$1+\frac{1+2}{2!}+\frac{1+2+3}{3!}+\frac{1+2+3+4}{4!}+\dots \infty =$

$\begin{array}{rcl} & & 1+\frac{1+2}{\overline{)\underline{2}}}+\frac{1+2+3}{\overline{)\underline{3}}}+\frac{1+2+3+4}{\overline{)\underline{4}}}+....\\ & =& \sum _1^{\infty } \frac{1+2+3+....+\mathrm{n}}{\overline{)\underline{\mathrm{n}}}}\\ & =& \sum _1^{\infty } \frac{\mathrm{n}(\mathrm{n}+1)}{2 \overline{)\underline{\mathrm{n}}}}\\ & =& \sum _1^{\infty } \frac{\mathrm{n}+1}{2 \overline{)\underline{\mathrm{n}-1}}}\\ & =& \frac{1}{2} \sum _1^{\infty } \frac{(\mathrm{n}-1)+2}{\overline{)\underline{\mathrm{n}-1}}}\\ & =& \frac{1}{2} \sum _1^{\infty } \left[\frac{1}{\overline{)\underline{\mathrm{n}-2}}}+\frac{2}{\overline{)\underline{\mathrm{n}-1}}}\right]\\ & =& \frac{1}{2}\left[\mathrm{e}+2\mathrm{e}\right]=\frac{3\mathrm{e}}{2}\end{array}$