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I'm encountering some difficulty on a question for finding the sum of the series $$\sum_{n=0}^\infty \dfrac{n}{(n+1)!}$$

The method I use to tackle this type of problem is generally to find a similar sum of a power series and algebraically manipulate it to match that of the original. I haven't found anything similar except for the summation of $e^x$ starting from $n=-1$, and subbing in $n^{\frac{1}{n}}$. Though, I'm not sure that will even work.

Thanks in advance for any help/advice!

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    $\begingroup$ Since $\frac{n}{(n+1)!} = \frac{1}{n!} - \frac{1}{(n+1)!}$, this is a telescoping series! $\endgroup$ Nov 20, 2013 at 5:55
  • $\begingroup$ See Dobinski's formula for Bell numbers : $$B_k=\frac1e\cdot\sum_{n=0}^\infty\frac{n^k}{n!}\in\mathbb{N}\quad\forall\ k\in\mathbb{N}$$ $\endgroup$
    – Lucian
    Nov 20, 2013 at 6:50
  • $\begingroup$ @achillehui, you should convert it to an answer. $\endgroup$ May 23, 2014 at 9:47

2 Answers 2

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hint:

$$ \sum_{n=0}^\infty \dfrac{n}{(n+1)!}= \sum_{n=1}^\infty \dfrac{n-1}{(n)!}$$

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We have $$\sum_{n=0}^\infty \dfrac{n}{(n+1)!} = \sum_{n=0}^\infty \left (\frac {1} {n!} - \frac {1} {(n + 1)!} \right) = \frac {1} {0!} - \frac {1} {\infty!} = 1 - 0 = 1.$$

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