jmc

By partial fractions, $$\frac{1}{m(m+n+1)}=\frac{1}{n+1}\left(\frac{1}{m}-\frac{1}{m+n+1}\right).$$Thus, $$\begin{array}{rl}\sum _{m=1}^{\infty }\frac{1}{m(m+n+1)}& =\sum _{m=1}^{\infty }\frac{1}{n+1}\left(\frac{1}{m}-\frac{1}{m+n+1}\right)\\ & =\frac{1}{n+1}\left(1-\frac{1}{n+2}\right)+\frac{1}{n+1}\left(\frac{1}{2}-\frac{1}{n+3}\right)\\ & +\frac{1}{n+1}\left(\frac{1}{3}-\frac{1}{n+4}\right)+\frac{1}{n+1}\left(\frac{1}{4}-\frac{1}{n+5}\right)+\cdots \\ & =\frac{1}{n+1}\left(1+\frac{1}{2}+\frac{1}{3}+\cdots +\frac{1}{n+1}\right).\end{array}$$Therefore, $$\begin{array}{rl}\sum _{m=1}^{\infty }\sum _{n=1}^{\infty }\frac{1}{mn(m+n+1)}& =\sum _{n=1}^{\infty }\frac{1}{n(n+1)}\left(1+\frac{1}{2}+\frac{1}{3}+\cdots +\frac{1}{n+1}\right)\\ & =\sum _{n=1}^{\infty }\frac{1}{n(n+1)}\sum _{k=1}^{n+1}\frac{1}{k}\\ & =\sum _{n=1}^{\infty }\sum _{k=1}^{n+1}\frac{1}{kn(n+1)}\\ & =\sum _{n=1}^{\infty }\left(\frac{1}{n(n+1)}+\sum _{k=2}^{n+1}\frac{1}{kn(n+1)}\right)\\ & =\sum _{n=1}^{\infty }\frac{1}{n(n+1)}+\sum _{n=1}^{\infty }\sum _{k=2}^{n+1}\frac{1}{kn(n+1)}.\end{array}$$The first sum telescopes as $$\sum _{n=1}^{\infty }\left(\frac{1}{n}-\frac{1}{n+1}\right)=1.$$For the second sum, we are summing over all positive integers $k$ and $n$ such that $2\le k\le n+1.$ In other words, we sum over $k\ge 2$ and $n\ge k-1,$ which gives us $$\begin{array}{rl}\sum _{k=2}^{\infty }\sum _{n=k-1}^{\infty }\frac{1}{kn(n+1)}& =\sum _{k=2}^{\infty }\frac{1}{k}\sum _{n=k-1}^{\infty }\frac{1}{n(n+1)}\\ & =\sum _{k=2}^{\infty }\frac{1}{k}\sum _{n=k-1}^{\infty }\left(\frac{1}{n}-\frac{1}{n+1}\right)\\ & =\sum _{k=2}^{\infty }\frac{1}{k}\cdot \frac{1}{k-1}\\ & =\sum _{k=2}^{\infty }\left(\frac{1}{k-1}-\frac{1}{k}\right)\\ & =1.\end{array}$$Therefore, $$\sum _{m=1}^{\infty }\sum _{n=1}^{\infty }\frac{1}{mn(m+n+1)}=\boxed{2}.$$