# The series $\sum_{n=1}^\infty \frac{\sin(\pi n / p)}{n}$ converges for each $p \in \mathbb{N}$

This problem showed up on UCLA's Spring 2018 basic exam for Math Ph.D. students. The problem asks to show that for each $p \in \mathbb{N}$, the infinite series $$\sum_{n=1}^{\infty}\frac{\sin(\pi n/p)}{n}$$ converges. I am curious to see what solutions people come up with. I tried to solve this using Fourier series but was unsuccessful.

• Summation by parts does the trick: it proves the pointwise convergence of the Fourier series of the sawtooth wave, such that the value of the series is $\frac{\pi}{2}\left(1-\frac{1}{p}\right)$. – Jack D'Aurizio Aug 18 '18 at 19:44

## 4 Answers

Hint: Show the partial sums of $\sum_{n=1}^{\infty}\sin(\pi n/p)$ are are uniformly bounded. You're then set up to use Dirchlet's convergence criterion.

This is an application of Summation by parts:

• $\sum_{k \le n} \sin(\frac{\pi k}{p})$ is bounded for $n \in \mathbb N$.
• $\sum \left(\frac{1}{n}-\frac{1}{n+1}\right)$ converges.

Since the community has pointed out that "Dirichlet's test" (or "summation by parts") provides a solution to this problem, I will now answer my own question.

Fix $p \in \mathbb{N}$. Let us define $a_n := \sin(\pi n / p)$. We first point out that the following relations hold for the $a_n$:

1. $a_{n + 2pk} = a_n$ for all $n, k \in \mathbb{N}^+$
2. $a_{n + p} = -a_n$ for all $n \in \mathbb{N}^+$

To see that (1) holds we note that $$a_{n + 2pk} = \sin\left(\frac{\pi(n + 2pk)}{p}\right) = \sin(\pi n /p + 2k \pi) = \sin(\pi n/p) = a_n.$$ To see that (2) holds we observe that $$a_{n + p}= \sin\left(\frac{\pi(n + p)}{p}\right) = \sin(\pi n/p + \pi) = - \sin(\pi n /p) = -a_n.$$ Using these relations, we will show that for all $k \in \mathbb{N}$ and $0 \leq l < 2p$ $$\sum_{n=1}^{2pk + l} a_n = \sum_{n=1}^l a_n$$ where the empty sum $\sum_{n=1}^0 a_n$ is defined to be zero. To show this we fix $l$ where $0 \leq l < 2p$ and induct on $k$.

The base case $k = 0$ holds trivially. Now suppose $$\sum_{n=1}^{2pk + l} a_n = \sum_{n=1}^l a_n$$ holds for some $k \in \mathbb{N}$. Then \begin{align*} \sum_{n = 1}^{2p(k + 1) + l}a_n &= \sum_{n=1}^{2pk + l}a_n + \sum_{n = 2pk + l + 1}^{2pk + 2p + l}a_{n} \\ &= \sum_{n = 1}^l a_n + \sum_{n = 2pk + l + 1}^{2pk + 2p + l}a_{n} \end{align*} We'd now like to show that $$\sum_{n = 2pk + l + 1}^{2pk + 2p + l}a_{n} = 0.$$ Well, $$\sum_{n = 2pk + l + 1}^{2pk + 2p + l}a_{n} = \sum_{j = 1}^{2p}a_{2pk + l + j} = \sum_{j = 1}^{2p}a_{l + j}$$ by relation (1). Then from relation (2) we can get $$\sum_{j = 1}^{2p}a_{l + j} = \sum_{j = 1}^{p}a_{l + j} + \sum_{j = 1}^p a_{l + j + p} = \sum_{j = 1}^{p}a_{l + j} + \sum_{j = 1}^p - a_{l + j} = 0$$ as desired. Thus we may conclude that $$\sum_{n = 1}^{2p(k + 1) + l}a_n = \sum_{n=1}^l a_n$$ and our inductive step is done.

Since every $N \in \mathbb{N}^+$ can be written in the form $N = 2pk + l$ with $k \in \mathbb{N}$ and $0 \leq l < 2p$, we can conclude that $$\left| \sum_{n=1}^N a_n \right| \leq \max \left\{\left|\sum_{n = 1}^l a_n \right| : 0 \leq l < 2p \ \right\} \in \mathbb{R}$$ for all $N \in \mathbb{N}$. Thus the partial sums $A_n := \sum_{k = 1}^n a_k$ form a bounded sequence. Now let $b_n := 1/n$. We note that $\{b_n\}_{n=1}^\infty$ is a monotonically decreasing sequence such that $\lim_{n \rightarrow \infty} b_n = 0$. It thus follows by Dirichlet's test that $$\sum_{n = 1}^\infty \frac{\sin(\pi n/p)}{n} = \sum_{n = 1}^\infty a_n b_n$$ converges.

Possibly too unsophisticated for UCLA:

Grouping the terms $p$ at a time produces an alternating series whose terms decrease absolutely towards $0$, which therefore converges. (It is decreasing because $a_{n+p} = -\frac{n}{n+p}a_n$, so successive groups of $p$ terms decrease absolutely by comparing term for term).

The partial sums that do not run over a multiple of $p$ terms differ by at most $p/n$ from the next partial sum that does, and this difference goes to $0$.