I was playing around with factorials on Wolfram|Alpha, when I got this amazing result :

$$\sum \limits_{n=0}^{\infty} \dfrac{n!}{(n+1)!+(n+2)!} = \dfrac{3}{4}.$$

Evaluating the first few partial sums makes it obvious that the sum converges to $\approx 0.7$. But I am not able to prove this result algebraically. I tried manipulating the terms and introducing Gamma Function, but without success.

Can anyone help me with this infinite sum ? Is there some well-known method of evaluating infinite sums similar to this ?

Any help will be gratefully acknowledged.

Thanks in advance ! :-)

EDIT : I realized that $(n!)$ can be cancelled out from the fraction and the limit of the remaining fraction as $n \to \infty$ can be calculated very easily to be equal to $0.75$. Very silly of me to ask such a question !!! Anyways you can check out the comment by @Did if this "Edit" section does not help.

  • 3
    $\begingroup$ I definitely haven't tried, but I wonder if you can factor $(n+1)!$ out of the bottom and write it as a pair of fractions so that the sum telescopes. $\endgroup$ – pjs36 Feb 17 '17 at 12:56
  • 15
    $\begingroup$ Perhaps not quite so amazing... $$\frac{n!}{(n+1)!+(n+2)!}=\frac1{(n+1)(n+3)}=\frac12\left(\frac1{n+1}-\frac1{n+3}\right)$$ hence the $n$th sum *telescopes* into $$\sum_{k=0}^n\frac{k!}{(k+1)!+(k+2)!}=\frac12\left(\frac11+\frac12-\frac1{n+2}-\frac1{n+3}\right)$$ which converges to $$\frac34$$ $\endgroup$ – Did Feb 17 '17 at 12:58
  • 1
    $\begingroup$ @Did Thanks for the useful comment ... but it is still amazing for me :) $\endgroup$ – user399078 Feb 17 '17 at 12:59
  • 1
    $\begingroup$ Make that an answer. $\endgroup$ – marty cohen Feb 17 '17 at 15:29
  • $\begingroup$ Some would upvote a comment and wont for an answer. $\endgroup$ – Zaid Alyafeai Feb 18 '17 at 1:58

Thanks to pjs36 and Did,

Notice that:


Thus, we get a telescoping series, leaving us with:



Your Answer

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