I am working on an alternating series stated below:
$$\sum_{n=0}^{\infty}\dfrac{(-1)^{n}(\log 2)^{n}}{n!}$$
The Wolfram-Alpha shows that this series converges and converges to $\dfrac{1}{2}$ without showing how it gets this limit.
The thought I have is trying to come up a Taylor expansion of some known function, but the closet I can get is $$e^{z}=\sum_{n=0}^{\infty}\dfrac{z^{n}}{n!},$$ then replace $z:=(\log 2)$.
However, this sum I found is not alternating.
Another closed one is binomial expansion with the power $\alpha=\dfrac{1}{2}$, that is $$(1+z)^{1/2}=1+\dfrac{1}{2}z-\dfrac{1}{8}z^{2}-\cdots,$$ however the sign is alternating in a wrong way...replacing $z:=-z$ cannot solve this problem since $(-z)^{2}=z^{2}$, so the sign is still wrong.
Is there any other alternatives?
Thank you!