$\frac{1.4}{0!}+\frac{2.5}{1!}+\frac{3.6}{2!}+\dots \dots =$

$\begin{array}{rcl} & & \frac{1.4}{\overline{)\underline{0}}}+\frac{2.5}{\overline{)\underline{1}}}+\frac{3.6}{\overline{)\underline{2}}}+\dots \dots \\ & =& \sum _1^{\infty } \frac{{\mathrm{n}}^2+3\mathrm{n}}{\overline{)\underline{\mathrm{n}-1}}}\\ & =& \sum _1^{\infty } \frac{({\mathrm{n}}^2-1)+1+3(\mathrm{n}-1)+3}{\overline{)\underline{\mathrm{n}-1}}}\\ & =& \sum _1^{\infty } \left(\frac{\mathrm{n}+1}{\overline{)\underline{\mathrm{n}-2}}}+\frac{4}{\overline{)\underline{\mathrm{n}-1}}}+\frac{3}{\overline{)\underline{\mathrm{n}-2}}}\right)\\ & =& \sum _1^{\infty } \left(\frac{1}{\overline{)\underline{\mathrm{n}-3}}}+\frac{3}{\overline{)\underline{\mathrm{n}-2}}}\right)+7\mathrm{e}\\ & =& \mathrm{e}+3\mathrm{e}+7\mathrm{e}=11\mathrm{e}\end{array}$