Beautiful Summation

Note that $\begin{array}{rcl} \displaystyle \sum_{m,n=1}^\infty \frac{1}{m^2n + mn^2 + 2mn} & = & \displaystyle \sum_{m,n=1}^\infty \frac{1}{mn(m+n+2)} \; = \; \sum_{n=1}^\infty \frac{1}{n}\sum_{m=1}^\infty \frac{1}{m(m+n+2)} \\ & = & \displaystyle \sum_{n=1}^\infty \frac{1}{n(n+2)}\sum_{m=1}^\infty \left(\frac{1}{m} - \frac{1}{m+n+2}\right) \\ & = & \displaystyle\sum_{n=1}^\infty \frac{1}{n(n+2)} \sum_{m=1}^{n+2}\frac{1}{m} \; = \; \frac32\sum_{n=1}^\infty\frac{1}{n(n+2)} + \sum_{m=3}^\infty \frac{1}{m}\sum_{n=m-2}^\infty \frac{1}{n(n+2)} \end{array}$ Since $\sum_{n=k}^\infty \frac{1}{n(n+2)} \; = \; \frac12\sum_{n=k}^\infty \left(\frac{1}{n} - \frac{1}{n+2}\right) \; = \; \frac12\left(\frac{1}{k} + \frac{1}{k+1}\right)$ we deduce that $\begin{array}{rcl} \displaystyle \sum_{m,n=1}^\infty \frac{1}{m^2n + mn^2 + 2mn} & = & \displaystyle \frac98 + \sum_{m=3}^\infty \frac{1}{2m}\left(\frac{1}{m-2} + \frac{1}{m-1}\right) \\ & = & \displaystyle \frac98 + \frac12\sum_{m=1}^\infty \frac{1}{m(m+2)} + \frac12\sum_{m=2}^\infty \frac{1}{m(m+1)} \\ & = & \displaystyle \frac98 + \frac38 + \frac12\sum_{m=2}^\infty \left(\frac{1}{m} - \frac{1}{m+1}\right) \; = \; \frac98 + \frac38 + \frac14 \\ & = & \boxed{\frac74}\;. \end{array}$