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Lambert W function evaluation error
I'm working on a homework assignment that asks me to evaluate $\sum\limits_{n=1}^\infty \frac{\log(n)}{1-2n^{ -2}+n^{ -4}}$
I'm not really sure how to go about this. I've been trying to use the substitution $u=1-2n^{ -2}+n^{ -4}$ and then use the power series expansion of $\log(u)$, but I couldn't find a way to simplify it.
This is a pretty simple problem, but I just can't wrap my head around it. I'm not sure if it is even possible to take the logarithm of $1-2n^{ -2}+n^{ -4}$?
Any help is appreciated
A:
One possibility is the Euler product,
$\log(1-2n^{ -2}+n^{ -4})=\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n\cdot 2n^2+n^4}=\sum_{n=1}^\infty\frac{n^2+n^4-(n^2-n^4)}{n(2n^2+n^4)}=\\
\sum_{n=1}^\infty\left(\frac1n-\frac1{n+2n^2}+\frac{1+2n^2}{n\cdot 2n^2+n^4}\right)$
which is analytic for $|z| 0b46394aab
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