$\frac{1+{\displaystyle \frac{2^2}{2!}}+{\displaystyle \frac{2^4}{3!}}+{\displaystyle \frac{2^6}{4!}}+\dots \dots \infty }{1+{\displaystyle \frac{1}{2!}}+{\displaystyle \frac{2}{3!}}+{\displaystyle \frac{2^2}{4!}}+\dots \dots \infty }=$

$\begin{array}{rcl} & & \frac{1+{\displaystyle \frac{2^2}{\overline{)\underline{2}}}}+{\displaystyle \frac{2^4}{\overline{)\underline{3}}}}+{\displaystyle \frac{2^6}{\overline{)\underline{4}}}}+\dots \dots \infty }{1+{\displaystyle \frac{1}{\overline{)\underline{2}}}}+{\displaystyle \frac{2}{\overline{)\underline{3}}}}+{\displaystyle \frac{2^2}{\overline{)\underline{4}}}}+\dots \dots \infty }\\ & =& \frac{{\displaystyle \frac{1}{4}}\left(2^2+{\displaystyle \frac{{\left(2^2\right)}^2}{\overline{)\underline{2}}}}+{\displaystyle \frac{{\left(2^2\right)}^3}{\overline{)\underline{3}}}}+{\displaystyle \frac{{\left(2^2\right)}^4}{\overline{)\underline{4}}}}+....\infty \right)}{{\displaystyle \frac{1}{4}}\left(2^2+{\displaystyle \frac{2^2}{\overline{)\underline{2}}}}+{\displaystyle \frac{2^3}{\overline{)\underline{3}}}}+{\displaystyle \frac{2^4}{\overline{)\underline{4}}}}+....\infty \right)}\\ & =& \frac{\left({\mathrm{e}}^{2^2}-1\right)}{({\mathrm{e}}^2+1)}=\frac{\left({\mathrm{e}}^2+1\right) \left({\mathrm{e}}^2-1\right)}{{\mathrm{e}}^2+1}=(\mathrm{e}+1)(\mathrm{e}-1)\end{array}$