Help me Prove $\prod_{k=1}^{2n}k=4^{n}\sec\left(\pi n\right)\left(\prod_{k=0}^{n-1}\left(-\frac{1}{2}-k\right)\right)\left(\prod_{k=1}^{n}k\right)$

Question

$$\prod_{k=1}^{2n}k=4^{n}\sec\left(\pi n\right)\left(\frac{1}{2n+1}\right)\left(\prod_{k=1}^{n}\left(-\frac{1}{2}-k\right)\right)\left(\prod_{k=1}^{n}k\right)$$

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Answer 1

Note that $\sec(\pi n) = \begin{cases}1 & n\text{ is even} \\ -1 & n\text{ is odd}\end{cases}$

Therefore, this simply cancels out with the negative from the $\prod_{k=1}^n (-\frac12-k))$. In addition, cancelling out $\prod_{k=1}^n k$ from both sides leaves us with:

$$\prod_{k=n+1}^{2n}k=4^n\left(\frac1{2n+1}\right)\left(\prod_{k=1}^n \left(\frac12+k\right)\right)$$

Now, we take one of the $2^n$ terms from $4^n$ and combine it with the last factor on the RHS to get $\prod_{k=1}^n\left(1+2k\right)$ and the $\frac1{2n+1}$ cancels out with the last term here, leaving us with

$$\prod_{k=n+1}^{2n}k=2^n\prod_{k=1}^{n-1}\left(2k+1\right)$$

Now, if we express every factor in the LHS as $2^ab$ for odd number $b$ (this is unique for every positive integer), we can see (via induction or $1$-$1$ correspondence) that the product of the $b$'s is the $\prod$ factor in the RHS, and that the product of all of the $2^a$ is the $2^n$ factor in the RHS, so everything cancels out nicely and we are done.

Answer 2

Consider $\sec(\pi n)$:

$$\sec(\pi n) = \cases{1, n\text{ is even}\\ -1, n\text{ is odd}}$$

To show: $$\prod_{k=1}^{n}\frac{-k-n}{4} = \frac1{2n+1}\prod_{k=1}^{n} (-\frac12-k)$$

\begin{align} LHS &= \prod_{k=1}^{2n}k \\ &= \prod_{k=1}^{n}k\prod_{k=n+1}^{2n}(k)\\ &= \sec(\pi n)(-1)^n\prod_{k=1}^{n}k\prod_{k=n+1}^{2n}(k)\\ &= \sec(\pi n)\prod_{k=1}^{n}k\prod_{k=n+1}^{2n}(-k)\\ &=4^n \sec(\pi n)\prod_{k=n}^{n}(\frac{-k-n}{4})\prod_{k=1}^{n}k\\ \end{align}

If $\frac1{2n+1}\prod_{k=1}^{n} (-\frac12-k)=\prod_{k=1}^{n}\frac{-k-n}{4} $, then

$$4^n \sec(\pi n)\prod_{k=n}^{n}(\frac{-k-n}{4})\prod_{k=1}^{n}k=4^{n}\sec\left(\pi n\right)\left(\frac{1}{2n+1}\right)\left(\prod_{k=1}^{n}\left(-\frac{1}{2}-k\right)\right)\left(\prod_{k=1}^{n}k\right)=RHS$$

To show $ \frac1{2n+1}\prod_{k=1}^{n} (-\frac12-k)=\prod_{k=1}^{n}\frac{-k-n}{4} $, I will use induction:

base case: $n=1$ $$LHS=\frac13 \cdot -\frac32=-\frac12=-\frac24=RHS$$ Assume true for some n,

At $n+1$,

\begin{align} LHS &= \frac1{2n+3}\prod_{k=1}^{n+1} (-\frac12-k)\\ &=\frac1{2n+3}(-\frac12-n-1)\prod_{k=1}^{n} (-\frac12-k)\\ &=-\frac12 (2n+1)\prod_{k=1}^{n} \frac{-k-n}{4}\\ &= -\frac{2n+1}2\prod_{k=0}^{n-1}\frac{-k-(n+1)}{4}\\ &= \left[-\frac{2n+1}2 \frac{-n-1}{4} \prod_{k=1}^{n+1}\frac{-k-(n+1)}{4} \right]\big/\left[\frac{-2n-1}4 \frac{-2n-2}4\right]\\ &=RHS \end{align}

Answer 3

First note that

$$ \prod_{k=1}^{n} k = n! = \Gamma(n+1)$$

$$\prod_{k=1}^{2n}k = (2n)! = \Gamma(2n+1)$$

Recall the definition of the Pochhammer symbol:

$$(x)_{n} = \prod_{j=0}^{n-1}(x+j) = \frac{\Gamma(x+n)}{\Gamma(x)}$$

Then

\begin{align*}\prod_{k=1}^{n} \left(-\frac{1}{2}-k\right)= &(-1)^n \prod_{k=1}^{n}\left( \frac{1}{2} + k \right)\\ =& (-1)^n \prod_{j=0}^{n-1}\left(\frac{3}{2}+j\right) \quad (k=j+1)\\ =& (-1)^n \left(\frac{3}{2}\right)_{n}\\ =& \frac{(-1)^n\Gamma\left(\frac{3}{2} + n\right)}{\Gamma\left(\frac{3}{2}\right)} \\ =& \frac{2(-1)^n\Gamma\left(\frac{3}{2} + n\right)}{\sqrt{\pi} } \end{align*}

$$\sec(\pi n) =(-1)^n$$

Then, we want to prove

$$\Gamma(2n+1) = \frac{4^{n+1}}{2(2n+1)\sqrt{\pi} }\Gamma\left(\frac{3}{2} + n\right)\Gamma(n+1)$$

using $\Gamma(x+1) = x\Gamma(x)$, this is equivalent to

$$\Gamma(2n+2) = \frac{4^{n+1}}{2\sqrt{\pi} }\Gamma\left(\frac{3}{2} + n\right)\Gamma(n+1)$$

This is just the famous duplication formula for the Gamma function

$$ \Gamma(2x) = \frac{4^x}{2\sqrt{\pi}} \Gamma(x)\Gamma\left(\frac{1}{2}+x\right)$$

with $x = n+1$

I will prove the duplication formula for your case using Pochhamer symbols. Note that

\begin{align*} (a)_{2n} = &a(a+1)(a+2)\cdots(a+2n-1) \\ =& \left[a(a+2)(a+4)\cdots(a+2n-2)\right]\left[(a+1)(a+3)\cdots(a+2n-1)\right]\\ =& \left[2^n \frac{a}{2}\left(\frac{a}{2}+1\right)\left(\frac{a}{2}+2\right)\cdots\left(\frac{a}{2}+n-1\right)\right]\left[ 2^n \frac{(a+1)}{2} \left(\frac{a+1}{2}+1\right)\cdots\left(\frac{a+1}{2} +n-1\right) \right]\\ = &2^{2n}\left(\frac{a}{2}\right)_{n} \left(\frac{a+1}{2}\right)_{n} \end{align*}

But this is equivalent to

$$\frac{\Gamma(a+2n)}{\Gamma(a)} = 2^{2n}\frac{ \Gamma\left(\frac{a}{2}+n\right)\Gamma\left(\frac{a+1}{2}+n\right)}{\Gamma\left(\frac{a}{2}\right) \Gamma\left(\frac{a+1}{2}\right)}$$

with $a=2$

$$\Gamma(2n+2) = \frac{4^{n+1}}{2\sqrt{\pi}} \Gamma\left(1+n\right)\Gamma\left(\frac{3}{2}+n\right)$$

which is what we wanted to prove.