# Show that $\sum_{n=0}^{\infty}\frac{(-1)^{n}(\log 2)^{n}}{n!}=\frac12$.

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!

• Hint: $$-\log{2}=\log{\frac{1}{2}}$$ Use this for your "first thought" of $e^z$... Jan 8, 2020 at 16:55
• @Eleven-Eleven OMG. you are right. I was being dumb. Thank you! Jan 8, 2020 at 16:57
• Its never about being dumb.... Sometimes the brightest days have a cloud here or there... Jan 8, 2020 at 17:01
• @Eleven-Eleven lol you are right. Jan 8, 2020 at 17:01
• @Eleven-Eleven thank you! Jan 8, 2020 at 17:02

The given series is just the Taylor series of the exponential function evaluated at $$-\log2=\log\frac12$$. Hence its value is $$e^{\log1/2}=\frac12$$.