Prove solution to $\sum\limits_{k=1}^{\infty}\frac{x^k}{x^{2k}-2x^k+1} = 0$.

Question

Numerically I found a solution to the following equation at about $x = -0.4112$.

$$\sum\limits_{k=1}^{\infty}\frac{x^k}{x^{2k}-2x^k+1} = 0 \quad x \in \mathbb{R}, -1 < x \leq 0$$

Now, I want to get the exact proven solution.

I know that I can reformulate the denominator as:

$$\sum\limits_{k=1}^{\infty}\frac{x^k}{(1-x^k)^2} = 0$$

But here I am stuck.

Where did this come from and why do you expect it has a closed-form? — Integrand, May 22 '20 at 18:25
Well, I didn't expect it, I kind of hoped for it because it looked like a nested geometric series. — thinkingeye, May 22 '20 at 18:37

Robert Israel · Accepted Answer · 2020-05-22T19:11:31.093

6

$$ \sum_{k=1}^\infty \frac{x^k}{(1-x^k)^2} = \sum_{k=1}^\infty\sum_{j=1}^\infty j x^{jk} = \sum_{m=1}^\infty \sum_{j \mid m} j x^m = \sum_{m=1}^\infty \sigma(m) x^m$$

where $\sigma(m)$ is the sum of the divisors of $m$. However, I don't think this has a "closed form" expression, nor do I think the root of this function has a closed form expression.

EDIT: Hmmm: it looks like the decimal expansion of the (negated) root is OEIS sequence A143441, which is "Decimal expansion of the (negated) value of q at which the q-Pochhammer symbol reaches a maximum along [ -1, 1]." I'm not sure of the relation between the q-Pochhammer symbol and your function, but evidently there is one.

edited May 22 '20 at 19:11

answered May 22 '20 at 18:26

Robert Israel

448,999

maybe you can explain more the second equality? – ILoveMath May 22 '20 at 18:34
1

@ILoveMath If the exponent is $m=jk$ then $j$ is a divisor of $m$. – saulspatz May 22 '20 at 18:47
Well, I hoped not to see the pseudo-random divisor function... – thinkingeye May 22 '20 at 19:08

Prove solution to $\sum\limits_{k=1}^{\infty}\frac{x^k}{x^{2k}-2x^k+1} = 0$.

1 Answers1