The limit of a sum: $\lim_{n\to \infty} \sum_{k=1}^{n} \left(\frac{k}{n^2}-\frac{k^2}{n^3}\right)$ 
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?
 A: With Riemann sums:
We have
$$
\sum_{k=1}^n \left(\frac{k}{n^2}-\frac{k^2}{n^3}\right)
= \frac{1}{n}\sum_{k=1}^n \left(\frac{k}{n}-\left(\frac{k}{n}\right)^2\right)
= \frac{1}{n}\sum_{k=1}^n \frac{k}{n}\left(1-\frac{k}{n}\right)
$$
which is a Riemann sum for $f\colon[0,1]\to\mathbb{R}$ defined by $f(x)=x(1-x)$. Therefore, we have
$$
\lim_{n\to\infty} \sum_{k=1}^n \left(\frac{k}{n^2}-\frac{k^2}{n^3}\right) = \int_0^1 f(x)dx = \left[\frac{x^2}{2}-\frac{x^3}{3}\right]^1_0 =\boxed{ \frac{1}{6}}\,.
$$
Without Riemann sums:
Here, you can directly use the facts that $\sum_{k=1}^n k = \frac{n(n+1)}{2}$ and $\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$ to compute the sum, and conclude afterwards.
$$
\sum_{k=1}^n \left(\frac{k}{n^2}-\frac{k^2}{n^3}\right)
= \frac{1}{n^2}\sum_{k=1}^n k-\frac{1}{n^3}\sum_{k=1}^n k^2 = \frac{n(n+1)}{2n^2}-\frac{n(n+1)(2n+1)}{6n^3} \xrightarrow[n\to\infty]{} \frac{1}{2} - \frac{1}{3} = \boxed{\frac{1}{6}}\,.
$$
A: Extra approach: by convolutions. By defining, for any $n\in\mathbb{N}$,
$$ a_n = \sum_{k=0}^{n}k(n-k) $$
we have that $a_n$ is the coefficient of $x^n$ in $\left(0+1x+2x^2+3x^3+\ldots\right)^2$, i.e.
$$ a_n = [x^n]\left(\frac{x}{(1-x)^2}\right)^2 = [x^{n-2}]\frac{1}{(1-x)^4}\stackrel{\text{stars and bars}}{=}[x^{n-2}]\sum_{k\geq 0}\binom{k+3}{3}x^k $$
so $a_n = \binom{n+1}{3}$ and 
$$ \lim_{n\to +\infty}\frac{a_n}{n^3} = \frac{1}{3!}=\color{red}{\frac{1}{6}}.$$
A: 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
$.
