What sequence is this $\prod_{i=1}^n \frac{2i}{2i + 1} = $?

Question

I'm trying to find a closed form solution for $\prod_{i=1}^n \frac{2i}{2i + 1} $, but I'm confused what kind of sequence this is. Does it have a closed form?

Answer 1

Closed Form $$ \begin{align} a_n &=\prod_{k=1}^n\frac{\color{#C00}{2k}}{\color{#090}{2k+1}}\tag{1a}\\ &=\frac{\color{#C00}{2^nn!}\,\color{#090}{2^nn!}}{\color{#090}{(2n+1)!}}\tag{1b}\\[3pt] &=\frac{4^n}{(2n+1)\binom{2n}{n}}\tag{1c} \end{align} $$

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Bounds $$ \begin{align} \frac{\sqrt{n+1}\ a_n}{\sqrt{n}\ a_{n-1}} &=\sqrt{\frac{n+1}{n}}\frac{2n}{2n+1}\tag{2a}\\ &=\sqrt{\frac{4n^3+4n^2}{4n^3+4n^2+n}}\tag{2b}\\[9pt] &\le1\tag{2c} \end{align} $$ Therefore, $\sqrt{n+1}\ a_n$ is decreasing. $$ \begin{align} \frac{\sqrt{n+\frac34}\ a_n}{\sqrt{n-\frac14}\ a_{n-1}} &=\sqrt{\frac{4n+3}{4n-1}}\frac{2n}{2n+1}\tag{3a}\\ &=\sqrt{\frac{16n^3+12n^2}{16n^3+12n^2-1}}\tag{3b}\\[12pt] &\ge1\tag{3c} \end{align} $$ Therefore, $\sqrt{n+\frac34}\ a_n$ is increasing.

Since $\sqrt{n+1}\ a_n$ is decreasing and greater than $\sqrt{n+\frac34}\ a_n$, which is increasing, and their ratio tends to $1$, they tend to a common limit, $L$. $$ \begin{align} L &=\lim_{n\to\infty}\sqrt{n+1}\prod_{k=1}^n\frac{2k}{2k+1}\tag{4a}\\ &=\lim_{n\to\infty}\frac{\sqrt{n+1}}{2n+1}\frac{4^n}{\binom{2n}{n}}\tag{4b}\\ &=\frac{\sqrt\pi}2\tag{4c} \end{align} $$ Explanation:
$\text{(4a)}$: definition of $L$
$\text{(4b)}$: apply $(1)$
$\text{(4c)}$: Theorem $1$ from this answer

Thus $(2)$, $(3)$, and $(4)$ yield $$ \bbox[5px,border:2px solid #C0A000]{\sqrt{\frac{\pi}{4n+4}}\le\prod_{k=1}^n\frac{2k}{2k+1}\le\sqrt{\frac{\pi}{4n+3}}}\tag5 $$

Answer 2

You can simplify it. \begin{align*} \prod_{i=1}^n \frac{2i}{2i + 1} &=\frac{2^n n!}{3\cdot5\cdot7\cdot\ldots\cdot(2n+1)}\\ &=\frac{2^{2n} (n!)^2}{(2n+1)!} \end{align*}

Answer 3

After @Lemoine's answer, you can also write $$P_n=\prod_{i=1}^n \frac{2i}{2i + 1}=\frac{\sqrt{\pi }\,\, \Gamma (n+1)}{2\,\, \Gamma \left(n+\frac{3}{2}\right)}$$ Take logarithms, use Stirling approximation twice and continue with Taylor series to obtain $$\log(P_n)=-\frac{1}{2} \log \left(\frac{4 n}{\pi }\right)-\frac{3}{8 n}+\frac{1}{8 n^2}+O\left(\frac{1}{n^3}\right) $$ $$P_n=e^{\log(P_n)}=\frac{\sqrt{\pi }}{2 \sqrt{n}}\,\,\exp\Big[-\frac{3}{8 n}+\frac{1}{8 n^2}+O\left(\frac{1}{n^3}\right) \Big]$$

Try it for $n=10$; the exact value is $$P_{10}=\frac{262144}{969969}\sim 0.270260$$ while the above truncated expansion gives $0.270272$.