$$ \lim_{x\to\infty}\prod_{n=1}^x \left(1+\frac1{xn(2n-1)}\right)^x=4 $$ I know this equals 4 because I derived this equation from the alternating harmonic series:
$1-\frac1{2} +\frac1{3} -\frac1{4} + \frac1{5}- \cdots=\ln2$
I could add a picture of my work but basically that series, s, yields this equation: $$ e^{2s}={\prod_{n=1}^\infty}e^{\frac1{2n^2-n}} $$
Then, we represent e using the limit definition: $$ \lim_{n\to\infty} \left(1+\frac x{n}\right)^n=e^x$$
The question is, how do we work out that infinite product that we know is equal to 4?
Like many problems with infinite products, take the logarithm. Then, $$ \log(P) = \log\Big(\prod_{n=1}^x \big(1+\frac{1}{x\,n(2n-1)}\big)^x \Big) =x\sum_{n=1}^x \log\big(1+\frac{1}{x\,n(2n-1)} \big)$$ $$ = x\sum_{n=1}^x \frac{1}{x\,n(2n-1)} -x \sum_{n=1}^x \frac{1}{ 2(x\,n(2n-1))^2} + ...$$ where the expansion for the logarithm $\log(1+z) = z -z^2/2+...$ has been used. The second sum goes to 0 as it has a 1/x factor. Thus, all that is left is to prove a formula that is pretty standard fare,
$$ \sum_{n=1}^\infty \frac{1}{n(2n-1)} = \log{4} . $$