It's well-known that \begin{array}{l}\frac1{1-x}=\sum_{n=0}^\infty x^n=\prod_{k=0}^\infty{\textstyle\sum_{i=0}^{m-1}}x^{im^k}.\\\end{array} By using the following "theorem about product and sum":
If $$f(x)=\prod_kf_k(x),$$then$$\frac{f'(x)}{f(x)}=\sum_k\frac{f_k'(x)}{f_k(x)}$$
we can get that \begin{array}{l}\forall m\in\mathbb{N}\backslash\{1\}\\\frac1{1-x}=\sum_{k=0}^\infty\frac{f_k'(x)}{f_k(x)},\\\text{where }f_k(x)={\textstyle\sum_{i=0}^{m-1}}x^{im^k}\\for\;x\in(0;1)\\\end{array} This identity interested me a lot. So, I was trying to find some other identities with similar construction. That intention motivated me to change the lower bound of summation from 0 to -infinity. That sum was too tough for me to express in terms of elementary functions, so I used a graphing calculator to find the function that is the closest function to this sum. After searching for roughly 2 hours I've finally found that function. It was -1/(x*ln(x)). Unfortunately, it's just a hypothesis, but who can stop me from writing it as a theorem? \begin{array}{l}\forall m\in\mathbb{N}\backslash\{1\}\\\frac{-1}{x\ln(x)}=\sum_{k=-\infty}^\infty\frac{f_k'(x)}{f_k(x)},\\where\;f_k(x)={\textstyle\sum_{i=0}^{m-1}}x^{im^k}\\for\;x\in(0;1)\\Note\;that\;\frac{\displaystyle-1}{\displaystyle x\ln(x)}=-\frac{{\displaystyle(}\ln{\displaystyle(}{\displaystyle x}{\displaystyle)}{\displaystyle)}{\displaystyle'}}{\ln{\displaystyle(}{\displaystyle x}{\displaystyle)}}=-(\ln\vert\ln(x)\vert)'\\\end{array} By using logarithmic rules, I rewrote the sum as a derivative of the product, but that didn't help me. That product diverges on the whole real axis. So, I was trying to find a normalization factor. Finally, using a graphing calculator, I've found it. Here is the final hypothesis I want to prove. Have you any ideas? \begin{array}{l}\forall m\in\mathbb{N}\backslash\{1\}:\\-\sqrt m\prod_{k=-\infty}^\infty\frac{\sum_{i=0}^{m-1}x^{im^k}}{\sqrt m}=\frac1{\ln\left(x\right)}\\for\;x\in(0;1).\\\end{array}
Let’s evaluate the corresponding finite product $\prod_{-N}^N$ first, and then take the limit.
First, $$\sum_{i=0}^{m-1}x^{im^k} = \frac{1-x^{m^{k+1}}}{1-x^{m^k}}$$
This is conveniently of the form $\frac{g(k+1)}{g(k)}$, and so the product of it will telescope: $$-\sqrt m \prod_{k=-N}^N \frac{1}{\sqrt m} \cdot \frac{1-x^{m^{k+1}}}{1-x^{m^k}} = -\sqrt m \cdot \frac{1}{(\sqrt m)^{2N+1}} \cdot \frac{1-x^{m^{N+1}}}{1-x^{m^{-N}}}=-\frac{1}{m^N} \frac{1-x^{m^{N+1}}}{1-x^{m^{-N}}} $$
Taking the limit as $N\to\infty$, the numerator just tends to $1$ and this is left: $$-\frac{1}{m^N(1-x^{m^{-N}})} $$ Let $m^N = T \to \infty$. The limit of the denominator is $$ T(1-x^{1/T} ) = \frac{1-x^{1/T}}{1/T} \to \frac{-x^{1/T} \ln x \cdot -1/T^2}{ -1/T^2}\to -\ln x $$ and so the original limit is $$ \frac{1}{\ln x} $$ as required.