If $1+\frac{1}{2!}+\frac{1.3}{4!}+\frac{1.3.5}{6!}+\dots \dots \infty =\mathrm{l},$

$1+\frac{2}{2!}+\frac{2+4}{3!}+\frac{2+4+6}{4!}+\dots \dots \infty =\mathrm{m},$

$\frac{1}{2!}+\frac{1+2}{3!}+\frac{1+2+3}{4!}+\dots \dots \infty =\mathrm{n}$ then the ascending order of l, m, n is

 

$\begin{array}{rcl}\mathrm{l}& =& 1+\frac{1}{\overline{)\underline{2}}}+\frac{1.3}{\overline{)\underline{4}}}+\frac{1.3.5}{\overline{)\underline{6}}}+\dots \dots \infty \\ & =& \sum _0^{\infty } \frac{1.3.5...(2\mathrm{n}-1)}{\overline{)2\mathrm{n}}}\\ & =& \sum _0^{\infty } \frac{1.3.5....(2\mathrm{n}-1)}{1.2.3.4 (2\mathrm{n}-1) \left(2\mathrm{n}\right)}\\ & =& \sum _0^{\infty } \frac{1}{2.4.6..... \left(2\mathrm{n}\right)}\\ & =& \sum _0^{\infty } \frac{1}{2^{\mathrm{n}}.(1.2.3.....\mathrm{n})}\\ & =& \sum _0^{\infty } \frac{{\left({\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right)}^{\mathrm{n}}}{\overline{)\underline{\mathrm{n}}}}={\mathrm{e}}^{1/2}\end{array}$

$\begin{array}{rcl}\mathrm{m}& =& \sum \frac{2+4+6+...+2\mathrm{n}}{\overline{)\underline{\mathrm{n}}}}\\ & =& 2\sum \frac{(1+2+3+...+\mathrm{n})}{\overline{)\underline{\mathrm{n}}}}\\ & =& 2\sum \frac{\mathrm{n}(\mathrm{n}+1)}{2\overline{)\underline{\mathrm{n}}}}\\ & =& \sum \frac{(1+1)}{\overline{)\underline{\mathrm{n}-1}}}\\ & =& \sum _1^{\infty } \frac{(\mathrm{n}-1)+2}{\overline{)\underline{\mathrm{n}-1}}}=\underset{1}{\overset{\infty }{\sum }}\left(\frac{1}{\overline{)\underline{\mathrm{n}-2}}}+\frac{2}{\overline{)\underline{\mathrm{n}-1}}}\right)\\ & =& (\mathrm{e}+2\mathrm{e})=3\mathrm{e}\end{array}$

$\begin{array}{rcl}\mathrm{n}& =& \sum _2^{\infty } \frac{1+2+3+....+(\mathrm{n}-1)}{\overline{)\underline{\mathrm{n}}}}\\ & =& \sum _2^{\infty } \frac{(\mathrm{n}-1)\mathrm{n}}{2 \overline{)\underline{\mathrm{n}}}}=\sum _2^{\infty }\frac{1}{2\overline{)\underline{\mathrm{n}-2}}}\\ & =& \frac{1}{2}\left(\sum _2^{\infty }\frac{1}{\overline{)\underline{\mathrm{n}-2}}}\right)\\ & =& \frac{1}{2}\left(\mathrm{e}\right)\end{array}$

Ascending order = $l, n, m$