Question

Prove the following statements with either induction, strong induction or proof by smallest counterexample. If \(n \in \mathbb{N},\) then \(\frac{1}{2 !}+\frac{2}{3 !}+\frac{3}{4 !}+\cdots+\frac{n}{(n+1) !}=1-\frac{1}{(n+1) !}\)

Short answer

The statement \(\frac{1}{2 !}+\frac{2}{3 !}+\cdots+\frac{n}{(n+1) !}=1-\frac{1}{(n+1) !}\) holds for all \(n \in \mathbb{N}\) and is proven by mathematical induction.

Step-by-step solution

Test the base case

The base case for this proof by induction is when \(n = 1\). The left side of the equation becomes \(\frac{1}{2!} = 0.5\). The right hand side of the equation: \(1-\frac{1} {2!} = 0.5\). Since both sides are equal, the statement holds for the base case.

Assume the induction hypothesis

Assume the statement is true for some arbitrary natural number \(k\), meaning assume: \(\frac{1}{2!} + \frac{2}{3!} + \cdots + \frac{k} {(k + 1)!} = 1 - \frac{1}{(k + 1)!}\). This is called the assumption or the induction hypothesis.

Prove the induction step

Next step is to prove that the statement holds for \(k + 1\). Start by adding \(\frac{k + 1} {((k + 1) + 1)!}\) to both sides of the induction hypothesis. The left hand side becomes \(\frac{1}{2!} + \frac{2}{3!} + \cdots + \frac{k} {(k + 1)!} + \frac{k + 1} {(k + 2)!}\) and the right hand side becomes \(1 - \frac{1}{(k + 1)!}+ \frac{k + 1} {(k + 2)!}\). By simplifying the right hand side, it becomes \(1- \frac{1}{(k+2)!}\) and the left hand side will also be \(1- \frac{1}{(k+2)!}\), so the statement holds for n = \(k + 1\).

Conclude the induction

Since the statement is correct for \(n=1\) (base case), and if it is true for \(n = k\), then it is also true for \(n = k + 1\), we can conclude through the principle of mathematical induction that the statement is true for all \(n \in \mathbb{N}\).