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

Question

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!

Answer 1

$\sum\limits_{i=1}^{\infty} \frac i {(i+1!)}=\sum\limits_{i=1}^{\infty} \frac {i+1} {(i+1)!} -\sum\limits_{i=1}^{\infty} \frac 1 {(i+1)!}=\sum\limits_{i=1}^{\infty} \frac 1 {i!} -\sum\limits_{i=1}^{\infty} \frac 1 {(i+1)!}$ If just write the terms you will see that all terms cancel out except $1$.