Prove $\prod_{n=1}^{\infty} (1+\frac{(-1)^{n+1}}{2n-1})=\sqrt2$

Question

This question came up while I was revising the Gamma function.

$$ \prod_{n=1}^{\infty} (1+\frac{(-1)^{n+1}}{2n-1})=\sqrt2 $$

Please prove me prove this infinite product, and explain the steps involved. I'm really having trouble with this one. Many thanks.

Answer 1

Let $P_m = \displaystyle\prod_{n=1}^{m}\left[1+\dfrac{(-1)^{n+1}}{2n-1}\right]$. Then, $P_{2m} = \dfrac{2}{1}\dfrac{2}{3}\dfrac{6}{5}\dfrac{6}{7} \cdots \dfrac{4m-2}{4m-3}\dfrac{4m-2}{4m-1}$.

This can be rewritten as $P_{2m} = \dfrac{2^{2m}(2m)!^3}{(4m)!(m!)^2} = \dfrac{2^{2m}\binom{2m}{m}}{\binom{4m}{2m}}$.

The Central Binomial Coefficient satisfies $\dbinom{2n}{n} \sim \dfrac{4^n}{\sqrt{\pi n}}$ for large $n$.

Thus, $P_{2m} = \dfrac{2^{2m}\binom{2m}{m}}{\binom{4m}{2m}} \to \dfrac{2^{2m}\frac{4^m}{\sqrt{\pi m}}}{\frac{4^{2m}}{\sqrt{2\pi m}}} = \sqrt{2}$. Also, $P_{2m-1} = \dfrac{4m-2}{4m-1}P_{2m} \to 1 \cdot \sqrt{2} = \sqrt{2}$.