Evaluate the following limit probably by Riemann Sum: $\lim_{n\to\infty} \sum_{k=1}^{n} \sin\left(\frac{(2k-1)a}{n^2}\right)$ I've tried solving the following limit and I think that it might be possible to transform the sequence into a Riemann Sum.
$$\lim_{n\to\infty} \sum_{k=1}^{n} \sin\left(\frac{(2k-1)a}{n^2}\right), \\ a \in \mathbb{R}$$
First, I've multiplied and divided by $n$ so I could use ${1\over n}$ as $\Delta x$ and turn it into an integral on $[0, 1]$. Since we have that $(2k - 1)$ in the limit, I thought it would be a good idea to use a midpoint Riemann Sum. So, $x_k^* = \frac{2k - 1}{2n}$
But I just have no idea what to do with the $n^2$ and the $n$ that I've multiplied.
Is this the way to go? 
 A: Try using complex numbers:
$$\sum\limits_{k=1}^{n} \sin\left(\frac{(2k-1)a}{n^2}\right)=\Im\left(\sum\limits_{k=1}^{n}z^{2k-1}\right)=
\Im\left(\frac{z (z^{2n} - 1)}{z^2 - 1}\right)= ...$$
where $z=e^{i\cdot \frac{a}{n^2}}$
$$...=\Im\left(\frac{\left(\cos{\frac{a}{n^2}} + i\sin{\frac{a}{n^2}}\right)\left(\cos{\frac{2a}{n}}-1 + i\sin{\frac{2a}{n}}\right)}{\cos{\frac{2a}{n^2}}-1 + i\sin{\frac{2a}{n^2}}}\right)=\\
\Im\left(\frac{\left(\cos{\frac{a}{n^2}} + i\sin{\frac{a}{n^2}}\right)
\left(\cos{\frac{2a}{n}}-1 + i\sin{\frac{2a}{n}}\right)
\color{red}{\left(\cos{\frac{2a}{n^2}}-1 - i\sin{\frac{2a}{n^2}}\right)}}{\left(\cos{\frac{2a}{n^2}}-1\right)^2 + \left(\sin{\frac{2a}{n^2}}\right)^2}\right)=\\
\Im\left(\frac{4\cdot\sin{\frac{a}{n^2}}\cdot\sin{\frac{a}{n}}\cdot\left(\cos{\frac{a}{n}} + i \sin{\frac{a}{n}}\right)}{2-2\cdot\cos{\frac{2a}{n^2}}}\right)=\\
\frac{2\cdot\sin{\frac{a}{n^2}}\cdot\sin{\frac{a}{n}}\cdot\sin{\frac{a}{n}}}{1-\cos{\frac{2a}{n^2}}}=
\frac{2\cdot\sin{\frac{a}{n^2}}\cdot\sin^2{\frac{a}{n}}}{2\cdot\sin^2{\frac{a}{n^2}}}=
\color{blue}{\frac{\sin^2{\frac{a}{n}}}{\sin{\frac{a}{n^2}}}}$$
Finally
$$\lim\limits_{n\to\infty}\frac{\sin^2{\frac{a}{n}}}{\sin{\frac{a}{n^2}}}=
\lim\limits_{n\to\infty}\frac{\sin^2{\frac{a}{n}}}{\left(\frac{a}{n}\right)^2} \cdot \frac{\frac{a}{n^2}}{\sin{\frac{a}{n^2}}}\cdot\frac{\left(\frac{a}{n}\right)^2}{\frac{a}{n^2}}=a$$

Some elaborations on the calculations above
$$\left(\cos{\frac{a}{n^2}} + i\sin{\frac{a}{n^2}}\right)
\left(\cos{\frac{2a}{n}}-1 + i\sin{\frac{2a}{n}}\right)
\color{red}{\left(\cos{\frac{2a}{n^2}}-1 - i\sin{\frac{2a}{n^2}}\right)}=\\
e^{i\cdot \frac{a}{n^2}} \left(e^{i\cdot \frac{2a}{n}}-1\right) \color{red}{\left(e^{-i\cdot \frac{2a}{n^2}}-1\right)}=
\left(e^{i\cdot \frac{2a}{n}}-1\right) \color{red}{\left(e^{-i\cdot \frac{a}{n^2}}-e^{i\cdot \frac{a}{n^2}}\right)}=...$$
which from $\sin{z}=\frac{e^{iz}-e^{-iz}}{2i}$ is 
$$...=\left(e^{i\cdot \frac{2a}{n}}-1\right)\cdot\color{red}{(-2i)\cdot\sin{\frac{a}{n^2}}}=
e^{i\cdot \frac{a}{n}}
\color{blue}{\left(e^{i\cdot \frac{a}{n}}-e^{-i\cdot \frac{a}{n}}\right)}
\cdot\color{red}{(-2i)\cdot\sin{\frac{a}{n^2}}}=\\
\left(\cos{\frac{a}{n}}+i\sin{\frac{a}{n}}\right)\color{blue}{(2i)\cdot\sin{\frac{a}{n}}}\cdot\color{red}{(-2i)\cdot\sin{\frac{a}{n^2}}}=\\
4\cdot\left(\cos{\frac{a}{n}}+i\sin{\frac{a}{n}}\right)\cdot\color{blue}{\sin{\frac{a}{n}}}\cdot\color{red}{\sin{\frac{a}{n^2}}}$$
Other results used are 


*

*$1-\cos{x}=2\sin^2{\frac{x}{2}}$ and 

*$\lim\limits_{x\to 0}\frac{\sin{x}}{x}=1$.

A: We have that by first order expansion for $\sin x=x+o(x)$
$$ \sin\left(\frac{(2k-1)a}{n^2}\right)=\frac{(2k-1)a}{n^2}+o\left(\frac{k}{n^2}\right)$$
and since $\sum _{k=1}^{n} k=\frac{n(n+1)}2$ we obtain 
$$\sum _{k=1}^{n} \frac{(2k-1)a}{n^2}=\frac{an(n+1)}{n^2}-\frac a n+o(1)=a+o(1)\to a$$

Edit
From the foundamental limit $\lim_{x\to 0} \frac{\sin x}x=1$ we have that by definition of little-o notation
$$\lim_{x\to 0} \frac{\sin x-x}x=0 \iff \sin x - x = o(x)$$
therefore indicating with $a_k(n)=\frac{(2k-1)a}{n^2} \to 0$ we have $\forall k \le n$
$$\sin (a_k(n)) - a_k(n) = o(a_k(n))$$
therefore
$$\sum_{k=1}^{n} \sin (a_k(n))=\sum_{k=1}^{n} \left[\sin (a_k(n))-a_k(n)\right]+\sum_{k=1}^{n} a_k(n)=\sum_{k=1}^{n}o(a_k(n))+\sum_{k=1}^{n} a_k(n)$$
and since $o( f(n) ) + o( g(n) ) = o( f(n) + g(n) )$ we have that
$$\sum_{k=1}^{n}o(a_k(n)))=o\left(\sum_{k=1}^{n} a_k(n)\right)=o(a)$$
we obtain
$$\sum_{k=1}^{n} \sin (a_k(n))=a+o(a)\to a$$
