Write the given product in closed form using the factorial operator:

Question

$$\prod_{k=1}^n \frac{2k}{2k-1}$$

I asked this earlier but am an idiot and wrote out the wrong question! Am still pretty confused on how to solve. Thanks for any tips/answers!!

Answer 1

$\begin{array}\\ \prod_{k=1}^n \frac{2k}{2k-1} &=\prod_{k=1}^n \frac{(2k)^2}{(2k)(2k-1)}\\ &= \dfrac{\prod_{k=1}^n(2k)^2}{\prod_{k=1}^n(2k)(2k-1)}\\ &= \dfrac{(\prod_{k=1}^n(2k))^2}{(2n)!}\\ &= \dfrac{(2^nn!)^2}{(2n)!}\\ \end{array} $

Answer 2

Look at what happens when $n=5$.

$$\prod_{k=1}^5\frac{2k}{2k-1}=\frac21\cdot\frac43\cdot\frac65\cdot\frac87\cdot\frac{10}9=\frac{2\cdot4\cdot6\cdot8\cdot10}{1\cdot3\cdot5\cdot7\cdot9}\;.$$

If we multiply the denominator by the numerator, we get $10!$:

$$\prod_{k=1}^5\frac{2k}{2k-1}=\frac{(2\cdot4\cdot6\cdot8\cdot10)^2}{10!}\;.$$

The problem now is to express that numerator using factorials. Notice that

$$2\cdot4\cdot6\cdot8\cdot10=(2\cdot1)(2\cdot2)(2\cdot3)(2\cdot4)(2\cdot5)=2^5\cdot5!\;,$$

so

$$\prod_{k=1}^5\frac{2k}{2k-1}=\frac{(2^5\cdot5!)^2}{10!}=\frac{2^{10}5!^2}{10!}\;.$$

I’ll leave it to you to check that this generalizes easily to arbitrary $n$.