Prove the following equality:$$\left(1-\frac{1}{n+1}\right)\left(1-\frac{1}{n+2}\right)\cdot\ldots\cdot\left(1-\frac{1}{2n}\right)=\frac{1}{2}$$
$$\text{for all }\;n\in\mathbb{N}\;.$$
Could you tell me if my proof is correct?
Is there another way to prove it?
Does it exist a simpler proof?
We are going to prove by induction the following equality:
$\left(1-\dfrac{1}{n+1}\right)\left(1-\dfrac{1}{n+2} \right)\cdot\ldots\cdot\left(1-\dfrac{1}{2n}\right)=\dfrac12\;.\quad\color{blue}{(*)}$
For $\;n=1\;$ we get that
$\left(1-\dfrac{1}{n+1}\right)\left(1-\dfrac{1}{n+2} \right)\cdot\ldots\cdot\left(1-\dfrac{1}{2n}\right)=$
$=\left(1-\dfrac12\right)=\dfrac12$.
Now we suppose that $(*)$ is true and prove that
$\left(1-\dfrac{1}{n+2}\right)\left(1-\dfrac{1}{n+3} \right)\cdot\ldots\cdot\left(1-\dfrac{1}{2n+2}\right)=\dfrac12\;.$
It results that
$\left(1-\dfrac{1}{n+2}\right)\left(1-\dfrac{1}{n+3} \right)\cdot\ldots\cdot\left(1-\dfrac{1}{2n+2}\right)=$
$=\dfrac{\left(1-\dfrac{1}{n+1}\right)\left(1-\dfrac{1}{n+2} \right)\cdot\ldots\cdot\left(1-\dfrac{1}{2n+2}\right)}{\left(1-\dfrac{1}{n+1}\right)}=$
$=\dfrac{\dfrac12\left(1-\dfrac{1}{2n+1}\right)\left(1-\dfrac{1}{2n+2}\right)}{1-\dfrac{1}{n+1}}=$
$=\dfrac{\dfrac12\left(\dfrac{2n}{2n+1}\right)\left(\dfrac{2n+1}{2n+2}\right)}{\dfrac{n}{n+1}}=$
$=\dfrac12\cdot\dfrac{\color{red}{2}\color{blue}{n}}{\color{green}{2n+1}}\cdot\dfrac{\color{green}{2n+1}}{\color{red}{2}(\color{brown}{n+1})}\cdot\dfrac{\color{brown}{n+1}}{\color{blue}{n}}=$
$=\dfrac12\;.$
Hence, by induction, the equality $(*)$ is true for all $n\in\mathbb{N}\;.$
Best Answer
HINT
Notice that each denominator cancels the next numerator, whence we get
$\begin{align} &\left(1 - \frac{1}{n+1}\right)\left(1 - \frac{1}{n+2}\right)\cdot\ldots\cdot\left(1 - \frac{1}{2n}\right) =\\ &=\left(\frac{n}{n+1}\right)\left(\frac{n+1}{n+2}\right)\cdot\ldots\cdot\left(\frac{2n-1}{2n}\right) = \frac{n}{2n}=\frac{1}{2}\;. \end{align}$