Putnam '69 B3

Let us substitute a few values to get an essence of the sequence -

$\begin{aligned} a_2 &= \dfrac1{a_1} \\ a_3 &= 2a_1 \\ a_4 &= \dfrac3{2a_1} \\ a_5 &= \dfrac{4 \cdot 2a_1}{3} \\ a_6 &= \dfrac{3 \cdot 5}{a_1 \cdot 2 \cdot 4} \\ a_7 &= \dfrac{a_1 \cdot 2 \cdot 4 \cdot 6}{3 \cdot 5} \\ a_8 &= \dfrac{3 \cdot 5 \cdot 7}{a_1 \cdot 2 \cdot 4 \cdot 6} \\ a_9 &= \dfrac{a_1 \cdot 2 \cdot 4 \cdot 6 \cdot 8}{3 \cdot 5 \cdot 7} \\ \text{So the general terms are } \\ a_{2n} &= \dfrac{3 \cdot 5 \cdot 7 \cdots (2n-1)}{a_1 \cdot 2 \cdot 4 \cdot 6 \cdots (2n-2)} \\ a_{2n+1} &= \dfrac{a_1 \cdot 2 \cdot 4 \cdot 6 \cdots (2n)}{3 \cdot 5 \cdot 7 \cdots (2n-1)} \end{aligned}$

$\begin{aligned} \implies \dfrac{a_{2n}}{a_{2n+1}} &= \dfrac{1 \cdot 3 \cdot 3 \cdot 5 \cdot 5 \cdots (2n-1)^2}{(a_1)^2 \cdot 2 \cdot 2 \cdot 4 \cdot 4 \cdot 6 \cdots (2n-2)^2 \cdot (2n)} \end{aligned}$

Using Wallis Product , as $n \to \infty$

$\begin{aligned} 1 &= \dfrac1{(a_1)^2} \cdot \dfrac2\pi \\ \implies a_1 &= \sqrt{\dfrac2\pi} \end{aligned}$

My latex experience is not good. So I typed in Word and screenshot it to here. I skipped the last limit calculation because I think it's well-known.