Calculating this Riemann sum limit 
Calculate the limit $$\lim_{n\to \infty} {\sum_{k=1}^{n} {\left(\frac{nk-1}{n^3}\right) \sin\frac{k}{n}}}$$

How exactly do we calculate this limit of the Riemann sum? I am never able to find what is the partition. I know that our $f(x)$ is $\sin(x)$.
 A: Rewrite the sum as
$$\frac{1}{n} \sum_{k=1}^n \left ( \frac{k}{n} - \frac{1}{n^2}\right ) \sin{\left ( \frac{k}{n}\right)}$$
As $n \to \infty$, the $1/n^2$ term vanishes and we are left with
$$\lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^n \frac{k}{n} \sin{\left ( \frac{k}{n}\right)}$$
which is the Riemann sum for the integral
$$\int_0^1 dx \, x \, \sin{x}$$
NB in general
$$\int_a^b dx \, f(x) = \lim_{n \to \infty} \frac{b-a}{n} \sum_{k=1}^n f\left (a + \frac{k}{n} (b-a) \right)$$
when the integral on the left exists.
ADDENDUM
I was asked to expand upon the claim that $1/n^2$ vanishes.  If we use this term, we see that its contribution is
$$\frac{1}{n^3} \sum_{k=1}^n \sin{\left ( \frac{k}{n}\right)}$$
which, in absolute value, is less than $(1/n^3) (n) = 1/n^2$, which obviously vanishes as $n \to \infty$.
A: Recall that if $f$ is integrable on $[a,b]$, then:

$$
\int_a^b f(x)~dx = \lim_{n\to \infty} \dfrac{b-a}{n}\sum_{k=1}^n f \left(a + k \left(\dfrac{b-a}{n}\right) \right)
$$

Notice that:
$$
\sum_{k=1}^{n} {\left(\frac{nk-1}{n^3}\right) \sin\frac{k}{n}}
= \sum_{k=1}^{n} {\left(\dfrac{k}{n^2} - \dfrac{1}{n^3}\right) \sin\frac{k}{n}}
= \dfrac{1}{n}\sum_{k=1}^{n} \dfrac{k}{n}\sin\frac{k}{n} - \dfrac{1}{n^3}\sum_{k=1}^{n} \sin\frac{k}{n}
$$
Hence, by letting $a=0$ and $b=1$ and considering the functions $f(x)=x \sin x$ and $g(x) = \sin x$, we obtain:
$$ \begin{align*}
\lim_{n\to \infty} {\sum_{k=1}^{n} {\left(\frac{nk-1}{n^3}\right) \sin\frac{k}{n}}}
&= \lim_{n\to \infty} \left[ \dfrac{1}{n}\sum_{k=1}^{n} \dfrac{k}{n}\sin\frac{k}{n} - \dfrac{1}{n^3}\sum_{k=1}^{n} \sin\frac{k}{n} \right] \\
&= \lim_{n\to \infty} \left[ \dfrac{1}{n}\sum_{k=1}^{n} \dfrac{k}{n}\sin\frac{k}{n} \right] - \lim_{n\to \infty}\left[\dfrac{1}{n^2} \right] \cdot \lim_{n\to \infty} \left[\dfrac{1}{n}\sum_{k=1}^{n} \sin\frac{k}{n} \right] \\
&= \int_0^1 x \sin x~dx - 0 \cdot \int_0^1 \sin x~dx \\
&= \int_0^1 x \sin x~dx\\
&= \left[\sin x - x\cos x \right]_0^1\\
&= \sin 1 - \cos 1\\
\end{align*} $$
