Double summation with $\zeta$

$\sum_{m=1}^{\infty}\sum_{n=1}^{\infty}\frac{1}{mn(m+n)}$ $\implies$ $\sum_{m=1}^{\infty}\frac{1}{m^2}\sum_{n=1}^{\infty}(\frac{1}{n}-\frac{1}{n+m})$ .Since $\psi$ $(\,m+1)\,$ $+$ $\gamma$ $=$ $\sum_{n=1}^{\infty}(\frac{1}{n}-\frac{1}{n+m})$ ,where $\psi$ $(\,s)\,$ denotes the diagamma function.Then substituting it to the original equation we have,

$\sum_{m=1}^{\infty}\frac{1}{m^2} (\gamma+\psi(m+1))$ $=$ $\alpha$ $\zeta(\beta)$ $\implies$ From the sum we can say it's exactly $-$ $\psi^2$ $(\,1)\,$ which is $2\zeta(3)$ hence $\alpha$ =2 and $\beta$ =3 then $\alpha$ + $\beta$ =5.The proof for why the sum is second derivative of diagamma function of 1 check out the discussion "The series representation of the Polygamma function".(Plus it is very nice that my own research matches Mr.Adhiraj'es question ,say we could be partners Mr. Adhiraj?) (Here $\psi^2$ $(\,s)\,$ denotes second derivative of diagamma function.)