# Evaluating $\lim_{k\to\infty}\sum_{n=1}^{\infty} \frac{\sin\left(\pi n/k\right)}{n}$

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?

• In a Riemann sum we would have both 'variables' tending to infinity simultaneously. In this case we have both $n$ and $k$ tending to infinity seperately and hence the Riemann sum approach should not be possible. This approach seems valid. – Peter Foreman Aug 12 at 21:48
• Nor can you (directly) apply dominated cvg theorem ... – Olivier Aug 12 at 21:51
• You effectively have $\lim_{k\to\infty}\lim_{N\to\infty}f(N,k)$ (where $N$ denotes the maximum summation bound). With a Riemann sum we would have $\lim_{N\to\infty}f(N)$ where $k$ may be used as an index such that the function $f(N)$ is of the form $\frac1N\sum_{k=1}^N g(k/N)\to\int_0^1g(x)\mathrm{d}x$. – Peter Foreman Aug 12 at 21:54
• (Rephrasing of what Peter says) You have $\lim_{k \to \infty} \frac{1}{k} \sum_{n=1}^{k} \frac{\sin(\pi n/k)}{(n/k)} \to \int_{0}^{1} \frac{sin(\pi x)}{x} dx$ by Riemann sum argument; but you have to manage a larger interval here... ($n\ge 1$, not just $n=1...k$) – Olivier Aug 12 at 22:03
• You can do a collection of Riemann sums to deal with n=1...k, then n=k+1...2k, n=2k+1...3k, but you would then need a further argument to put these sums altogether. – Olivier Aug 12 at 22:16

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 $$\lim_{k\to\infty}\sum_{n=1}^\infty\frac{\sin\left(\frac{\pi n}k\right)}{n}=\frac\pi2\tag{12}$$

• Riemann sums works on a finite intervals; what's the additional argument used here to extend this to (0,\infty) ? – Olivier Aug 12 at 22:11
• I've gotta say, that is one sexy solution and elegant approach! +1 :) – Sanjoy Kundu Aug 12 at 22:16
• @Olivier: In this case, we can take the sum from $1$ to $mk$ to get the integral from $0$ to $m$. Then carefully let $m\to\infty$. – robjohn Aug 12 at 22:27
• @Olivier: I believe one can use Abel's Test to estimate the remainder. – robjohn Aug 12 at 22:42
• @Olivier: I have added a section showing one method to control the remainders. It is a bit long, to include a lot of detail, but quite typical for this kind of Riemann Sum. – robjohn Aug 13 at 8:55