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I want to find out the answer to sum of reciprocals of multiplication of $n!$ and $2^n$.

$$\sum_{n=1}^{\infty} \frac{1}{n!(2^n)}.$$

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    your sum should be $$\sqrt{e}-1$$ – 2017-02-23
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    Can you please give me a hint how did you calculate it ? – 2017-02-23
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    should be ?well this is math – 2017-02-23

2 Answers 2

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It is, hopefully, a well known result that $$ \mathrm{e}^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}.$$ Your series can be rewritten as \begin{align} \sum_{n=1}^{\infty} \frac{1}{n! 2^n} &= \sum_{n=1}^{\infty} \frac{ \left(\frac{1}{2} \right)^n }{n!} \\ &= \left[ \frac{ \left(\frac{1}{2}\right)^0 }{0!} \color{red}{- \frac{ \left(\frac{1}{2}\right)^0 }{0!}}\right] + \sum_{n=1}^{\infty} \frac{ \left(\frac{1}{2} \right)^n }{n!} && (\text{add zero}) \\ &= \color{red}{-1} + \sum_{n=0}^{\infty} \frac{ \left(\frac{1}{2} \right)^n }{n!} \\ &= -1 + \mathrm{e}^{\frac{1}{2}} && (\text{use the well known result}) \\ &= \sqrt{\mathrm{e}} -1. \end{align}

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    Can't believe it took 9 months for someone to give a good answer. Poor OP! Nice answer though. – 2017-11-09
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    Oh, man, I didn't even notice how old the question was. Heh. – 2017-11-09
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Maclauren exspansion of sqrt e is sum(1/(n! * 2n) , 0 < n < infinity. This is an identity. You specified starting at one, the sum of terms lack one. hence, sqrt e -1 is sum(1/(n! * 2n) , 1 < n < infinity. ref. and google https://oeis.org/wiki/Sqrt(e) and Maclauren exspansion.