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How do they change the denominator from the equation into $2^{n}n!$ in this case? Isn't it just $2n$ as $n$ approaches infinity?

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1 Answer

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$$\begin{align*} 2\cdot 4\cdot 6\cdot\ldots\cdot(2n)&=(2\cdot\color{crimson}1)\cdot(2\cdot\color{crimson}2)\cdot(2\cdot\color{crimson}3)\cdot\ldots\cdot(2\cdot \color{crimson}n)\\ &=(\underbrace{2\cdot2\cdot2\cdot\ldots\cdot2}_{n\text{ factors}})\cdot(\color{crimson}{1\cdot2\cdot3\cdot\ldots\cdot n})\\ &=2^n\cdot\color{crimson}{n!} \end{align*}$$

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