Is it possible to write Catalan's product for e as a product of a sequence?

Question

Catalan found a product for $e$: $$e=\dfrac{2}{1}\left(\dfrac{4}{3}\right)^{\frac{1}{2}}\left(\dfrac{6\cdot 8}{5\cdot 7}\right)^{\frac{1}{4}}\left(\dfrac{10\cdot 12\cdot 14\cdot 16}{9\cdot 11\cdot 13\cdot 15}\right)^{\frac{1}{8}}\cdots$$

Is it possible to write this as an infinite product in the form: $$e = \prod_{n=0}^\infty a_n$$ I know that it would likely be some sequence of factorials to the $\frac{1}{2^n}$ power, but I cannot think of what the sequence would be.

Thanks!

Answer 1

Look at that fourth term:

$$\left(\frac{10\cdot 12\cdot 14\cdot 16}{9\cdot 11\cdot 13\cdot 15}\right)^{1/8}$$

Factor out the $2$s in the numerator:

$$10\cdot 12\cdot 14\cdot 16=2^4\cdot5\cdot6\cdot7\cdot 8=2^4\frac{8!}{4!}$$

The denominator is $$\begin{align}9\cdot 11\cdot 13\cdot 15 &= \frac{16!}{2^8\cdot 8!\cdot (1\cdot 3\cdot 5\cdot 7)}\\ 1\cdot 3\cdot 5\cdot 7&=\frac{8!}{2^4 4!} \end{align}$$

So:

$$9\cdot 11\cdot 13\cdot 15 = \frac{16!\cdot 2^4\cdot 4!}{2^8\cdot 8!\cdot 8!}=\binom{16}{8}\frac{4!}{2^4}$$

So $$a_4^{8}=\frac{2^4\frac{8!}{4!}}{\binom{16}{8}\frac{4!}{2^4}}=\frac{2^8\binom{8}{4}}{\binom{16}{8}}$$ of $$a_4=2\left(\frac{\binom{8}{4}}{\binom{16}{8}}\right)^{1/8}$$

Thus, for $n>1$, we might guess:

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

Somewhat tricky, because $n=1$ doesn't work.

If you want to expand it as factorials:

$$a_n==2\left(\frac{\left(2^{n-1}!\right)^3}{2^n!\left(2^{n-2}!\right)^2}\right)^{1/2^{n-1}}$$

If you define $$b_k=2\left(\frac{\left((2k)!\right)^3}{(4k)!(k!)^2}\right)^{\frac{1}{2k}}$$

Then $a_n=b_{2^{n-2}}$ for $n\geq2$ and $a_1=1$.

Answer 2

Using factorials and cancellations in considering the partial product, you may just obtain $$ a_n=2^{\large\frac{n+1}2}\left(\frac{(2^n!)^2}{2^{n-1}!\: 2^{n+1}!} \right)^{\large 1/2^n} $$ proved here (p.5-6) with others interesting Catalan type infinite products.