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

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

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