Explicit summation of a cosine series

Question

I was wondering if there is a simple representation of the series $$ \sum_{n \ge 1} \frac{1}{3n - K} \cos(3 n x)$$ where $K$ is some integer less than $3$. Mathematica was able to evaluate the sum using some combination of Hypergeometric functions but I am curious if this can be done using elementary functions.

Answer 1

I will show how to derive the Hypergeometric form for this series.

First, a remark: the series is conditionally convergent, it doesn't converge absolutely, which can be seen by comparison to harmonic series. This is all I will say about the convergence issues.

Now, the most direct way would be to represent the cosine as:

$$\cos 3nx=\frac{1}{2} \left(e^{3 i n x}+e^{-3 i n x} \right)$$

I won't provide the rigorous proof that we can separate the series in two parts, which may lead to problems because of the conditional convergence. But I will do it nontheless.

$$\sum_{n \ge 1} \frac{1}{3n - K} \cos(3 n x)=\frac{1}{6} \sum_{ n \geq 1} \frac{e^{3 i n x}}{n - K/3}+\frac{1}{6} \sum_{ n \geq 1} \frac{e^{-3 i n x}}{n - K/3}$$

Setting:

$$z=e^{3 i x},\qquad K/3=a$$

We write the first series as:

$$\sum_{ n \geq 1} \frac{z^n}{n -a }$$

The second series can be represented in exactly the same way.

This can be written in terms of Hurwitz-Lerch Transcendent, but I would prefer giving the Hypergeometric form.

$$\sum_{ n \geq 1} \frac{z^n}{n -a }=z \sum_{ n \geq 0} \frac{z^n}{n+1 -a }$$

$$c_0=\frac{1}{1-a}$$

$$\frac{c_{n+1}}{c_n}=\frac{n+1-a}{n+2-a} z=\frac{(n+1-a)(n+1)}{n+2-a} \frac{z}{n+1}$$

Which by definition makes the series:

$$\sum_{ n \geq 0} \frac{z^n}{n+1 -a }=\frac{1}{1-a} {_2 F_1} (1,1-a;2-a;z)$$

And the original series:

$$\sum_{n \ge 1} \frac{\cos(3 n x)}{3n - K} = \\ = \frac{1}{6-2K} \left(e^{3 i x}{_2 F_1} \left(1,1-\frac{K}{3};2-\frac{K}{3};e^{3 i x} \right)+e^{-3 i x}{_2 F_1} \left(1,1-\frac{K}{3};2-\frac{K}{3};e^{-3 i x}\right) \right)$$