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.
 A: $$1-\frac{1}{(k+1)!}+\frac{k+1}{(k+2)!}$$
Note that to simplify this we need a common denominator. Let it be $(k+2)!$. Recall that $(k+1)! = (k+1)(k)(k-1)(k-2) \cdots$ So to get a $(k+2)!$ in the denominator of the fraction we must multiply the numerator and denominator by $k+2$ and get:
\begin{align*}
1-\frac{k+2}{(k+2)!} + \frac{k+1}{(k+2)!} &=1-(\frac{k+2}{(k+2)!} - \frac{k+1}{(k+2)!}) \\
&=1-(\frac{k+2-k-1}{(k+2)!}) \\
&= 1-\frac{1}{(k+2)!} \\
\end{align*}
A: HINT: we have to show that $$\frac{(k+1)}{(k+2)!}-\frac{1}{(k+1)!}=-\frac{1}{(k+2)!}$$ this is true since $$\frac{(k+1)}{(k+1)!(k+2)}-\frac{1}{(k+1)!}=\frac{1}{(k+1)!}\left(\frac{k+1}{k+2}-1\right)=\frac{1}{(k+1)!}\left(\frac{k+1-k-2}{k+2)}\right)=-\frac{1}{(k+1)!(k+2)}=-\frac{1}{(k+2)!}$$
A: For $n=1,2,3$ we have:
\begin{eqnarray}
\frac{1}{2!}&=&\frac12=1-\frac12=1-\frac{1}{2!}\\
\frac{1}{2!}+\frac{2}{3!}&=&1-\frac{1}{2!}+\frac13=1-\frac16=1-\frac{1}{3!}\\
\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}&=&1-\frac16+\frac{1}{8}=1-\frac{1}{24}=1-\frac{1}{4!}.
\end{eqnarray}
If we assume that up to $n=3$, we have
$$
\sum_{k=1}^n\frac{k}{(k+1)!}=1-\frac{1}{(n+1)!}.
$$
Then
\begin{eqnarray}
\sum_{k=1}^{n+1}\frac{k}{(k+1)!}&=&\sum_{k=1}^n\frac{k}{(k+1)!}+\frac{n+1}{(n+2)!}=1-\frac{1}{(n+1)!}+\frac{n+1}{(n+2)!}\\
&=&1-\frac{n+2-(n+1)}{(n+2)!}=1-\frac{1}{(n+2)!}.
\end{eqnarray}
