n>1$ and $S_n=\sum\limits_{k=1}^{n} \frac{kn}{k+n^3}$ 
Calculate $\lim\limits_{n \to \infty}  S_n  $  
Well its obvious that $\frac{n^2}{n^3+n} \leq S_n\leq \frac{n^3}{n^3+1}$
$S_n$ converge to a limit $l$ such that $0\leq l \leq 1$
How can we determine the value of $l$ ?

You can also get the result if you only estimate the denominator in order to get better bounds:
$$\frac{1}{2} \stackrel{n\to\infty}{\longleftarrow}\frac{n(n+1)}{2(1+n^2)} = \frac{n}{n+n^3} \sum \limits_{k=1}^n k \leq S_n \leq \frac{n}{1+n^3} \sum \limits_{k=1}^n k = \frac{n^2(n+1)}{2(1+n^3)} \stackrel{n\to\infty}{\longrightarrow} \frac{1}{2}\, . $$
