Convergence of series and summation methods for divergent series

Question

I would like to know what is the sum of this series:

$$\sum_{k=1}^\infty \frac{1}{1-(-1)^\frac{n}{k}}$$ with $$ n=1, 2, 3, ...$$

In case the previous series is not convergent, I would like to know which are the conditions that would have been required in order for it to be convergent. I can understand that there could be a set of values of $n$ for which the series is not convergent, but this does not directly prove that there are no values of $n$ for which, instead, it is.
In case the previous series is not convergent in the “classical” sense, I would like to know if it can be associated to it a sum, employing those summation methods used to assign a value to a divergent series; like, for example, the Ramanujan summation method which associates to the following well known divergent series

$$\sum_{k=1}^\infty k $$

the value $ -\frac{1}{12} $.

https://en.wikipedia.org/wiki/1_%2B_2_%2B_3_%2B_4_%2B_%E2%8B%AF

https://en.wikipedia.org/wiki/Divergent_series#Examples

Note that, in general, the argument of the sum considered can assume complex values!

Answer 1

For the terms with $k\nmid2n$ you take a fractional power of $-1$; this is not uniquely defined and so it is not clear what these terms of your sum should be. For the terms with $k\mid n$ you are dividing by $0$, so these terms in your sum are not defined at all. That leaves the term $k=2n$ which equals $\frac{1}{4n}$. All other terms are undefined.

Before worrying about convergence, make sure that each of the terms in the sum is well-defined.

Answer 2

HINT:

$$S = \sum_{k=2}^\infty \frac{1}{k} \frac{1}{1-\color{blue}{(-1)^\frac{2n}{k}}}$$

You should note, that every even number is of the form $2n, \forall n \in \mathbb{N}$, in this way:

$$(-1)^\frac{2n}{k} = \sqrt[k]{(-1)^{2n}} = \sqrt[k]{1} = 1^{\frac{1}{k}}$$

And $$\lim_{k\to\infty}\left( \frac{1}{k\left(1-1^\frac{1}{k}\right)}\right) = \tilde{\infty}$$

The series diverges.