On average, how many uniformly random real numbers $u$ are needed for their sum to exceed $1$, if $u_1$ is in $(0,1)$ and $u_k$ is in $(0,eu_{k-1})$? A well-known question is: on average, how many uniformly random real numbers in $(0,1)$ are needed for their sum to exceed $1$? The answer is $e$.
Let's tweak this question by making each random number $u_k$, after the first, to be in $(0,eu_{k-1})$.

On average, how many uniformly random real numbers $u$ are needed for their sum to exceed $1$, if $u_1$ is in $(0,1)$ and $u_k$ is in $(0,eu_{k-1})$?

Why $e$? Because anything less than $e$ would make the expectation  to be infinite, and anything greater than $e$ would make the expectation to be finite. But with $e$, I do not know what the expectation would be.
(Here is a more general version of this question on Math SE, currently unanswered.)
EDIT
My statement that "anything less than $e$ would make the expectation to be infinite, and anything greater than $e$ would make the expectation to be finite" was based on invalid reasoning. But it may still be true, based on other reasoning. In any case, my question remains the same as before.
My reasoning (which I should have shown originally) was as follows.
Replace the $e$ with real number $a$, and let $x_k$ be i.i.d $\text{Uniform}(0,1)$-variables. So for $k>1$, we have:
$u_k=u_1 \prod\limits_{i=1}^{k-1}ax_i$
$\log u_k = \log{u_1} + \sum\limits_{i=1}^{k-1}(\log{a}+\log{x_i})$
$\color{red}{E(\log{u_k})}=-1+(k-1)(\log{a}-1)\color{red}{=k(\log{a}-1)-\log{a}}$
$a>e \implies \lim\limits_{k\to\infty}E(\log{u_k})=\infty \implies \lim\limits_{k\to\infty}E(u_k)=\infty$. I thought we could conclude that the expected number of $u$'s for their sum to exceed $1$, is finite. But this is an invalid conclusion, as the following counter-example demonstrates. Consider random variable $v$, where $P(v_1=0)=P(v_1=1)=\frac12$, and $v_{k}=2v_{k-1}$ for $k>1$. We have $\lim\limits_{k\to\infty}E(v_k)=\infty$, but the expected number of $v$'s for their sum to exceed $1$, is not finite.
$a<e \implies E(\log{u_k})<\log{p^k}$ for some $p<1$ and sufficiently large $k$. I thought we could conclude that, since $\sum\limits_{k=s}^\infty p^k$ is less than $1$ for sufficiently large $s$, so $\sum\limits_{k=1}^\infty u_k$ might be less than $1$, so the expected number of $u$'s for their sum to exceed $1$, must be infinite. But to make such a conclusion, what I really needed was $\log{u_k}<\log{p^k}$ for some $p$ and sufficiently large $k$, which I didn't have.
Certainly, if $a\le 1$ then the expectation is infinite.
I thank @ChristianRemling for not taking my claims for granted.
This question seems to be much more difficult than I had thought.
EDIT 2
Answers so far have established that the expectation is indeed finite for $a>e$, and infinite for $a<e$. One approach that I haven't seen yet, would be to use computer simulations to numerically approximate the expectation with values of $a$ that approach $e$ from above, e.g. $3, 2.9, 2.8, 2.75, 2.72,$ etc. This might suggest what the expectation is when $a=e$. (I don't have adequate technology to perform such simulations.)
 A: We can try the method from the second answer to the first question you linked to. I originally thought that would clarify everything, but that was based on a miscalculation. I can now finally prove something (namely, that the integral equation for the expectation that I derive below has a finite solution for $c>e$), though this isn't very close to what the OP asked.
Let $X_j$ be iid and uniformly distributed on $[0,1]$, and
$$
S_n = X_1 +cX_1X_2+ \ldots + c^{n-1} X_1X_2\cdots X_n .
$$
Define $N(x)=\min\{ n\ge 1: S_n\ge x\}$ and $f(x)=EN(x)$. We are then interested in $f(1)$.
Notice that $S_n=X_1(1+cT_{n-1})$; here $T_{n-1}$ is independent of $X_1$ and has the same distribution as $S_{n-1}$. Condition on $X_1$:
$$
E(N(x)|X_1=y)=\begin{cases} 1 & y\ge x\\ 1+ f((x-y)/(cy)) & y<x\end{cases} \quad\quad (1)
$$
We then have
$$
f(x) = \int_0^1 E(N(x)|X_1=y)\, dy ,
$$
and now (1) and a substitution yield
$$
f(x) = 1 + cx \int_{\max \{ 0,(x-1)/c\} }^{\infty} \frac{f(t)}{(1+ct)^2}\, dt . \quad\quad (2)
$$
We can prove that (2) has a solution $f(x)<\infty$ for $c>e$; as Iosif pointed out in a comment below, the significance of this is unclear.
We argue as follows: View (2) as an equation $f=1+Lf$ in the space $X=C^{b}_w([0,\infty))$, that is, functions $g=wh$ with $h$ bounded and continuous and with weight $w(x)=(1+cx)^{\alpha}$ and norm $\|g\|=\|h\|_{\infty}$. Then, for $x\ge 1$,
\begin{align*}
\frac{|(Lg)(x)|}{\|g\|_X} & \le \frac{cx}{(1+cx)^{\alpha}}\int_{(x-1)/c}^{\infty} (1+ct)^{\alpha-2}\, dt \\ & = \frac{x^{\alpha}}{(1-\alpha)(1+cx)^{\alpha}} \le
\frac{1}{(1-\alpha)c^{\alpha}} .
\end{align*}
This can be made $<1$ for $c>e$ by taking $\alpha\to 0+$. A similar calculation works for $0\le x\le 1$. So if $c>e$, we are dealing with a contraction, and thus (2) has a  solution satisfying $f(x)=O(x^{\alpha})$ for all $\alpha>0$, which we can find by iterating (2).
A: Here is a probabilistic argument for OP's statement, showing that the expectation is finite for $a>e$ and infinite for $a<e$.
Setting. Let $(\tau_k)_{k\geq 1}$ denote a sequence of i.i.d. $\text{Exp}(1)$ variables. Let $\mu=\log a$ so that $a = e^{\mu}$, and consider the sum
$$ S_n = \sum_{k=1}^{n} e^{(k-1)\mu - X_k}, \qquad X_k = \sum_{j=1}^{k} \tau_j. $$
Noting that $e^{-\tau_k}\sim\text{Uniform}(0,1)$, this is the same setting as in OP. Then we are interested in the expectation of the stopping time
$$ N(x) = \inf\{ n \geq 1 : S_n > x \}. $$
Case $a < e$. In this case, SLLN tells that
$$ \sqrt[k]{e^{(k-1)\mu - X_k}} \longrightarrow e^{\mu-1} < 1 \qquad\text{a.s.,} $$
hence $S_n$ converges a.s., say, to $S_{\infty}$. Then by noting that
$$ S_{\infty} = e^{-\tau_1}( 1 + e^{\mu} \tilde{S}_{\infty} ) $$
for some $\tilde{S}_{\infty} \stackrel{d}= S_{\infty}$ independent of $\tau_1$, it is easy to show that
\begin{align*}
\mathbf{P}(N(1) = \infty)
&= \mathbf{P}(S_{\infty}\leq 1) \\
&= \int_{0}^{\infty} e^{-t} \mathbf{P}\biggl(S_{\infty} \leq \frac{e^t - 1}{e^{\mu}} \Biggr) \, \mathrm{d}t > 0.
\end{align*}
Therefore, $\mathbf{E}[N(1)] = \infty$.
Case $a > e$. In this case, $\mu > 1$. Define the stopping time $T$ by
$$ T = \inf\{k \geq 1: (k-1)\mu > X_k\}. $$
Then SLLN shows that $T$ is finite a.s. Also, since $S_T > 1$ holds, we have $N(1) \leq T$. Hence,
\begin{align*}
\mathbf{P}(N(1) > n)
&\leq \mathbf{P}(T > n) \\
&\leq \mathbf{P}((n-1)\mu \leq X_n).
\end{align*}
Using $X_n \sim \text{Gamma}(n)$ and applying the Chernoff bound, we can show that the last line is further bounded by $e^{\mu-1-nI(\mu)}$, where $I(\mu)=\mu-1-\log\mu>0$.1) Consequently,
$$ \mathbf{E}[N(1)] = \sum_{n=0}^{\infty} \mathbf{P}(N(1) > n) \leq e^{\mu-1}\sum_{n=0}^{\infty} e^{-nI(\mu)} < \infty. $$

