# Finding a closed form for $\sum_{k=1}^{\infty}\frac{1}{(2k)^5-(2k)^3}$

I'm looking for a closed form for the expression $$\sum_{k=1}^{\infty}\frac{1}{(2k)^5-(2k)^3}$$ I know that Ramanujan gave the following closed form for a similar expression $$\sum_{k=1}^{\infty}\frac{1}{(2k)^3-2k}= \ln(2)-\frac{1}{2}$$ I wonder if it is possible to find such a similarly simple and nice closed form for the above case.

Thanks.

$$\frac{1}{(2k)^5 - (2k)^3} + \frac{1}{(2k)^3} = \frac{1 + (2k)^2 - 1}{(2k)^5 - (2k)^3} = \frac{1}{(2k)^3 -2k}$$

So by Ramanujan's result:

$$\sum_{k=1}^{\infty} \frac{1}{(2k)^5 - (2k)^3} = \ln(2) - \frac{1}{2} - \frac{1}{8}\zeta(3)$$

Not an answer, but if you want to look for different solutions in the future in the same vein as these you posed, this might be a general method.

For an alternate representation for $$:$$\frac{1}{a-b}=\frac{1}{a}+\frac{b}{a^2}+\frac{a^2}{b^3}+\frac{b^3}{a^4}+\cdots\qquad(1)$$Plugging in a=(2k)^{5} and b=(2k)^3$$\frac{1}{(2k)^3}+\frac{(2k)^3}{(2k)^{10}}+\frac{(2k)^6}{(2k)^{15}}+\cdots=\sum_{n\geqslant 1}\frac{1}{(2k)^{2n+1}}$$And thus your sum can be evaluated as$$\sum_{k\geqslant 1}\sum_{n\geqslant 1}\frac{1}{(2k)^{2n+1}}

I am not entirely convinced if this was helpful to you. I am sorry if it was not.

$(1)$ Reason