Let $X_1, \ldots, X_n$ be independent binary uniform random variable taking values in $\{0,1\}$. For $a>1$, we define the random variable $Y_n$ as $$Y_n := \sum_{i=1}^n X_i \frac{a^i}{a^{n+1}-1}. $$ We denote the distribution subject to $Y_n$ by $p_n$. I would like to know the limiting distribution $\lim_{n\to \infty} p_n$.
Indeed, when $a=2$, $(2^{n+1}-1)Y_n$ is subject to the uniform distribution on the set $\{0,1, \ldots, 2^{n+1}-1\}$. Hence, the limiting distribution $\lim_{n\to \infty} p_n$ is the uniform distribution on the interval $[0,1]$. However, it is not so easy to solve this for general $a>1$.
More precisely, I would like to know the probability density function of the limiting distribution when it has the limiting distribution.
In particular, I would like to know the answer when $1<a<2$.
Since $a^{n+1}-1$ is close to $a^{n+1}$, we can consider the following variable instead of $Y_n$; $$Z_n := \sum_{i=1}^n X_i a^{i-n-1}=\sum_{i=1}^n X_i a^{-i}. $$
In fact, I would like to know the form of the derivative of the probability density function.
Notice that the support of $Z_n$ is contained in $[0,1/(a-1)]$. Hence, the the support of the limiting distribution is also contained in $[0,1/(a-1)]$. I am interested in the value of the probability density function at the boundary $0$ and $1/(a-1)$.
If $a$ is an integer, then I think $Y_n$ is uniform among "$n$-digit decimals" in base $a$ consisting of only $0$ and $1$.