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$$x + \frac{x^3}{1\cdot 3} + \frac{x^5}{1\cdot 3\cdot 5}+...$$ I was wondering how should I move ahead to try to figure out the sum of this series. I will appreciate any hints.

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    What number is 1.3.5? – 2017-01-11
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    @PrinceM Likely the double factorial $$(5)!!=5\cdot 3\cdot 1$$ is what is intended. – 2017-01-11
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    Some places use a lowered dot $.$ instead of the centered dot $\cdot$ for multiplication. – 2017-01-11
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    Hint (assuming $a.b$ is $a \times b$). Differentiate the sum and see if you can manipulate the result algebraically to get the original power series back. – 2017-01-11
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    Is the series $\sum_{j=0}^\infty \frac{x^{2j+1}}{(2j+1)!}j!\cdot 2^j$? – 2017-01-11
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    @canseeker, don't forget that when you are satisfied with an answer you are encouraged to [accept it](http://math.stackexchange.com/help/someone-answers). – 2017-01-11
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    This is very famous. See http://math.stackexchange.com/q/833920/72031 – 2017-02-06

2 Answers 2

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Look at the series expansions for the Error function

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    No, those have complete factorials in the denominators. – 2017-01-11
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    You are right. Errorfunctions a better hint? – 2017-01-11
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    Ok, so now this is where I say "Please see identity $(9-10)$ in the following link" – 2017-01-11
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    @SimpleArt Please don't condone such an answer, nor attempt to translate and edit it into a form the answerer probably doesn't understand (but has googled) and in doing so, essentially endorsing such actions. This is, and remains, a link only answer. If you feel you must intervene for the sake of "Truth" (with a capital "T), then vote for the correct answer, not one that even the asker could have written. – 2017-01-14
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Note that $f'(x)=1+x·f(x)$ with $f(0)=0$, so that one gets the series expression as solution to this initial value problem.

$$ \frac{d}{dx}(e^{-x^2/2}f(x))=e^{-x^2/2}\\~\\ e^{-x^2/2}f(x)-0=\int_0^xe^{-s^2/2}\,ds\\~\\ f(x)=\int_0^xe^{(x^2-s^2)/2}\,ds $$

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    Wonderful argument. Note that this can also be expressed in the form $\frac {\sqrt{2 \pi}}{2} e^{x^2/2}erf(\frac{x}{\sqrt{2}})$ – 2017-01-11