Deduce limit of sequence from another sequence We have a sequence $(a_n)$ defined by the formula $$a_n = \int_0^1 (1 - x^2)^{\frac n2} \mathrm dx$$ We are also asked to prove, which I managed to do, the following recursion formula $$a_n = \frac{n + 3}{n + 2} \cdot a_{n + 2}$$ We are then asked to deduce with this information the limit of the series $$\frac 21 \cdot \frac 23, \frac 21 \cdot \frac 23 \cdot \frac 43 \cdot \frac 45, \frac 21 \cdot \frac 23 \cdot \frac 43 \cdot \frac 45 \cdot \frac65 \cdot \frac 67 \dots$$ But I cannot seem to find a relation between these sequences. I am quite sure that the limit should be $\frac \pi 2$, which is $2 \cdot a_1$, but I am not really sure if this is a coincidence and, in case it's not, why.
If someone could me maybe help to figure out how to proceed, I would be very thankful!
 A: Hint. Note that $0\le  (1-x^2)^{(n+2)/2} \le (1-x^2)^{(n+1)/2} \le (1-x^2)^{n/2}\le 1$ when $0\le x \le1$. This gives $0 \le a_{n+2} \le a_{n+1} \le a_n \le 1$. So, $\dfrac{a_{n+2}}{a_n} \le \dfrac{a_{n+1}}{a_n} \le 1$. Can you simplify $\dfrac{a_{n+2}}{a_n}$? What can you say about the limit $\lim_{n\to \infty} \frac{a_{n+1}}{a_n} $? How can you use this limit for the given infinite product?

Edit. I came up with a full solution.
As above, $ \frac{a_{n+2}}{a_n} = \frac{n+2}{n+3}\le \frac{a_{n+1}}{a_n} \le 1 $, and taking limit yields $\lim \frac{a_{n+1}}{a_n} = 1$.
$$a_{2k}= \left( \frac{2k}{2k+1}\right) a_{2k-2} =
\left( \frac{2k}{2k+1}\right) \left( \frac{2k-2}{2k-1}\right) a_{2k-4} = \dots = \frac{(2k)(2k-2)\dots 2 }{(2k+1)(2k-1) \dots  3} a_0,$$
$$a_{2k-1}= \left( \frac{2k-1}{2k}\right) a_{2k-3} =
\left( \frac{2k-1}{2k}\right) \left( \frac{2k-3}{2k-2}\right) a_{2k-3} = \dots = \frac{(2k-1)(2k-3)\dots 3 }{(2k)(2k-2) \dots  4} a_1.$$
We can write the second equation as
$$\frac{a_{2k-1}}{2} = \frac{(2k-1)(2k-3)\dots 3 \cdot 1 }{(2k)(2k-2) \dots  4 \cdot 2} a_1.
$$
(This adjustment makes the multiplicants in numerators and denominators to be all $k$ in the fractions $a_{2k}$ and $a_{2k-1}/2$.)
So taking quotient yields
\begin{align*}
\frac{2 \cdot a_{2k}}{a_{2k-1}} &= \frac{(2k)(2k-2)\dots 2 }{(2k+1)(2k-1) \dots  3} \cdot
\frac{(2k)(2k-2) \dots  4 \cdot 2}{(2k-1)(2k-3)\dots 3 \cdot 1 }\cdot 
\frac{a_0}{a_1}\\ 
& = \frac{(2k)(2k)}{(2k+1)(2k-1)} \dots \frac{2\cdot 2}{3\cdot 1} \frac{a_0}{a_1}.
\end{align*}
By taking limit $k\to\infty$, we have
$$ L\cdot \frac{a_0}{a_1} = 2,$$
$$ L = \frac{2a_1}{a_0} = \frac{\pi}{2},$$
where $L$ is the desired infinite product. This proves that the infinite product converges, and the limit is $\pi/2$.
