<s>[INST] Double series convergent to $2\zeta(4)$? Using a computer I found the double sum
$$S(n)= \sum_{j=1}^n\sum_{k=1}^n \frac{j^2 + jk + k^2}{j^2(j+k)^2k^2}$$
has values
$$S(10) \quad\quad= 1.881427206538142 \\ S(1000) \quad= 2.161366028875634 \\S(100000) =  2.164613524212465\\$$
As a guess I compared with fractions $\pi^p/q$ where $p,q$ are positive integers and it appears 
$$\lim_{n \to \infty} S(n) = \frac{\pi^4}{45} =  2\zeta(4) \approx 2.164646467422276 $$
I'd be interested in seeing a proof if true.  
 [/INST] So before I start, I've never even attempted to evaluate a double sum before, so there could very well have been an easier way.
$$\sum_{j=1}^\infty\sum_{k=1}^\infty \frac{j^2+jk+k^2}{j^2k^2(j+k)^2} =  \sum_{j=1}^\infty\sum_{k=1}^\infty \frac{1}{k^2(j+k)^2} +\sum_{j=1}^\infty\sum_{k=1}^\infty \frac{1}{jk(j+k)^2} + \sum_{j=1}^\infty\sum_{k=1}^\infty \frac{1}{j^2(j+k)^2}= $$
$$ 2\sum_{j=1}^\infty\sum_{k=1}^\infty \frac{1}{k^2(j+k)^2} +\sum_{j=1}^\infty\sum_{k=1}^\infty \frac{1}{jk(j+k)^2}$$
Through partial fraction decomposition the above equals:
$$\sum_{j=1}^\infty\sum_{k=1}^\infty \frac{1}{j^3k} -\frac{1}{j^3(j+k)}-\frac{1}{j^2(j+k)}+2\sum_{j=1}^\infty\sum_{k=1}^\infty \frac{2}{j^3(j+k)}-\frac{2}{j^3k}+\frac{1}{j^2k^2}+\frac{1}{j^2(j+k)^2} $$
Collecting like-terms:
$$3\sum_{j=1}^\infty\sum_{k=1}^\infty \frac{1}{j^3(j+k)}-3\sum_{j=1}^\infty\sum_{k=1}^\infty \frac{1}{j^3k} +\sum_{j=1}^\infty\sum_{k=1}^\infty \frac{1}{j^2(j+k)^2} + 2\sum_{j=1}^\infty\sum_{k=1}^\infty \frac{1}{j^2k^2} =   $$
The final sum clearly equals $2\zeta(2)^2$ or $\frac{\pi^4}{18}$. I then evaluate the first two sums by combining them to get:
$$3\sum_{j=1}^\infty\sum_{k=1}^\infty \frac{1}{j^3}(\frac{1}{j+k}-\frac{1}{k}) $$
Interchanging j and k to and averaging the two sums to get:
$$\frac{3}{2}\sum_{j=1}^\infty\sum_{k=1}^\infty \frac{1}{j^3}(\frac{1}{j+k}-\frac{1}{k})+\frac{1}{k^3}(\frac{1}{j+k}-\frac{1}{j}) $$
This can be rewritten as:
$$-\frac{3}{2}\sum_{j=1}^\infty\sum_{k=1}^\infty \frac{1}{j^2k(j+k)}+\frac{1}{k^2j(j+k)}= -\frac{3}{2}\sum_{j=1}^\infty\sum_{k=1}^\infty \frac{1}{j^2k^2} = -\frac{3}{2}\zeta{(2)}^2 = -\frac{\pi^4}{24}$$
So putting that back into the original problem:
$$\sum_{j=1}^\infty\sum_{k=1}^\infty \frac{1}{j^2(j+k)^2} + \frac{\pi^4}{18} -\frac{\pi^4}{48} = \sum_{j=1}^\infty\sum_{k=1}^\infty \frac{1}{j^2(j+k)^2} + \frac{\pi^4}{72}  $$
This is all I got to. I couldn't evaluate the other sum the way I did before. Using a calculator there is a very good chance it equals $\frac{\pi^4}{120}$.
Just for fun I was able to write the remaining sum as:
$$\sum_{j=1}^\infty\sum_{k=1}^\infty \frac{1}{j^2(j+k)^2} = \sum_{n=1}^\infty \frac{\zeta(2,n+1)}{n^2}$$
Where $\zeta(x,y)$ is the Hurwitz Zeta Function. Wolfram Alpha was able to calculate the sum as $\frac{\pi^4}{120}$ as desired.
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