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$$\sum^\infty_{n=0} \frac{n}{(n+1)!}$$

I don't even know where to begin. I posted another question about a topic like this, but with the factorial thrown into the mix, it is dubious if the same methodology would work.

How do I evaluate this?

Thanks.

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    $\begingroup$ Write out the terms, list them out one by one. $\endgroup$ – IAmNoOne Mar 1 '15 at 2:22
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    $\begingroup$ Indeed telescoping. The $n$-th term is $\frac{n+1-1}{(n+1)!}$, which is $\frac{1}{n!}-\frac{1}{(n+1)!}$. $\endgroup$ – André Nicolas Mar 1 '15 at 2:23
  • $\begingroup$ How did you get to that last conclusion? EDIT: figured it out. Thanks for the hint! $\endgroup$ – louie mcconnell Mar 1 '15 at 2:24
  • $\begingroup$ Add the terms, modified to look like above, and observe the mass cancellations. Added: good! $\endgroup$ – André Nicolas Mar 1 '15 at 2:25
  • $\begingroup$ Why don't you post that as an answer, @AndréNicolas? $\endgroup$ – hjhjhj57 Mar 1 '15 at 3:46
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Write $n/(n+1)!=(n+1-1)/(n+1)!=1/n!-1/(n+1)!$

Then the sum from 0 to infinity is trivially $1$!

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