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!

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

1 Answer 1


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

  • $\begingroup$ Yes you are definitely right.. I was being really dumb. Thank you so much!!! $\endgroup$ Jan 8, 2020 at 16:58

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