Mathematical Induction Regarding Factorials

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

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 the process of mathematical induction the statement is true for all integers \( n \ge 1 \).


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