How can i put a log inside an infinite sum?

Question

iI would like to turn this

$$ e^z-1 = \sum^\infty_1 \frac {z^n}{n!} $$

into this

$$ \text{something} =\sum^\infty_1 \frac {z^n}{\log_e(n!)} $$

Is this at all possible? thank you very much in advance!

Ps- i apologize of title not adequate , i did not know how to word it.

Answer 1

This is exactly same as your first sum as you said anything would work

$$something =\sum^\infty_1 \frac {z^n}{\log e^{n!}}$$

Answer 2

I'm not even sure what you're asking, but $\displaystyle \sum_{2}^{\infty}\frac{z^n}{\ln(n!)}>\sum_{2}^{\infty}\frac{z^n}{\ln(n^n)}=\sum_{2}^{\infty}\frac{z^n}{n\ln(n)}>\sum_{2}^{\infty}\frac{z^n}{n^2}$, the last of which does not converge (ratio between consecutive terms > $1$).

Therefore, $\displaystyle \sum_{2}^{\infty}\frac{z^n}{\ln(n!)}$ does not exist, but can a series have a log in the denominator? sure.