The sum of the infinite series $1+\left(1+\frac{{\displaystyle 1}}{{\displaystyle 2}}\right)\left(\frac{{\displaystyle 1}}{{\displaystyle 3}}\right)+\left(1+\frac{{\displaystyle 1}}{{\displaystyle 2}}+\frac{{\displaystyle 1}}{{\displaystyle 2^2}}\right)\left(\frac{{\displaystyle 1}}{{\displaystyle 3^2}}\right)+....\mathrm{\infty }$

$\begin{array}{l}\mathrm{S}=1+\left(1+\frac{1}{2}\right)\cdot \frac{1}{3}+\left(1+\frac{1}{2}+\frac{1}{2^2}\right)\cdot \frac{1}{3^2}+\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}\right)\cdot \frac{1}{3^3}+......\\ \frac{\frac{\mathrm{S}}{3}=\frac{1}{3}+\left(1+\frac{1}{2}\right)\cdot \frac{1}{3^2}+\left(1+\frac{1}{2}+\frac{1}{2^2}\right)\cdot \frac{1}{3^3}+\dots \dots }{\frac{2\mathrm{S}}{3}=1+\frac{1}{2\cdot 3}+\frac{1}{2^2\cdot 3^2}+\frac{1}{2^3\cdot 3^3}+\dots \dots \dots \mathrm{\infty }}\\ \frac{2\mathrm{S}}{3}=1+\frac{1}{6}+\frac{1}{6^2}+\frac{1}{6^3}+\dots \dots \mathrm{\infty }\\ \frac{1}{1-\frac{1}{6}}=\frac{6}{5}\Rightarrow \mathrm{S}=\frac{9}{5}\\ \end{array}$