Recently, I was asked by a friend to compute the limit of the following series
$$\displaystyle{\lim_{k\to\infty}}\sum_{n=1}^{\infty} \frac{\sin\left(\frac{\pi n}{k}\right)}{n}$$
Having seen a similar problem to this before, Difficult infinite trigonometric series, I used the same complex argument approach as seen in that problem.
Ultimately, for this problem, I obtained $\frac{\pi}{2}$ as my answer. However, this limit can also be interpreted as a Riemann Sum, except the answer to the Riemann Sum differs from what I obtained as my answer, and according to Wolfram Alpha, the answer is expressed in terms of $Si$, where $Si$ is the sine integral.
I'm wondering, does the limit invalidate the argument approach, or is there something else I'm missing, because this limit if I'm not mistaken is a Riemann Sum after all?
The Riemann Sum would be $$ \begin{align} \lim_{k\to\infty}\sum_{n=1}^\infty\frac{\sin\left(\frac{\pi n}k\right)}{n} &=\lim_{k\to\infty}\sum_{n=1}^\infty\frac{\sin\left(\frac{\pi n}k\right)}{n/k}\frac1k\\ &=\int_0^\infty\frac{\sin(\pi x)}x\,\mathrm{d}x\\ &=\int_0^\infty\frac{\sin(x)}x\,\mathrm{d}x\\[3pt] &=\frac\pi2\tag1 \end{align} $$ However, a cleaner way is to note that $$ \begin{align} \sum_{n=1}^\infty\frac{\sin\left(\frac{\pi n}k\right)}{n} &=-\mathrm{Im}\!\left(\log\left(1-e^{i\pi/k}\right)\right)\\ &=\frac\pi2-\frac\pi{2k}\tag2 \end{align} $$ and the limit is easy.
Take Care
One must be careful with the convergence of the Riemann Sum. Here is one method to control the remainders.
Because $|\sin(\pi x)|\le1$, we have $$ \int_m^{m+1}\left|\frac{\sin(\pi x)}x\right|\,\mathrm{d}x \le\frac1m\tag3 $$ Furthermore, $\int_m^{m+2}\sin(\pi x)\,\mathrm{d}x=0$, thus, $$ \begin{align} \left|\int_m^{m+2}\frac{\sin(\pi x)}x\,\mathrm{d}x\right| &=\left|\int_m^{m+2}\sin(\pi x)\left(\frac1x-\frac1{m+1}\right)\mathrm{d}x\right|\\ &\le\frac1{m(m+1)}+\frac1{(m+1)(m+2)}\\[6pt] &=\frac1m-\frac1{m+2}\tag4 \end{align} $$ Therefore, for any $N\ge m$, $$ \left|\int_m^N\frac{\sin(\pi x)}x\,\mathrm{d}x\right| \le\frac1m\tag5 $$ Because $|\sin(\pi x)|\le1$, we have $$ \sum_{n=mk}^{(m+1)k}\left|\frac{\sin\left(\frac{\pi n}k\right)}{n}\right| \le\frac1m\tag6 $$ Furthermore, $\sum\limits_{n=mk}^{(m+2)k}\sin\left(\frac{\pi n}k\right)=0$, thus, $$ \begin{align} \left|\sum_{n=mk}^{(m+2)k}\frac{\sin\left(\frac{\pi n}k\right)}{n}\right| &=\left|\sum_{n=mk}^{(m+2)k}\sin\left(\frac{\pi n}k\right)\left(\frac1n-\frac1{(m+1)k}\right)\right|\\ &\le\frac1{m(m+1)}+\frac1{(m+1)(m+2)}\\[6pt] &=\frac1m-\frac1{m+2}\tag7 \end{align} $$ Therefore, for any $M\ge mk$, $$ \left|\sum_{n=mk}^M\frac{\sin\left(\frac{\pi n}k\right)}{n}\right|\le\frac1m\tag8 $$ For any $\epsilon\gt0$, let $m\ge\frac4\epsilon$. Then Riemann Sums allow us to choose a $k$ large enough so that $$ \left|\int_0^m\frac{\sin(\pi x)}x\,\mathrm{d}x-\sum_{n=1}^{mk}\frac{\sin\left(\frac{\pi n}k\right)}{n/k}\frac1k\right|\le\frac\epsilon2\tag9 $$ Inequalities $(5)$ and $(8)$ show that for any $N\ge m$ and $M\ge mk$, $$ \left|\int_m^N\frac{\sin(\pi x)}x\,\mathrm{d}x\right|\le\frac\epsilon4 \quad\text{and}\quad \left|\sum_{n=mk}^M\frac{\sin\left(\frac{\pi n}k\right)}{n/k}\frac1k\right|\le\frac\epsilon4\tag{10} $$ Inequalities $(9)$ and $(10)$ show that, for the $k$ chosen for $(9)$, $$ \left|\sum_{n=1}^\infty\frac{\sin\left(\frac{\pi n}k\right)}{n}-\frac\pi2\right|\le\epsilon\tag{11} $$ Since $\epsilon\gt0$ was arbitrary, $(11)$ says that $$ \lim_{k\to\infty}\sum_{n=1}^\infty\frac{\sin\left(\frac{\pi n}k\right)}{n}=\frac\pi2\tag{12} $$