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I have to solve this sum: $$\sum_{n=1}^{\infty} \frac{n 3^n}{ (n+3)!}$$ but I really don't have any idea.According to wolfram-alpha I should get $\frac{1}{2}$. Any help will be greatly appreciated.

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Hints:

$$\sum_{n=1}^\infty\frac{n3^n}{(n+3)!}=\frac1{27}\left(\sum_{n=1}^\infty\frac{(n+3)3^{n+3}}{(n+3)!}-3\sum_{n=1}^\infty\frac{3^{n+3}}{(n+3)!}\right)$$

$$e^x=\sum_{n=0}^\infty\frac{x^n}{n!}\implies e^x=(e^x)'=\sum_{n=1}^\infty\frac{nx^{n-1}}{n!}\;\;,\;\;\;\forall\,x\in\Bbb R\;\text{(in fact}\,,\;\forall\,x\in\Bbb C)$$

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    $\begingroup$ Good catch, @user88595. Thanks. $\endgroup$ – DonAntonio Apr 27 '14 at 12:33
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    $\begingroup$ I managed to finish from here. Thanks for the tips! $\endgroup$ – user137209 Apr 27 '14 at 12:36
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Write as $$a_n=\frac{n3^n}{(n+3)!}=\frac{(n+3-3)3^n}{(n+3)!}=\frac{3^n}{(n+2)!}-\frac{3^{n+1}}{(n+3)!}=t_n-t_{n+1}$$ a telescoping series.

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  • $\begingroup$ Shouldn't one of the (n+2)s be (n+3)? $\endgroup$ – steven gregory Apr 17 '16 at 22:52
  • $\begingroup$ Yes. Thanks. I fixed it. $\endgroup$ – student forever Apr 17 '16 at 22:54

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