4
$\begingroup$

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.

enter image description here

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?

$\endgroup$
  • 1
    $\begingroup$ 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. $\endgroup$ – Peter Foreman Aug 12 at 21:48
  • 1
    $\begingroup$ Nor can you (directly) apply dominated cvg theorem ... $\endgroup$ – Olivier Aug 12 at 21:51
  • 1
    $\begingroup$ 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 $\frac1N\sum_{k=1}^N g(k/N)\to\int_0^1g(x)\mathrm{d}x$. $\endgroup$ – Peter Foreman Aug 12 at 21:54
  • 2
    $\begingroup$ (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 $n=1...k$) $\endgroup$ – Olivier Aug 12 at 22:03
  • 1
    $\begingroup$ 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. $\endgroup$ – Olivier Aug 12 at 22:16
6
$\begingroup$

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

$\endgroup$
  • 2
    $\begingroup$ Riemann sums works on a finite intervals; what's the additional argument used here to extend this to (0,\infty) ? $\endgroup$ – Olivier Aug 12 at 22:11
  • 2
    $\begingroup$ I've gotta say, that is one sexy solution and elegant approach! +1 :) $\endgroup$ – Sanjoy Kundu Aug 12 at 22:16
  • $\begingroup$ @Olivier: In this case, we can take the sum from $1$ to $mk$ to get the integral from $0$ to $m$. Then carefully let $m\to\infty$. $\endgroup$ – robjohn Aug 12 at 22:27
  • 1
    $\begingroup$ @Olivier: I believe one can use Abel's Test to estimate the remainder. $\endgroup$ – robjohn Aug 12 at 22:42
  • 1
    $\begingroup$ @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. $\endgroup$ – robjohn Aug 13 at 8:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.