Infinite sum of cosecant, $\frac k{\sin(a+ck)}$

Question

My question

Does S converge or diverge?

If it converged, ideally I intend to find its explicite expression as a function of c.

$S=\sum_{k=1}^{\infty}\frac{k}{\sin(a+ck)}$ where $c\in R^+$

...within the problem's context the constant c is typically $\approx10^4$ and $a$ is a number $>0$, typically $<10$

The sum S is part of an anisotropic 3D heat transfer modelling in a cristall like structure. The impact of the sum to diverge would massively alter the overall solution of heat transport.

What I know

For $\forall c^*=n\pi$ (with $n\in N$) the sum converges and becomes a Harmonic Number and can for that special case be calculated.

$S^*=\pm \frac{1}{\sin(a^*)} \sum_{k=1}^{\infty}(\pm 1)^k\cdot k$, where the $\pm$ alters wether n is pair or impair

so that

$S^*=\mp \frac{1}{12\sin(a^*)}$ for pair $n$ and $S^*=\mp\frac{1}{4\sin(a^*)}$ for impair $n$

Still, in the context of the problem c is not not necessarily a multiple of $\pi$.

What I have tried so far (without success)

In the below several ansatzes which I have considered so far without anyone of them allowing to proceed further:

Searching through publications and textbooks

The most extensive list of infinite sums I saw was in Table of Infinite Products (I am adding the link here as the list there is remarkably extensive). Still alike other sources, not mentioned here, seemingly the only published infinite sums of cosecant contain the square of csc and multiples of $\pi$.

Approximating the cosecant

Any approach alike the one described by user Claude seems not feasible for the impossibility to express infinitely continuing oscillations by polynomials.

Mathematica

I aborted Mathematica, after having run the sum using Dirichlet or Borel regularization for 24 hours each. I could let it run longer, but as one cannot see what that program is doing and wether it is "going anywhere", chances are rather slim to succeed with that press-a-button-method.

Laplace Transform

The Laplace (nor Fourier) Transform of a cosecant doesn't seem to exist.

Intuition

Albeit I risk to get downvoted severely there is a cowboy-mathematics or intuitive (at best) approach. The absolute value of the $\csc$-function $\|\csc(ck)\|$ is repeating once every $ck=\pi$. The value of the sum S can then be thought of taking the sum on only their average values

$S=\sum_{k=1}^{\infty}(-1)^k\frac{k}{\|\csc(ck)\|}$ _where $(-1)^k$ "simulates the alternating once every $\pi$

For the cosecant average it is

$\|\csc(ck)\|=\frac{1}{b-a}\int_{a}^{b}\frac{1}{\sin(x)}dx$

As the average must be taken over the range in between 2 consecutive singularities let $\epsilon$ be a very small, positive number close to zero so that the average can be written as (avoiding to say limit)

$\|\csc(ck)\|=\frac{1}{\pi-2\epsilon}\int_{\epsilon}^{\pi-\epsilon}\frac{1}{\sin(x)}dx$

which yields after some transformation and using the first summand of Taylor expansion of the trigonometric terms for $x\rightarrow \frac{\epsilon}{2}$

$\|\csc(ck)\|=-\frac{2}{\pi}\ln(\epsilon)$ and plugged into the sum $S=-\frac{\pi}{2}\ln(\epsilon)\sum_{k=1}^{\infty}(-1)^kk=\frac{\pi}{8}\ln(\epsilon)$

As $\epsilon$ goes towards zero the above indicates that the sum diverges. Still, this is too intuitive for being reliable, but it is the only indication I have found so far.

Identities

The last ansatz I can think of is the use of trignometric identities, but none of them "triggers eagerness" to explore the sum S by applying anyone of these identities.

Add-on to orignal post, taking into account comments

Thanks to Igor and Dave for your replies. Still, I have a doubt that the denominator's singularities necessarily bring the infinite sum out of bound; allow me to give 3 examples where the doubt stems from:

There is a similar sum with a finite (and explicit) result

• $\sum_{k=1}^{\infty}\frac{\csc(k\pi\sqrt2)}{k^3}=-\frac{13\pi^3}{360\sqrt2}$

and another example, an integral taken over one of the singularities (here at x=0)

• $\int_{0}^{\frac{\pi}{2}}x\csc(x)=C$ where C is the Catalan constant

both examples can be found in Introduction to Cosecant.

Using Mathematica I find

• $\sum_{k=1}^{m}\csc(k+a)=\frac{1}{2}(-1+(-1)^m)\csc(a\pi)$

which is supported by one of the numerous Dieckmann results

Answer 1

I am not sure I know what you mean. For $c = n \pi,$ all terms in the sum are undefined (since $\sin kn \pi = 0.)$ In general, if $c$ is not a rational multiple of $\pi,$ the terms will be arbitrarily large, so the sum cannot possibly converge.

UPDATE adding a constant to the argument of $\sin$ does not materially change things - if $c$ is an irrational multiple of $\pi,$ the denominator will come arbitrarily close to zero, so the terms will blow up.

Answer 2

There is a necessary condition for series convergence, which is that if $\sum_{n=1}^\infty a_n$ is to converge then $\lim_{n\to\infty}a_n=0$. You should check $\lim_{n\to\infty}\frac{n}{\sin(cn)}$ first, and see if it goes to $0$. If the limit is not $0$, then this series must diverge.

Also, certain values of $c$ will make the series undefined, as the denominator could be $0$ at certain points in the summation.