Let $(a_n)_{n\ge1}$ be a sequence of strictly positive numbers defined by the recursion $$a_{n+1}=\ln\sum_{k=1}^n a_k$$ The problem asks to calculate $$\lim_{n\to\infty} \bigg(\frac{a_{n+1}}{a_n}-1\bigg)e^{a_n}$$ What I managed to do is to prove that $$\lim_{n\to\infty}a_n=\infty$$, which is easy to see because $(a_n)_{n\ge2}$ is strictly increasing and we can thus show by contradiction that $(a_n)_n$ cannot have a finite limit. One would do that by adding $\ln\frac{1}{n}$ in the recursion formula and by taking $\lim_{n\to\infty}$ in the whole equation to get $$l-\infty=\ln l$$, where $l$ is the supposed limit. This however would only be possible for $l=0$ or $l=\infty$, but because of the properties of our sequence the only left option is $l=\infty$.
I know that $$\lim_{n\to\infty}\frac{a_{n+1}}{a_n}=1$$, so the indeterminate form of the original limit would be $0\cdot\infty$, but I have no idea how to prove it. I tried with the Stolz–Cesàro theorem but no luck.
I'm kind of new to this type of mathematics and got this problem at a math contest, destroyed me. I'd prefer an elementary solution, regarding the context. Thank you in advance.
We observe that for $n\ge2$, \begin{align*} a_{n+1}-a_n &= \ln\sum_{k=1}^n a_k - \ln\sum_{k=1}^{n-1} a_k \\ &= \ln \biggl( \sum_{k=1}^n a_k \bigg/ \sum_{k=1}^{n-1} a_k \biggr) \\ &= \ln \biggl( 1 + a_n \bigg/ \sum_{k=1}^{n-1} a_k \biggr) = \ln\Bigl(1+\frac{a_n}{e^{a_n}} \Bigr). \end{align*} Therefore $$ \biggl(\frac{a_{n+1}}{a_n}-1\biggr)e^{a_n} = \frac{e^{a_n}}{a_n} (a_{n+1}-a_n) = \frac{e^{a_n}}{a_n}\ln\Bigl(1+\frac{a_n}{e^{a_n}} \Bigr) = \frac{\ln(1+t_n)}{t_n} $$ where we have set $t_n = \dfrac{a_n}{e^{a_n}}$. Given that $\lim\limits_{n\to\infty} a_n=\infty$, we deduce that $\lim\limits_{n\to\infty} t_n=0$ and hence $$ \lim_{n\to\infty} \biggl(\frac{a_{n+1}}{a_n}-1\biggr)e^{a_n} = \lim_{n\to\infty} \frac{\ln(1+t_n)}{t_n} = \lim_{x\to0} \frac{\ln(1+x)}x = 1 $$ by l'Hôpital's rule.