<s>[INST] Prove that $ \frac{1}{2!} + \frac{2}{3!} + \frac{3}{4!}+\cdots + \frac{n}{(n+1)!} = 1 - \frac{1}{(n+1)!}$ for $n\in \mathbb N$ I want to prove that if $n \in \mathbb N$ then $$\frac{1}{2!} + \frac{2}{3!} + \frac{3}{4!}+ \cdots+ \frac{n}{(n+1)!} = 1 - \frac{1}{(n+1)!}.$$ 
I think I am stuck on two fronts. First, I don't know how to express the leading terms on the left hand side before the $\dfrac{n}{(n+1)!}$ (or if doing so is even necessary to solve the problem). I am also assuming that the right high side should initially be expressed $1 - \dfrac{1}{(n+2)!}$. But where to go from there. 
I'm actually not sure if I'm even thinking about it the right way. 
 [/INST] Hint.  If, for somespecific $n$, we have
$$\frac{1}{2!} + \frac{2}{3!} + \frac{3}{4!}+ \cdots+ \frac{n}{(n+1)!} = 1 - \frac{1}{(n+1)!}\ ,$$
then
$$\eqalign{\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\cdots+\frac{n+1}{(n+2)!}
  &=1-\frac{1}{(n+1)!}+\frac{n+1}{(n+2)!}\cr
  &=1-\frac{n+2}{(n+2)!}+\frac{n+1}{(n+2)!}\cr
  &=1-\frac{1}{(n+2)!}\ .\cr}$$
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