I want to find the value of $$\sum_{r=0}^\infty \frac{1}{(2r+1)^{n}}\sum_{l=0}^\infty\frac{1}{(2l+1)^{n+2}}.$$ For this I know that we can use the cauchy product of two infinte series and then we have the new series as $$\sum_{k=0}^{\infty}c_{k}$$ where $c_{k}= \sum_{j=0}^ka_{j}b_{k-j}$.
And so the double series will convert into the series $$ \sum_{k=0}^\infty \sum_{j=0}^k \frac{1}{(2j+1)^n}\frac{1}{(2(k-j)+1)^{n+2}} .$$
However this doesn't seem to be helpful. Any ideas/help is highly appreciated.
For integer $n>1$ we have that
$$\zeta(n) = \sum_{k=1}^\infty \frac{1}{k^n} \implies \sum_{k\text{ odd}} \frac{1}{k^n} = \zeta(n) - \sum_{m=1}^\infty \frac{1}{(2m)^n} = (1-2^{-n})\zeta(n)$$
The double sum you have is a product of two sums of odd terms, which evaluates to
$$(1-2^{-n})(1-2^{-n-2})\zeta(n)\zeta(n+2)$$