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

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?


  • 3
    $\begingroup$ Write out the terms, list them out one by one. $\endgroup$ – IAmNoOne Mar 1 '15 at 2:22
  • 3
    $\begingroup$ 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

Write $n/(n+1)!=(n+1-1)/(n+1)!=1/n!-1/(n+1)!$

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.