Arithmetic Series on Factorials

$\begin{aligned} S & = \frac 1{2!}+\frac 1{3!}+\frac 1{4!}+\frac 1{5!}+\frac {202}{6!}+\frac {7}{7!}+\frac {22}{8!}+\frac {304}{9!}+\frac {13}{10!}+\frac {43}{11!}+\frac {406}{12!}+\cdots \\ & \because a_n=1+6\cdot(n-1)=2\cdot(3n+1)-7 \\ & b_n=1+21\cdot(n-1)=7\cdot(3n+2)-34 \\ & c_n=202+102\cdot(n-1)=34\cdot(3n+3)-2 \\ \therefore S & = \frac 1{2!}+\frac {3-2}{3!}+\frac {4\times2-7}{4!}+\frac {5\times7-34}{5!}+\frac {6\times34-2}{6!}+\frac {7\times2-7}{7!}+\frac {8\times7-34}{8!}+\frac {9\times34-2}{9!}+\frac {10\times2-7}{10!}+\frac {11\times7-34}{11!}+\frac {12\times34-2}{12!}+\cdots \\ & = \frac 1{2!}+\frac 1{2!}-\frac 2{3!}+\frac 2{3!}-\frac 7{4!}+\frac 7{4!}-\frac {34}{5!}+\frac {34}{5!}-\frac 2{6!}+\frac 2{6!} -\frac 7{7!}+\frac 7{7!}-\frac {34}{8!}+\cdots \\ & = \boxed{1} \end{aligned}$

Challenge: Why applying the above procedure and the answer we obtain is correct , but $e-2=1$ obtained using the above procedure is incorrect.

See $e-2=1$