Evaluate the infinite product

By expanding prior to taking the limit, we find:

exp(1/1 + 1/3 + 1/5 + ... + 1/(2*m-1)) / exp(1/2 + 1/4 + ... + 1/2m) =

exp(1 - 1/2 + 1/3 - 1/4 + .... - 1/2m).

As m-> infinity, the quantity in parenthesis is well known to approach ln(2) (Google "sum of alternating harmonic series for an explanation"). Therefore, the limit is:

exp(ln(2)) = 2

$\prod_{n=1}^∞ e^{\frac{1}{(2n-1)}} = e^{\sum_{n=1}^∞ \frac{1}{(2n-1)} } =\alpha$ $\prod_{n=1}^∞ e^{\frac{1}{2n}} = e^{\sum_{n=1}^∞ \frac{1}{2n} } =\beta$ So $\frac{\alpha}{\beta} = e^{\sum_{n=1}^∞ \frac{1}{(2n-1)}-\frac{1}{2n} } = e^{\log(2)}=2$

$\color{#20A900}\textrm{Note}$

$\sum_{n=1}^∞ \frac{1}{2n-1} -\frac{1}{2n} = 1-\frac{1}{2} +\frac{1}{3} -.. =\log(2)$