<s>[INST] How do I evaluate $\sum_{m,n\geq 1}\frac{1}{m^2n+n^2m+2mn}$ I saw a problem here  which state to evalute $$\sum_{m=1}^{\infty}\sum_{n=1}^{\infty}\frac{1}{m^2n+n^2m+2mn}$$
My attempt
Let $$f(m,n)=\sum_{m=1}^{\infty}\sum_{n=1}^{\infty}\frac{1}{m^2n+n^2m+2mn}$$   and interchanging $m,n$ as $n,m$ we have $$f'(n,m) = \sum_{m=1}^{\infty}\sum_{n=1}^{\infty}\frac{1}{mn^2+nm^2+2mn}$$ then I add $f(m,n)+f'(n,m)$ which gives the same which gives me the twice the original series.
I failed to evaluate the series with the process I have applied.  How do I evaluate the series?  Any help will be appreciated.
 [/INST] Using integration trick
Note that $$\dfrac{1}{nm^2+n^2m+2mn} =\dfrac{1}{mn(m+n+2)}=\dfrac{1}{mn}\int_0^1 x^{m+n+1}\,dx$$ then we have $$\sum_{n=1}^{\infty}\sum_{m=1}^{\infty}\dfrac{1}{mn} \int_0^1 x^{m+n+1}\,dx=\int_0^1x\left(\sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \dfrac{x^{m}\cdot x^n}{mn}\right)\,dx \\= \int_0^1 x\ln^2(1-x)\,dx =\int_0^1 (1-x)\ln^2 x\,dx $$ hence integrating (by parts) we have the result $\dfrac{7}{4}$.

Without integration trick
Expanding the inner summation gives us $$\sum_{n\geq 1} \left(\frac{1}{n(n+3)}+\frac{1}{2n(n+4)}+\frac{1}{3n(n+15)}+\cdots\right)$$ making partial fraction of each summand we have $$\sum_{n\geq 1}\left(\frac{1}{3}\left(\frac{1}{n}-\frac{1}{n+3}\right)+\frac{1}{8}\left(\frac{1}{n}-\frac{1}{n+4}\right)+\cdots\right)=\sum_{n\geq 3} \frac{H_n}{n(n-2)}\cdots(1)$$ from here we can carry on our work by using the generation function of harmonic number, ie $$\sum_{n\geq 1} x^nH_n=\frac{\ln(1-x)}{x-1}$$

If we dont wish to make repeated  integration and multiplying process then let's continue our work from $(1)$ we have above $$\sum_{n\geq 3}\frac{H_n}{2}\left(\frac{1}{n-2}-\frac{1}{n}\right)=\frac{1}{2}\left(H_3+\frac{H_4}{2}\right)+\frac{1}{2}\sum_{n\geq 3}\left(\frac{H_{n+2}}{n}-\frac{H_n}{n}\right)=\frac{1}{2}\left(1+\frac{1}{2}+\frac{1}{3}\right)+\frac{1}{4}\left(1+\frac{1}{2}+\cdots +\frac{1}{4}\right)+\sum_{n\geq 3}\frac{1}{2}\left(\frac{H_n}{n}-\frac{H_n}{n}+\frac{1}{n}\left(\frac{1}{n+1}+\frac{1}{n+2}\right)\right)\\=\frac{1}{2}\left(\frac{11}{6}+\frac{25}{24}\right)+\sum_{n\geq 3}\left(\frac{2}{(n+2)(n+3)}+\frac{3}{n(n+1)(n+2)}\right)=\frac{1}{2}\left(\frac{69}{24}+\frac{1}{2}+\sum_{n\geq 1}\frac{1}{n(n+1)(n+2)}-\frac{3}{6}-\frac{3}{24}\right)=\frac{1}{2}\left(\frac{69}{24}+\frac{1}{2}+\frac{3}{2\cdot  2!}-\frac{5}{8}\right)=\frac{7}{4}$$
Recall that  for all $ p>0$ $$\sum_{n\geq 1}\frac{1}{n(n+1)(n+2)\cdots(n+p)}=\frac{1}{p\cdot  p!}$$
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