Addendum. Here is a numerical simulation of $\mathbf{E}[N(1)]$ as a function of $\frac{1}{\mu-1}$, where $\mu = \log a$ and $a > e$, using $10^5$ samples:

This seems hinting that $\mathbf{E}[N(1)] \sim (\mu - 1)^{-1}$ as $\mu \to 1^+$ and hence $\mathbf{E}[N(1)] = \infty$ for $a = e$, although I have no idea how to justify this result.

1) My original bound was of the form $\mathcal{O}(e^{-\varepsilon\sqrt{n}})$, which was a consequence of CLT-like estimate. However, large deviation estimates gives better result. Thanks @Pierre PC for pointing this out!
A: The integral equation in Christian Remling's answer can be written:
$$\frac{f(x)-1}{x}=c \int_{\max\{0,(x-1)/c\}}^\infty \frac{f(t)}{(1+ct)^2}dt$$
Differentiating and then multiplying by $x^2$ gives
$$x f'(x)+1-f(x)=-f\left(\frac{x-1}{c}\right)\text{ for }x>1$$
This has a near-solution of
$$f(x)\sim a + \frac{\log(x+\frac{1}{c-2})}{\log c - 1}\ \text{ for }c>e,\ x>1$$
in the sense that it makes the differential equation accurate to order $O(1/x^2)$, and that for $c<e$ the value of $f$ would be negative and not meaningful. We can also get more accuracy with more terms, e.g. $x+\frac{1}{c-2}+\frac{1}{2(c-2)^2(c^2-3)}x^{-1}$ for $O(1/x^3)$.
So the equation as a whole has a near-solution for $c>e$ of
$$f(x)\sim 1+\Big(\!\min\{x,1\}\Big)\Big(a - 1 + \frac{\log(\max\{x,1\}+\frac{1}{c-2})}{\log c - 1}\Big)$$
At $x=1$, this gives
$$f(1)=a+\frac{\log(1+\frac{1}{c-2})}{\log c -1}$$
which suggests that one additional boundary condition might be enough to give a nice approximate answer to the question.
