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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!

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  • $\begingroup$ It's a rational number. Where did this come from? $\endgroup$ – Matt Samuel Mar 14 '18 at 2:25
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    $\begingroup$ Hint: the Taylor series for the hyperbolic cosine function is $\cosh(x) =\sum_{n=0}^\infty \frac{x^{2n}}{(2n)!}$. $\endgroup$ – Nick Mar 14 '18 at 2:28
  • $\begingroup$ Further hint: $\cosh(x) = \frac{1}{2}(e^x + e^{-x})$. $\endgroup$ – Cameron Williams Mar 14 '18 at 2:45
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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.

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

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  • $\begingroup$ Suggested topics for reading: Euler's Formula $(e^{ix}=\cos x +i\sin x$)......Hyper-trigonometric functions. ($\sinh, \cosh, \tanh$ etc.) $\endgroup$ – DanielWainfleet Mar 14 '18 at 7:39
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$s=\frac{1}{2}∑ _{m ∈ A} \frac{(\log2)^m}{m !}$

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:

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

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  • $\begingroup$ $e^{\log 2}=2$................ $\endgroup$ – DanielWainfleet Mar 14 '18 at 7:21
  • $\begingroup$ @DanielWainfleet,thank you for reminding me. I edited my answer , I think symbol 'Ln' is better for natural logarithm, when you use 'log' you have to mention the base too, in case you do not mention it would mean the base is 10, this is what I was told 45 years ago, may be some definitions are changed and I am not aware of. $\endgroup$ – sirous Mar 14 '18 at 12:29
  • $\begingroup$ It is now common usage that the "default" meaning of $\log$ is the natural logarithm......BTW $\frac {1}{4}(2^{5/2}+2^{1/2})2^{-3/2})$ $ =\frac {1}{4}(2^{5/2-3/2}+2^{1/2-3/2})=$ $\frac {1}{4}(2+1/2)=$ $\frac {5}{8}.$ $\endgroup$ – DanielWainfleet Mar 14 '18 at 17:36

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