What does this series converges to, if it does?

Question

The series is $$ \frac12 \sum_{m \in A} \frac{(\log 2)^m}{m!}. $$ The subset $A$ is $\{0,2,4,6,\cdots,\infty\}$. According to Wikipedia, if the subset $A$ contains all natural numbers, then this series would be $e^x$ but it did not help me much. Can anyone give me a hand? Thank you!

Answer 1

As a hint, consider the series

$$e^x=\sum_{m=0}^\infty \frac{x^m}{m!}$$

This is almost your series, but we need to delete the odd terms. Notice that the corresponding series for $e^{-x}$ flips the sign of the odd terms. If we expand the sum $e^x+e^{-x}$, the even terms will be doubled and the odd terms will cancel.

Answer 2

$s=\frac{1}{2}∑ _{m ∈ A} \frac{(\log2)^m}{m !}$

We use following relation*:

$(e^{5x}+e^x)e^{-3x}=2[1+\frac{(2x)^2}{2 !}+\frac{(2x)^4}{4 !}+\frac{(2x)^6}{6 !}+ . . .+\frac{(2x)^m}{m!}+. . .]$

Where $m ∈ 2k; k ∈ N$

So plugging $(2x=\log2 ⇒ x=\frac{1}{2}\log 2)$ in above relation gives:

$s=\frac{1}{2}∑ _{m ∈ A} \frac{(\log2)^m}{m !}=\frac{1}{4}(e^{(5/2)\log 2}+ e^{(1/2)\log 2}) e^{(- 3/2)\log2}=\frac{1}{4}(2^{5/2}+2^{1/2})2^{-3/2}$

Answer 3

$$e^x+e^{-x}=\lim_{N\to \infty}\sum_{j=0}^{2N}x^j/j! + \lim_{N\to\infty}\sum_{j=0}^{2N}(-x)^j/j!=$$ $$=\lim_{n=0}^{N}2x^{2n}/(2n)! =\sum_{n\in A}2x^{2n}/(2n)!$$ because the terms of odd degree in $x$ all subtract out. When $x=\log 2$ this gives $$5/2=2+2^{-1}=e^{\log 2}+e^{-\log 2}=$$ $$=2\sum_{n\in A}(\log 2)^n/n!.$$