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 $l$ ?
 A: This problem can be tackled by squeeze theorem plus riemann sum.
Note that $$S_n=\sum^n_{k=1}\frac{\frac{k}n}{\frac{k}{n^3}+1}\frac1n$$
We can obtain the following inequality easily
$$\sum^n_{k=1}\frac{\frac{k}n}{\frac{n}{n^3}+1}\frac1n\le S_n\le\sum^n_{k=1}\frac{k}n\frac1n$$
Taking the limit $n\to\infty$, the lower bound approaches 
$$\lim_{n\to\infty}\left(\frac1{\frac1{n^2}+1}\right)\int^1_0xdx=\frac12$$
The upper bound approaches $$\int^1_0xdx=\frac12$$ as well.
By squeeze theorem, $$\color{red}{\lim_{n\to\infty}S_n=\frac12}$$
(This agrees with your expectation that the limit is between $0$ and $1$.)
A: You can also get the result if you only estimate the denominator in order to get better bounds:
$$\frac{1}{2} \stackrel{n\to\infty}{\longleftarrow}\frac{n(n+1)}{2(1+n^2)} = \frac{n}{n+n^3} \sum \limits_{k=1}^n k \leq S_n \leq \frac{n}{1+n^3} \sum \limits_{k=1}^n k = \frac{n^2(n+1)}{2(1+n^3)} \stackrel{n\to\infty}{\longrightarrow} \frac{1}{2}\, . $$
A: If you are familiar with harmonic numbers$$S_n=\sum_{k=1}^{n} \frac{kn}{k+n^3}=n^4 \left(H_{n^3}-H_{n^3+n}\right)+n^2$$ Now, using the asymptotics
$$H_p=\gamma +\log \left({p}\right)+\frac{1}{2 p}-\frac{1}{12
   p^2}+O\left(\frac{1}{p^4}\right)$$ and applying would give
$$S_n=\frac{1}{2}+\frac{1}{2 n}-\frac{1}{3 n^2}-\frac{1}{2
   n^3}+O\left(\frac{1}{n^4}\right)$$ which shows the limit and how it is approached.
Moreover, this gives a good approximation. For example, the exact value
$$S_{10}=\frac{102209264496236445266975}{187134762733574325443631}\approx 0.546180$$ while the above expansion would give
$$ \frac{3277}{6000}\approx 0.546167$$
