# Convergence of series and summation methods for divergent series

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!

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.

• For $k\nmid2n$ the fractional power of $-1$ is just a complex number, because $-1=e^{i\pi}$: $$(-1)^\frac{2n}{k}=(e^{i\pi})^\frac{2n}{k}=e^{i\frac{2n}{k}\pi}$$ The fact that for $k\mid n$ there is a division by $0$ and that for $k=2n$ the sum is $$\frac{1}{2}\sum_{k=2}^\infty \frac{1}{k}$$ that is clearly not convergent, would just mean that there is a set of values of $n$ for which the series is not convergent (in the classical sense). And what about the summation methods used to assign a value to a divergent series? Are there any useful in this case? Thank you for your remarks. – Joe Mar 3 at 1:53
• To your first remark; if $k\nmid2n$ then there are multiple solutions to $x^{\frac{k}{2n}}=-1$ in the complex numbers, and indeed $e^{i\frac{2n}{k}\pi}$ is one of them. Choosing this value makes these terms well-defined, so that's a good start. – Servaes Mar 3 at 10:12
• To your second remark; for every value of $n$ there are terms that are not defined; these are the terms for the divisors of $n$. – Servaes Mar 3 at 10:12

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.

• Assume that in the expression $$(-1)^\frac{2n}{k}$$ $$n=5$$ and $$k=3$$ Following your reasoning the expression would be $$(-1)^\frac{2n}{k}=(-1)^\frac{10}{3}=\sqrt{(-1)^{10}}=\sqrt{1}=1^{\frac{1}{3}}=1$$ But that’s not the case! In fact $$-1=e^{i\pi}$$ Hence $$(-1)^\frac{2n}{k}=(e^{i\pi})^\frac{2n}{k}=e^{i\frac{10}{3}\pi}$$ which is just the trivial complex number $$e^{i\frac{10}{3}\pi}=-\frac{1}{2}-i\frac{\sqrt{3}}{2}$$ en.wikipedia.org/wiki/… . I look forward to hear from you about it. – Joe Mar 2 at 23:48