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} $.



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.

  • $\begingroup$ 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. $\endgroup$ – Joe Mar 3 at 1:53
  • $\begingroup$ 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. $\endgroup$ – Servaes Mar 3 at 10:12
  • $\begingroup$ To your second remark; for every value of $n$ there are terms that are not defined; these are the terms for the divisors of $n$. $\endgroup$ – Servaes Mar 3 at 10:12


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

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.

  • 1
    $\begingroup$ Assume that in the expression $$ (-1)^\frac{2n}{k} $$ $$ n=5 $$ and $$ k=3 $$ Following your reasoning the expression would be $$ (-1)^\frac{2n}{k}=(-1)^\frac{10}{3}=\sqrt[3]{(-1)^{10}}=\sqrt[3]{1}=1^{\frac{1}{3}}=1 $$ 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. $\endgroup$ – Joe Mar 2 at 23:48

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