This is not a recurring decimal even though it is a repeating pattern.
This decimal can be represented by the sum of two series. First write the number as 0.10110011100011110000... + 0.0200220002220000...
We can write this sum as:
(⅟₉)(0.9099009990009999... + 2×(0.090099000999...)).
We can write this as:
(⅟₉)∑[(10ⁿ-1)/10^n²)(1+2/10ⁿ)] where ∑ means “sum of” and is applied to the terms in the square brackets where n is an integer greater than 0.
From this we can see that the nth term is expressed by the quantity in the square brackets. To see this more clearly put n=1 to give us the first term:
(⅟₉)(10-1)/10)(1.2)=0.12=3/25.
Now, n=2:
(⅟₉)(100-1)/10000)(1.02)=0.001122=561/500000;
and n=3:
(⅟₉)(1000-1)/10⁹)(1.002)=0.000000111222=55611/(5×10¹¹).
Add these together: 0.121122111222, a finite decimal.
A recurring decimal always represents one unique fraction, but it’s clear that the fraction given by the formula produces a finite decimal (with n²+n decimal places) and a changing fraction as n increases. Therefore, the given decimal cannot be a repeating (recurring) decimal and it has an infinite number of decimal places as n➝∞.