$\frac{1}{1!}+\frac{1+3}{2!}+\frac{1+3+5}{3!}+\dots \dots \infty =$

$\begin{array}{rcl} & & \frac{1}{\overline{)\underline{1}}}+\frac{1+3}{\overline{)\underline{2}}}+\frac{1+3+5}{\overline{)\underline{3}}}+\dots \dots \\ & =& \sum _1^{\infty } \frac{1+3+5+...+(2\mathrm{n}-1)}{\overline{)\underline{\mathrm{n}}}}\\ & =& \sum _1^{\infty } \frac{{\mathrm{n}}^2}{\overline{)\underline{\mathrm{n}}}}=\sum _1^{\infty } \frac{\mathrm{n}}{\overline{)\underline{\mathrm{n}-1}}}\\ & =& \sum _1^{\infty } \frac{(\mathrm{n}-1)+1}{\overline{)\underline{\mathrm{n}-1}}}\\ & =& \sum _1^{\infty }.\left(\frac{1}{\overline{)\underline{\mathrm{n}-2}}}+\frac{1}{\overline{)\underline{\mathrm{n}-1}}}\right)\\ & =& \mathrm{e}+\mathrm{e}=2\mathrm{e}\end{array}$