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
