If $\mathrm{x}=\frac{2}{5}+\frac{1.3}{2!}{\left(\frac{2}{5}\right)}^2+\frac{1.3.5}{3!}{\left(\frac{2}{5}\right)}^3+\dots \dots$ then $\mathrm{x}+\frac{1}{\mathrm{x}}=$

$\begin{array}{rcl}1+\mathrm{x}& =& 1+\left(\frac{2}{5}\right)+\frac{1.3}{2!}{\left(\frac{2}{5}\right)}^2+...\end{array}$

$\begin{array}{rcl} & =& {(1-\mathrm{x})}^{-\mathrm{p}/\mathrm{q}} \mathrm{where} \mathrm{p}=1, \mathrm{q}=2,\frac{\mathrm{x}}{\mathrm{q}}=\frac{2}{5}\end{array}$

$\begin{array}{rcl} & =& {\left(1-\frac{4}{5}\right)}^{-1/2} \Rightarrow \mathrm{x}=\frac{4}{5}\end{array}$

$\begin{array}{rcl}1+\mathrm{x}& =& \sqrt{5}\end{array}$

$\begin{array}{rcl}\mathrm{x}& =& \sqrt{5}-1\end{array}$

$\begin{array}{rcl}\frac{1}{\mathrm{x}}& =& \frac{1}{\sqrt{5}-1}=\frac{\sqrt{5}+1}{4}\end{array}$

$\begin{array}{rcl}\mathrm{x}+\frac{1}{\mathrm{x}}& =& \sqrt{5}-1+\frac{\sqrt{5}+1}{4}=\frac{5\sqrt{5}-3}{4}\end{array}$