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I came across this result randomly and am quite sure it's right. Is there any way to prove it rigorously? The numerator always seems to be one less than the denominator. Thanks!

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marked as duplicate by Theo Bendit, RRL, Sil, Eevee Trainer, José Carlos Santos sequences-and-series Mar 25 at 0:06

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  • $\begingroup$ This is so obviously false… $\endgroup$ – José Carlos Santos Mar 21 at 8:57
  • $\begingroup$ Are you sure you wrote the exact equation you wanted to write down? $\endgroup$ – Jakobian Mar 21 at 8:58
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    $\begingroup$ Do you maybe mean $n = \infty$? $\endgroup$ – Dirk Mar 21 at 8:58
  • $\begingroup$ Indeed n = infinity $\endgroup$ – 3684 Mar 21 at 9:00
  • $\begingroup$ Since $\frac{i}{(i+1)!} = \frac{(i+1)-i}{(i+1)!} = \frac{1}{i!} - \frac{1}{(i+1)!}$, your sum is a telescoping sum. $\endgroup$ – achille hui Mar 21 at 9:03
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$\sum\limits_{i=1}^{\infty} \frac i {(i+1!)}=\sum\limits_{i=1}^{\infty} \frac {i+1} {(i+1)!} -\sum\limits_{i=1}^{\infty} \frac 1 {(i+1)!}=\sum\limits_{i=1}^{\infty} \frac 1 {i!} -\sum\limits_{i=1}^{\infty} \frac 1 {(i+1)!}$ If just write the terms you will see that all terms cancel out except $1$.

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