Recurring Styles #15 - Continued Product!

Solving the recurrence relation certainly gives $a_n \; = \; \frac{(n-1)!!}{(n-2)!!} \qquad, \qquad \qquad n\ge 2 \;.$ but it is useful to note this alternative expression for the sequence: $a_{2n} \; = \; 2nc_n \qquad \qquad a_{2n+1} \; = \; c_n^{-1}$ where $c_n \; = \; 2^{-2n}{2n \choose n} \;.$ Stirling's approximation tells us that $c_n \,\sim\, \frac{1}{\sqrt{\pi n}}$ as $n \to \infty$ . If we put $b_n \; =\; \frac{a_{n+1} - a_n}{\sqrt{n}} \;,$ then it is easy to use the asymptotic form for $c_n$ to see that $\lim_{n \to\infty} b_{2n} \; = \; \sqrt{\tfrac{\pi}{2}} - \sqrt{\tfrac{2}{\pi}} \; = \; -\lim_{n\to\infty}b_{2n+1} \;,$ making $L \; = \; \sqrt{\tfrac{\pi}{2}} - \sqrt{\tfrac{2}{\pi}} \,$ and the answer $1 \times 2 \times 2 \times -1 \,=\, \boxed{-4}$ .