TITLE: The limit of a sum: $\lim_{n\to \infty} \sum_{k=1}^{n} \left(\frac{k}{n^2}-\frac{k^2}{n^3}\right)$
QUESTION [5 upvotes]: Evaluate the following limit:
  $$
\lim_{n\to \infty} \sum_{k=1}^{n} \left(\frac{k}{n^2}-\frac{k^2}{n^3}\right)
$$

I haven't ever taken the limit of the sum... Where do I start?
Do I start taking the sum?

REPLY [2 votes]: Since Riemann sums
seem to do the trick,
let's look at
$d(u)
=\lim_{n\to \infty} \sum_{k=1}^{n} \frac{k^u}{n^{u+1}}
$
where
$u \ge 0$.
Doing what was done before,
$\begin{array}\\
d(u)
&=\lim_{n\to \infty} \dfrac1{n}\sum_{k=1}^{n} \frac{k^u}{n^{u}}\\
&=\lim_{n\to \infty} \dfrac1{n}\sum_{k=1}^{n} \left(\frac{k}{n}\right)^u\\
&=\int_0^1 x^u dx\\
&=\dfrac1{u+1}\\
\end{array}
$
Therefore,
if
$d(u, v)
=\lim_{n\to \infty} \sum_{k=1}^{n} \left(\frac{k^u}{n^{u+1}}-\frac{k^v}{n^{v+1}}\right)
$,
$d(u, v)
=\dfrac1{u+1}-\dfrac1{v+1}
$.
If
$u=1, v=2$,
this gives
$d(1, 2)
=\dfrac12-\dfrac13
=\dfrac16
$.