Mathematical Induction Regarding Factorials

Prove by mathematical induction that for all integers $n \ge 1$ ,
$$\dfrac{1}{2!} + \dfrac{2}{3!} + \dfrac{3}{4!} + \cdots + \dfrac{n}{(n+1)!} = 1-\dfrac{1}{(n+1)!}$$

Step 1

Show it is true for $n=1$.
\begin{align} \displaystyle \text{LHS } &= \dfrac{1}{2!} = \dfrac{1}{2} \\ \text{RHS } &= 1-\dfrac{1}{2!} \\ &= 1-\dfrac{1}{2} \\ &= \dfrac{1}{2} \end{align}
Thus, the statement is true for $n=1$.

Step 2

Assume the statement is true for $n=k$, that is;
$\dfrac{1}{2!} + \dfrac{2}{3!} + \dfrac{3}{4!} + \cdots + \dfrac{k}{(k+1)!} = 1-\dfrac{1}{(k+1)!}$

Step 3

Show the statement is true for $n=k+1$, that is;
$\dfrac{1}{2!} + \dfrac{2}{3!} + \dfrac{3}{4!} + \cdots + \dfrac{k}{(k+1)!} + \dfrac{k+1}{(k+2)!} = 1-\dfrac{1}{(k+2)!}$
\begin{align} \displaystyle \text{LHS } &= \dfrac{1}{2!} + \dfrac{2}{3!} + \dfrac{3}{4!} + \cdots + \dfrac{k}{(k+1)!} + \dfrac{k+1}{(k+2)!} \\ &= 1-\dfrac{1}{(k+1)!} + \dfrac{k+1}{(k+2)!} \\ &= 1-\dfrac{k+2}{(k+2)(k+1)!} + \dfrac{k+1}{(k+2)!} \\ &= 1-\dfrac{k+2}{(k+2)!} + \dfrac{k+1}{(k+2)!} \\ &= 1-\bigg[\dfrac{k+2}{(k+2)!}-\dfrac{k+1}{(k+2)!}\bigg] \\ &= 1-\dfrac{1}{(k+2)!} \\ &= \text{RHS} \end{align}
Therefore, by mathematical induction, the statement is true for all integers $n \ge 1$.