Establish a formula for the product (1-1/2)(1-1/3)...(1 - 1/n)

Claim:

$\left( 1 - \frac{1}{2} \right) \left(1 - \frac{1}{3} \right) \cdots \left( 1 - \frac{1}{n} \right) = \frac{1}{n}.$

Proof. If $n = 2$ on the left we have $(1-1/2) = 1/2$ and on the right we have $(1/n) = 1/2$ . Thus, the formula is true for the case $n=2$ .
Assume then that the formula is true for some $k = n \in \mathbb{Z}_{\geq 2}$ . So,

$\begin{align*} && \left(1 - \frac{1}{2}\right) \left(1 - \frac{1}{3} \right) \cdots \left(1 - \frac{1}{k} \right) &= \frac{1}{k} & (\text{Ind. Hyp.}) \\ \implies && \left( 1 - \frac{1}{2} \right) \left(1 - \frac{1}{3} \right) \cdots \left(1 - \frac{1}{k+1}\right) &= \left(\frac{1}{k} \right) \left(1- \frac{1}{k+1} \right) & (\text{Multiplying by } 1 - \frac{1}{k+1} )\\ &&&= \frac{1}{k} - \frac{1}{k+1} \\ &&& = \frac{k+1-1}{k(k+1)} \\ &&& = \frac{1}{k+1}. \end{align*}$

Thus, if the formula is true for $k$ then it is true for $k+1$ . Since we have established that it is true for $n =2$ , we have that is true for all $n \in \mathbb{Z}_{\geq 2}. \qquad \blacksquare$

Stumbling Robot

Establish a formula for the product (1-1/2)(1-1/3)…(1 – 1/n)

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