Proof that $\sum\limits_{i=1}^\infty \frac{i}{(i+1)!} = 1$ [duplicate]

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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• Do you maybe mean $n = \infty$? – Dirk Mar 21 at 8:58
• Since $\frac{i}{(i+1)!} = \frac{(i+1)-i}{(i+1)!} = \frac{1}{i!} - \frac{1}{(i+1)!}$, your sum is a telescoping sum. – achille hui Mar 21 at 9:03
$$\sum\limits_{i=1}^{\infty} \frac i {(i+1!)}=\sum\limits_{i=1}^{\infty} \frac {i+1} {(i+1)!} -\sum\limits_{i=1}^{\infty} \frac 1 {(i+1)!}=\sum\limits_{i=1}^{\infty} \frac 1 {i!} -\sum\limits_{i=1}^{\infty} \frac 1 {(i+1)!}$$ If just write the terms you will see that all terms cancel out except $$1$$.