Evaluating an infinite product (Putnam 2017) I've been trying to figure out how I could evaluate the following infinite product.
I am apparently supposed to find the exact value, but I don't see any way I could do this.
$$\prod_{n=0}^\infty \frac{{(4n+2)}^{\frac{3}{4n+2}}}{{(4n+3)}^{\frac{1}{4n+3}}{(4n+4)}^{\frac{1}{4n+4}}{(4n+5)}^{\frac{1}{4n+5}}}$$
I don't wish to know the solution I just want to know what techniques I should learn to be able to find the solution.
(edit)
I forgot to write that the top of the fraction is cubed. Of course this problem would be simple if the power was 1.
 A: The original version of your infinite product can be written in the following form:
$$ \exp\sum_{n\geq 0}\underbrace{\left[\frac{\log(4n+2)}{4n+2}-\frac{\log(4n+3)}{4n+3}-\frac{\log(4n+4)}{4n+4}-\frac{\log(4n+5)}{4n+5}\right]}_{\approx -2\frac{\log(4n+4)}{4n+4}}$$
the series is quite blatantly divergent to $-\infty$, hence the associated product is convergent to zero.

Interesting things happen if we change a sign:
$$ \exp\sum_{n\geq 0}\underbrace{\left[\frac{\log(4n+2)}{4n+2}-\frac{\log(4n+3)}{4n+3}+\frac{\log(4n+4)}{4n+4}-\frac{\log(4n+5)}{4n+5}\right]}_{\approx 2\frac{\log(4n+4)}{(4n+4)^2}}$$
equals ${\large\prod_{n\geq 0}}\frac{(4n+2)^{\frac{1}{4n+2}}(4n+4)^{\frac{1}{4n+4}}}{(4n+3)^{\frac{1}{4n+3}}(4n+5)^{\frac{1}{4n+5}}}$ or
$$\exp\sum_{n\geq 1}\frac{(-1)^{n}\log n}{n}=\exp\eta'(1)=\exp\left[\gamma\log 2-\frac{1}{2}\log^2 2\right]=2^{\gamma-\frac{\log 2}{2}}. $$

Similarly,
$$ \exp\sum_{n\geq 0}\underbrace{\left[3\cdot\frac{\log(4n+2)}{4n+2}-\frac{\log(4n+3)}{4n+3}-\frac{\log(4n+4)}{4n+4}-\frac{\log(4n+5)}{4n+5}\right]}_{\approx 6\frac{\log(4n+4)}{(4n+4)^2}}$$
can be written as
$$\exp\underbrace{\left[\sum_{n\geq 1}\frac{(-1)^n \log n}{n}+2\sum_{n\geq 1}\frac{(-1)^{n+1} \log(2n)}{2n}\right]}_{\text{massive cancellation!}}=\color{red}{\exp\log^2(2)}=2^{\log 2}. $$
