Non-recursive, explicit rational sequence that converges to $\sqrt{2}$

Following up on J. W. Tanner’s suggestion (now moved to chat, sigh) to use ratios of Pell numbers, you can take $$a_n=2^n\frac{\bigl((1+2^n)^n+(1-2^n)^n\bigr)\bmod(4^n-2)}{\bigl((1+2^n)^n-(1-2^n)^n\bigr)\bmod(4^n-2)}.$$ To see this, let $p_n,q_n$ be the unique natural numbers such that $$\tag{$*$}(1\pm\sqrt2)^n=q_n\pm p_n\sqrt2$$ (so $p_n$ are the Pell numbers, and $q_n$ half the Pell–Lucas numbers). On the one hand, $(*)$ implies $$\begin{align*} q_n&=\frac{(1+\sqrt2)^n+(1-\sqrt2)^n}{2},\\ p_n&=\frac{(1+\sqrt2)^n-(1-\sqrt2)^n}{2\sqrt2}, \end{align*}$$ whence $$\frac{q_n}{p_n}\sim\frac{(1+\sqrt2)^n/2}{(1+\sqrt2)^n/(2\sqrt2)}=\sqrt2.$$ On the other hand, if we put $r=2^n$ and $m=4^n-2$, we have $$r^2\equiv2\pmod m,$$ hence there is a ring homomorphism $\mathbb Z[\sqrt2]\to\mathbb Z/m\mathbb Z$ that reduces each integer modulo $m$, and maps $\sqrt2$ to $r$. Applying it to $(*)$, we obtain $$(1\pm r)^n\equiv q_n\pm p_nr\pmod m,$$ whence $$\begin{align*} (1+r)^n+(1-r)^n&\equiv 2q_n\phantom r\pmod m,\\ (1+r)^n-(1-r)^n&\equiv 2rp_n\pmod m. \end{align*}$$ But also $0<p_n<q_n<2^{n-1}$, i.e., $2rp_n,2q_n<m$, thus $$\begin{align*} \bigl((1+r)^n+(1-r)^n\bigr)\bmod m&=2q_n,\\ \bigl((1+r)^n-(1-r)^n\bigr)\bmod m&=2rp_n. \end{align*}$$ That is, $$a_n=\frac{q_n}{p_n}.$$ Note that nothing special about $r=2^n$ was used in the derivation above; you can take any integer $r$ larger than $(1+\sqrt2)^n$ or so, and $m=r^2-2$.

A straightforward way is to use Stirling approximation $n! \sim \sqrt{2 \pi n}\left(\frac{n}{e}\right)^n$ which for $2n$ gives $(2n)! \sim \sqrt{2 \pi \cdot2n}\left(\frac{2n}{e}\right)^{2n}$. Then $\frac{(2n)!}{n!}=\sqrt{2} \frac{(2n)^{2n}}{n^n e^n}$ which one can solve for $\sqrt2$. Since $e$ is not rational we need to replace it by a rational approximation. Here $(1+\frac{1}{n})^n$ would not work because it does not converge to $e$ fast enough, but one can use $n^2$ in place of $n$ and replace $e$ by $(1+\frac1{n^2})^{n^2}$ and this does converge to $e$ fast enough.

We have

$$\sqrt 2 = \left(1-\frac{1}{2}\right)^{-1/2}=\sum_{k=0}^\infty\binom{-1/2}{k}\left(-\frac{1}{2}\right)^k$$

So a sequence of rationals converging to $\sqrt 2$ is

$$a_n=\sum_{k=0}^n\binom{-1/2}{k}\left(-\frac{1}{2}\right)^k$$

I am not sure whether an infinite product whose factors are fractions qualifies under the restriction "explicit rational sequence". In case it does, the following should be relevant.

A useful resource for identifying relevant sequences is the OEIS (On-line Encyclopedia of Integer Sequences). In this case, one could search it for sqrt(2), which finds myriad sequences somehow related to or involving $\sqrt{2}$. Since I had an idea what I might be looking for, I relatively quickly identified an infinite product whose factors are fractions, with the numerators specified by sequence A016826 and the denominators specified by sequence A001539. I then used Wolfram Alpha to verify that

$$\prod_{n=0}^{\infty}\frac{(4n+2)^{2}}{(4n+1)(4n+3)} = \prod_{n=1}^{\infty}\frac{1}{1-\frac{1}{4(2n-1)^{2}}} = \frac{4}{3}\cdot\frac{36}{35}\cdot\frac{100}{99}\cdot\frac{196}{195}\cdot \ldots\ = \sqrt{2}$$

An alternate approach to identifying similar infinite products would be to search the literature with Google Scholar (or just plain Google) for Wallis-type infinite products.