How to evaluate the product:
$$\prod _{n=2}^\infty\left(1+\dfrac{1}{n^2}+\dfrac{1}{n^4}+\dfrac{1}{n^6}+\dfrac{1}{n^8}+\ldots\right)$$
$$\left(1+\dfrac{1}{n^2}+\dfrac{1}{n^4}+\dfrac{1}{n^6}+\dfrac{1}{n^8}+\ldots\right)=\dfrac{1}{1-\dfrac{1}{n^2}}$$
How to proceed now?
Proceeding $\scr{r9m's} $ suggestion: $$\require{cancel}\frac1{1-1/n^2}=\frac{n\cdot n}{(n-1)(n+1)}\\\prod_{n=2}^{\infty}\sum_{k=0}^{\infty}\frac1{n^{2k}}=\prod_{n=2}^{\infty}\frac{n\cdot n}{(n-1)(n+1)}\\=\lim_{n\to\infty}\underbrace{\frac21\cdot\frac {\cancel{\color{blue}{2}}}{\cancel{\color{red}{3}}}}_{t_1}\cdot\underbrace{\frac{\cancel{\color{red}{3}}}{\cancel{\color{blue}{2}}}\cdot\frac {\cancel{\color{fuchsia}{3}}}{\cancel{\color{green}{4}}}}_{t_2}\cdot\underbrace{\frac{\cancel{\color{green}{4}}}{\cancel{\color{fuchsia}{3}}}\cdots}_{t_3}\cdots\underbrace{\frac {\cancel{\color{maroon}{n}}}{\cancel{\color{orange}{n-1}}}\frac n{n+1}}_{t_n} =\lim_{n\to\infty}\frac{2n}{n+1}=2$$