What is the value of- $$\frac{1}{3!}+\frac2{5!}+\frac3{7!}+\frac{4}{9!}+\cdots$$ I wrote it as general term $\sum\frac{n}{(2n+1)!}$. As the series converges it should be telescopic (my thought). But i dont know how to proceed. I also know $\sum\frac{1}{n!}=e$ Any help /hints appreciated. Thanks!

  • $\begingroup$ There are many converging series that are not telescoping, I guess yours is one. $\endgroup$ – Henrik supports the community Jul 22 '16 at 17:06
  • $\begingroup$ My logic is like we will get something like $1/k!-1/(k+1)!$ $\endgroup$ – Archis Welankar Jul 22 '16 at 17:11
  • $\begingroup$ I always find it helpful to put a sequence into WolframAlpha and see whether the answer is something that looks possible to reach by hand. $\endgroup$ – Arthur Jul 22 '16 at 17:12

One may write $$ \begin{align} \sum_{n=0}^\infty\frac{n}{(2n+1)!}&=\frac12\sum_{n=0}^\infty\frac{(2n+1)-1}{(2n+1)!} \\\\&=\frac12\sum_{n=0}^\infty\frac{1}{(2n)!}-\frac12\sum_{n=0}^\infty\frac{1}{(2n+1)!} \\\\&=\frac12\left(\frac{e+e^{-1}}2 \right)-\frac12\left(\frac{e-e^{-1}}2 \right) \\\\&=\frac12\cdot e^{-1} \end{align} $$

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    $\begingroup$ Much more than just a hint. $\endgroup$ – Simply Beautiful Art Jul 22 '16 at 17:15
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    $\begingroup$ To me a complete answer would justify all rearrangements of terms and explain the penultimate step :) $\endgroup$ – Olivier Oloa Jul 22 '16 at 17:20
  • $\begingroup$ I think its quite understandable though. But it doesn't hurt to explain more. $\endgroup$ – Simply Beautiful Art Jul 22 '16 at 17:23
  • $\begingroup$ Can you tell me how $1/(2n)!=1/2(e+e^{-1})$ and similarly for other one $\endgroup$ – Archis Welankar Jul 22 '16 at 17:26
  • $\begingroup$ One has $e+e^{-1}=\sum\frac{1+(-1)^p}{p!}=\sum\frac{2}{(2n)!}=2\sum\frac1{(2n)!}$ since all terms $1+(-1)^p$ with an odd integer $p$ cancel. Are you Ok with this? Only terms with an even $p$ remain. Thanks. $\endgroup$ – Olivier Oloa Jul 22 '16 at 17:30

We know that $$\frac{\sinh x}{x}=\sum_{n=0}^{\infty }\frac{x^{2n}}{(2n+1)!}$$

let $x\rightarrow \sqrt{x}$ $$\frac{\sinh \sqrt{x}}{\sqrt{x}}=\sum_{n=0}^\infty \frac{x^n}{(2n+1)!}$$ $$\left(\frac{\sinh \sqrt{x}}{\sqrt{x}}\right)'=\sum_{n=1}^\infty \frac{nx^{n-1}} {(2n+1)!}$$ let $x=1$ to get what do you want

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    $\begingroup$ I have qualms about the appropriateness of the word "hint". $\qquad$ $\endgroup$ – Michael Hardy Jul 22 '16 at 18:04
  • $\begingroup$ @MichaelHardy I have edited the answer. Thanks $\endgroup$ – E.H.E Jul 22 '16 at 18:07
  • $\begingroup$ Thanks . Your answer is nice but since I am not familiar with hyperbolic functions (my bad) soI accepted the other answer still (+1):) $\endgroup$ – Archis Welankar Jul 23 '16 at 5:03
  • $\begingroup$ @ArchisWelankar you are welcome and thanks for your nice question $\endgroup$ – E.H.E Jul 23 '16 at 5:08

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