Yes, every decimal expansion corresponds to a real number.
Specifically, if we have
- a finite sequence $a_1 a_2 a_3 {\tiny \ldots} a_n$ of decimal digits, and
- an infinite sequence $b_1 b_2 b_3 {\tiny \ldots}$ of decimal digits,
then the decimal expansion $a_1 a_2 a_3 {\tiny \ldots} a_n . b_1 b_2 b_3 {\tiny \ldots}$ is the decimal expansion of a real number. Specifically, the decimal expansion $a_1 a_2 a_3 {\tiny \ldots} a_n . b_1 b_2 b_3 {\tiny \ldots}$ is defined as the real number
$$\left (\sum_{i=1}^n a_i 10^{n - i} \right) + \left (\sum_{i=1}^\infty b_i 10^{- i} \right).$$
The series on the right must converge because its terms are bounded by the exponentially decreasing sequence $0.9, 0.99, 0.999, \ldots$.
If you have a number with a finite sequence of digits after the decimal point, then you can turn that into an infinite sequence by appending infinitely many $0$s on the end.
Note that it's not permissible to have infinitely many digits before the decimal point. It's also not permissible to have digits that are "infinitely far beyond the decimal point"—in other words, there may be infinitely many digits after the decimal point, but each individual digit has only finitely many digits before it. (Hence why there is no "last digit of $\pi$", or "infinitieth digit of $\pi$", and why there's no such number as $0.999\ldots9$, with infinitely many $9$s between the first $9$ and the last one.)