Are these two series really equal to each other? If so, why?
Solution 1:
They are in fact equal.
$$\sum_{n=1}^{\infty} \frac{\ln(x)^n}{(2^n-1)n!} = \sum_{n=1}^{\infty} \frac{\left(\frac{\ln(x)}{2} \right)^n}{n!} \cdot \frac{1}{1-2^{-n}}= \sum_{n=1}^{\infty} \frac{\left(\frac{\ln(x)}{2} \right)^n}{n!} \cdot \sum_{k=1}^\infty(2^{-n})^{k-1} $$ $$=\sum_{k=1}^\infty \sum_{n=1}^{\infty} \frac{\left(\frac{\ln(x)}{2^{k}} \right)^n}{n!} = \sum_{k=1}^\infty (e^{\frac{\ln(x)}{2^{k}}}-1)= \sum_{k=1}^\infty (x^{\frac{1}{2^k}}-1).$$