Let $u_1$ be a random real number uniformly distributed between $0$ and $1$.
Let $u_k$ be a random real number uniformly distributed between $0$ and $a\times{u_{k-1}}$, for $k>1$ and fixed positive real number $a$.
What is the expected minimum value of $n$ such that $\sum\limits_{k=1}^n u_k>1$, in terms of $a$?
(In other words, on average how many $u$'s are needed for their sum to exceed $1$?)
I did simulations using Excel. For small values of $a$ (e.g. $4$), sometimes the sequence of $u$'s gets "stuck" around very low values, so that many $u$'s are required for their sum to exceed $1$. I suspect the expectation is infinity for low values of $a$, and possibly for all values of $a$.
This is a variation of a simpler question: On average, how many uniformly random real number between $0$ and $1$ are needed for their sum to exceed $1$? The answer to that question is $e$. I tried to apply similar methods to my question, to no avail.
Let $X_i$ be i.i.d. Uniform[0,1], and $S_n=X_1+aX_2+...+a^{n-1}X_n$. Your question is: when $\mathbb{E}\tau<\infty$ where $$ \tau=\min\{n: S_n\ge 1\}. $$ Claim: if $0\le a<1$, then $S=\lim_n S_n<\infty$ with a positive probability; hence $\mathbb{E}\tau=\infty$.
Proof: let (non-random) $N$ be so large that $a^{N}+a^{N+1}+a^{N+2}+\dots=\frac{a^N}{1-a}<1/2$; such an $N$ exists since $a<1$. Now, with some positive probability $S_N<1/2$; and at the same time $$ S-S_N=a^{N} X_{N+1}+a^{N+1} X_{N+2}+...\le a^{N}+a^{N+1}+a^{N+2}+\dots<1/2 $$ As a result, $S<1$.
Next, for $a\ge 1$, $\mathbb{E}\tau<\infty$ as $S$ is clearly monotone in $a$ and we know the answer for $a=1$. In fact, one can compute $\mathbb{E}\tau$ in the same spirit as it is done for $a=1$: indeed, $$ \mathbb{E}\tau=1+\sum_{n=1}^\infty \mathbb{P}(\tau>n)=1+\sum_{n=1}^\infty \mathbb{P}(S_n<1) $$ The collection of $(x_1,...,x_n)$ such that $x_1+ax_2+...+a^{n-1}x_n$, $x_i\ge 0$, is a stretched simplex (by $a^{k-1}$ at side $k=1,2,\dots,n$), hence its volume is $$ \frac{1}{n!}\cdot 1 \cdot \frac{1}{a} \cdot \frac{1}{a^2} ... \cdot \frac{1}{a^{n-1}} $$ so $$ \mathbb{E}\tau=1+\sum_{n=1}^\infty \frac{a^{-n(n-1)/2}}{n!} $$