How to calculate the series?

Question

How can we calculate the series: $$ F(x)=\sum_{n=1}^{\infty}\frac{(-1)^n}{1-x^n} $$ I found that $$ F(x)=\sum_{n=1}^{\infty}(-1)^n\Big(\sum_{m=0}^{\infty}(x^n)^m\Big)=\sum_{m=0}^{\infty}\Big(1-\frac{1}{1+x^m}\Big). $$ Any suggestion?

EDIT: I once set if $\vert x\vert<1$, but the series doesn't converge at all. So In what occasion does the series converge?

Answer 1

For every $|x|\gt1$, the series converges since $|x^n-1|\geqslant\frac12|x|^n$ for every $n$ large enough, and $$F(x)=\sum_{n\geqslant1}\frac{(-1)^{n+1}}{x^n-1}=\sum_{n\geqslant1}(-1)^{n+1}x^{-n}\frac1{1-x^{-n}}=\sum_{n\geqslant1}(-1)^{n+1}x^{-n}\sum_{k\geqslant0}x^{-nk},$$ that is, $$F(x)=\sum_{n\geqslant1}\alpha_nx^{-n},\qquad\alpha_n=\sum_{d\mid n}(-1)^{d+1}.$$ Note that $$\alpha_n=\sum_{d\mid n}1-2\sum_{d\mid n,\ d\ \mathrm{even}}1=\sigma_0(n)-2\sigma_0^{(e)}(n)=\sigma_0(n)-2\sigma_0(n/2)\mathbf 1_{n\ \text{even}},$$ see here, where $\sigma_0(n)$ is the number of divisors of $n$. The generating function of the sequence $(\sigma_0(n))$ is known, hence all this yields, for every $|z|\lt1$, $z\ne0$, $$F(z^{-1})=\frac{\psi_z(1)+\log(1-z)}{\log z}-2\frac{\psi_{z^2}(1)+\log(1-z^2)}{\log(z^2)},$$ and, finally, $$F(z^{-1})=\frac{\psi_z(1)-\psi_{z^2}(1)-\log(1+z)}{\log z}.$$ As usual, this formula, based on the so-called q-polygamma function (see also (5)-(6)-(7) there), is, as given, a mere rewriting of the series one started with.

Answer 2

The answer between the two horizontal lines was to the original question, which specified $|x| <1$.

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This doesn't converge. If the sum is

$$ \sum_{n=1}^{\infty}a_n $$

with

$$a_n = \frac{(-1)^n}{1-x^n} $$

then, since $|x|<1$ we have $$\lim_{n\to\infty} x^n = 0 \qquad\Rightarrow\qquad \lim_{n\to\infty} 1-x^n = 1$$

This gives

$$\lim_{n\to\infty}a_n \,\,= \,\,\lim_{n\to\infty} \frac{(-1)^n}{1-x^n}\,\,\, =\,\,\, \lim_{n\to\infty} (-1)^n$$

Clearly, this limit doesn't exist. The terms don't go to zero, so the sum cannot possibly converge.

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In response to your edited question, the condition on $x$ for which the series converges is exactly $|x|>1$.

When $x>1$ we get an alternating series in which the terms $a_n$ go to zero, so it must converge.

If $x<-1$, we get a non-alternating series in which the terms go to zero as well. This converges by comparison: $$\sum\limits_{n=1}^{\infty} \frac{1}{(-x)^n-1} < 1 + \sum\limits_{n=1}^{\infty} \frac{1}{(\sqrt{-x})^n}$$

The right hand side above converges because it's a geometric series with a common ratio that is greater than zero and less than one.