The equivalent resistance of the infinite network given below is

The above infinite network can be considered as
 

where R is the equivalent resistance of remaining network. As it is an infinite series, the equivalent resistance of the above circuit also will be R.
Here, R and 1 Ω are in parallel and combination is in series with other two 1 Ω resistance.
So, diagram will be

Total resistance,
$\begin{array}{l}1+1+\frac{\mathrm{R}}{\mathrm{R}+1}=\mathrm{R}\\ \frac{\mathrm{R}}{\mathrm{R}+1}=\mathrm{R}-2\\ \mathrm{R}={\mathrm{R}}^2-\mathrm{R}-2\\ {\mathrm{R}}^2-2\mathrm{R}-2=0\end{array}$
Solving R=(√3+1)Ω, 

so the equivalent resistance is (√3+1)Ω.