jmc

First, we can write $$\begin{array}{rl}& \frac{1}{(i+j+1)(i+j+2)\cdots (i+j+6)(i+j+7)}\\ & =\frac{1}{6}\cdot \frac{(i+j+7)-(i+j+1)}{(i+j+1)(i+j+2)\cdots (i+j+6)(i+j+7)}\\ & =\frac{1}{6}\left(\frac{1}{(i+j+1)(i+j+2)\cdots (i+j+6)}-\frac{1}{(i+j+2)\cdots (i+j+6)(i+j+7)}\right).\end{array}$$Thus, the following sum telescopes: $$\begin{array}{rl}& \sum _{j=0}^{\infty }\frac{1}{(i+j+1)(i+j+2)\cdots (i+j+6)(i+j+7)}\\ & =\sum _{j=0}^{\infty }\frac{1}{6}\left(\frac{1}{(i+j+1)(i+j+2)\cdots (i+j+6)}-\frac{1}{(i+j+2)\cdots (i+j+6)(i+j+7)}\right)\\ & =\frac{1}{6}\left(\frac{1}{(i+1)\cdots (i+6)}-\frac{1}{(i+2)\cdots (i+7)}\right)\\ & +\frac{1}{6}\left(\frac{1}{(i+2)\cdots (i+7)}-\frac{1}{(i+3)\cdots (i+8)}\right)\\ & +\frac{1}{6}\left(\frac{1}{(i+3)\cdots (i+8)}-\frac{1}{(i+4)\cdots (i+9)}\right)+\cdots \\ & =\frac{1}{6(i+1)(i+2)\cdots (i+5)(i+6)}.\end{array}$$We can then write $$\begin{array}{rl}& \frac{1}{6(i+1)(i+2)\cdots (i+5)(i+6)}\\ & =\frac{1}{5}\cdot \frac{(i+6)-(i+1)}{6(i+1)(i+2)\cdots (i+5)(i+6)}\\ & =\frac{1}{30}\left(\frac{1}{(i+1)(i+2)(i+3)(i+4)(i+5)}-\frac{1}{(i+2)(i+3)(i+4)(i+5)(i+6)}\right).\end{array}$$We obtain another telescoping sum: $$\begin{array}{rl}& \sum _{i=0}^{\infty }\frac{1}{6(i+1)(i+2)\cdots (i+5)(i+6)}\\ & =\sum _{i=0}^{\infty }\frac{1}{30}\left(\frac{1}{(i+1)(i+2)(i+3)(i+4)(i+5)}-\frac{1}{(i+2)(i+3)(i+4)(i+5)(i+6)}\right)\\ & =\frac{1}{30}\left(\frac{1}{(1)(2)(3)(4)(5)}-\frac{1}{(2)(3)(4)(5)(6)}\right)\\ & +\frac{1}{30}\left(\frac{1}{(2)(3)(4)(5)(6)}-\frac{1}{(3)(4)(5)(6)(7)}\right)\\ & +\frac{1}{30}\left(\frac{1}{(3)(4)(5)(6)(7)}-\frac{1}{(4)(5)(6)(7)(8)}\right)+\cdots \\ & =\frac{1}{30}\cdot \frac{1}{(1)(2)(3)(4)(5)}=\boxed{\frac{1}{3600}}.\end{array}$$