Not an answer, but if you want to look for different solutions in the future in the same vain 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

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