The sequence $a_n$ is closely related to the Wallis product $$a'_n = \prod_{i = 1}^\infty \left(\frac{2i}{2i - 1} \frac{2i}{2i + 1}\right),$$ which converges to $\pi/2$ as $n$ goes to infinity. Namely, we have $$a'_n = a_{n + 1} \cdot \frac{2n}{2n + 1}$$. This could be proven by induction or maybe more easily by defining $b_n = a_n a_{n - 1}$ and noticing that the recursion for $a_n$ implies the (very simple) recursion $$b_{n + 1} = 1 + b_n$$ for $b_n$ and expressing $a_n$ in terms of the $b_n$.

For more general values of $a_0$ one gets similar formulas for $a_n$ as (up to a factor converging to 1) a Wallis product or inverse of a Wallis product where a few of the lower terms in the product are missing.

The sequence $a_n$ is closely related to the Wallis product $$a'_n = \prod_{i = 1}^n \left(\frac{2i}{2i - 1} \frac{2i}{2i + 1}\right),$$ which converges to $\pi/2$ as $n$ goes to infinity. Namely, we have $$a'_n = a_{n + 1} \cdot \frac{2n}{2n + 1}$$. This could be proven by induction or maybe more easily by defining $b_n = a_n a_{n - 1}$ and noticing that the recursion for $a_n$ implies the (very simple) recursion $$b_{n + 1} = 1 + b_n$$ for $b_n$ and expressing $a_n$ in terms of the $b_n$.

For more general values of $a_0$ one gets similar formulas for $a_n$ as (up to a factor converging to 1) a Wallis product or inverse of a Wallis product where a few of the lower terms in the product are missing.

The sequence $a_n$ is closely related to the Wallis product $$a'_n = \prod_{i = 1}^\infty \left(\frac{2i}{2i - 1} \frac{2i}{2i + 1}\right),$$ which converges to $\pi/2$ as $n$ goes to infinity. Namely we have $$a'_n = a_{n + 1} \cdot \frac{2n}{2n + 1}$$. This could be proven by induction or maybe more easily by defining $b_n = a_n a_{n - 1}$ and noticing that the recursion for $a_n$ implies the (very simple) recursion $$b_{n + 1} = 1 + b_n$$ for $b_n$ and expressing $a_n$ in terms of the $b_n$.

For more general values of $a_0$ one gets similar formulas for $a_n$ as (up to a factor converging to 1) a Wallis product or inverse of a Wallis product where a few of the lower terms in the product are missing.

The sequence $a_n$ is closely related to the Wallis product $$a'_n = \prod_{i = 1}^\infty \left(\frac{2i}{2i - 1} \frac{2i}{2i + 1}\right),$$ which converges to $\pi/2$ as $n$ goes to infinity. Namely, we have $$a'_n = a_{n + 1} \cdot \frac{2n}{2n + 1}$$. This could be proven by induction or maybe more easily by defining $b_n = a_n a_{n - 1}$ and noticing that the recursion for $a_n$ implies the (very simple) recursion $$b_{n + 1} = 1 + b_n$$ for $b_n$ and expressing $a_n$ in terms of the $b_n$.

For more general values of $a_0$ one gets similar formulas for $a_n$ as (up to a factor converging to 1) a Wallis product or inverse of a Wallis product where a few of the lower terms in the product are missing.