Induction Proof using factorials Recall that for $n \in N$, $n! = 1 \cdot 2 \cdots n$.
Prove the following for each $n \in N$:
$$\frac{1}{2!} + \frac{2}{3!} + \frac{3}{4!} + \cdots + \frac{n}{(n+1)!} = 1 - \frac{1}{(n+1)!}$$
I understand how to do the proof, but in the inductive step I am facing some difficulty proving the left-hand side is equivalent to the right-hand side.To be direct I am facing some difficulty with the algebra required to make LHS = RHS.
Here is what I have done so far:
1) Base Case
$n = 1$
LHS:
$1/2$
and RHS is $1/2$ $\checkmark$
2) Inductive Step
For $k \geq 1$, Assume $n = k$
$$\frac{1}{2!} + \frac{2}{3!} + \frac{3}{4!} + \cdots + \frac{k}{(k+1)!} = 1 - \frac{1}{(k+1)!}$$
$$n = k + 1$$
$$\frac{1}{2!} + \frac{2}{3!} + \frac{3}{4!} + \cdots + \frac{n}{(n+1)!} + \frac{k+1}{(k+2)!} = 1 - \frac{1}{(n+1)!}$$
$$\implies 1-\frac{1}{(k+1)!} + \frac{k+1}{(k+2)!} = 1 - \frac{1}{(k+2)!}$$
Here is  where i do not know how to make the LHS = RHS.

HINT: we have to show that $$\frac{(k+1)}{(k+2)!}-\frac{1}{(k+1)!}=-\frac{1}{(k+2)!}$$ this is true since $$\frac{(k+1)}{(k+1)!(k+2)}-\frac{1}{(k+1)!}=\frac{1}{(k+1)!}\left(\frac{k+1}{k+2}-1\right)=\frac{1}{(k+1)!}\left(\frac{k+1-k-2}{k+2)}\right)=-\frac{1}{(k+1)!(k+2)}=-\frac{1}{(k+2)!}$$
