Can you write a Geometric random variable as some combination of Bernoulli random variables? Background
Given $Y \sim \text{Binomial(n,p)}$, we can write $Y = \sum_{i=1}^{n} X_i$ where $X_1,X_2,...,X_n$ are iid $\text{Bernoulli}(p)$. This is useful in, for example, determining the mean of a binomial random variable:
$$E(Y)=E\left(\sum X_i\right) = \sum E(X_i) = np$$
Question
If we are given $Y \sim \text{Geometric(p)}$, can we similarly write $Y$ as some combination of Bernoulli random variables?
 A: For clarity, I am going to look at the version of the geometric distribution with support on the non-negative integers, with expected value $\mathbb{E}(Y) = (1-p)/p$.  Now, suppose we have a sequence of Bernoulli random variables $X_1,X_2,X_3,... \sim \text{IID Bern}(p)$ to use for the construction.
The geometric random variable $Y$ can be interpreted as the number of "failures" that occur before the first "success", so it can be written as:
$$\begin{align}
Y 
&\equiv \max \ \{ y = 0,1,2,... | X_1 = \cdots = X_{y} = 0 \} \\[12pt]
&= \max \Bigg\{ y = 0,1,2,... \Bigg| \prod_{\ell = 1}^{y} (1-X_\ell) = 1 \Bigg\} \\[6pt]
&= \sum_{i=1}^\infty \prod_{\ell = 1}^{i} (1-X_\ell). \\[6pt]
\end{align}$$
This is probably the "simplest" you can write the expression, since it accords with the descriptive intuition of what the random variable represents.  As whuber notes in the comments, every discrete distribution can be written as a mixture of a countable number of shifted or scaled Bernoulli random variables (and indeed, there are an infinite number of ways to do this).
(Note: For the other version of the geometric distribution, with support on the positive integers, the random variable can be interpreted as the number of trials that occur by the time of the first "success", which is one more than the number of "failures".  In this case you just add one to the above expression to get construct the random variable.)

Confirming the result: To see that this expression is adequate, note that:
$$\begin{align}
F_Y(y) \equiv \mathbb{P}(Y \leqslant y)
&= 1 - \mathbb{P}(Y \geqslant y+1) \\[12pt]
&= 1 - \mathbb{P} \Bigg( \prod_{\ell = 1}^{y+1} (1-X_\ell) = 1 \Bigg) \\[6pt]
&= 1 - \prod_{\ell = 1}^{y+1} \mathbb{P}(1-X_\ell = 1) \\[6pt]
&= 1 - \prod_{\ell = 1}^{y+1} \mathbb{P}(X_\ell = 0) \\[6pt]
&= 1 - \prod_{\ell = 1}^{y+1} (1-p) \\[6pt]
&= 1 - (1-p)^{y+1}, \\[6pt]
\end{align}$$
which is the CDF of the (chosen version of the) geometric distribution.
