Prove that the series is convergent and calculate the sum.

Let's say that $$x_{1} \gt 0$$ We define the sequence by the formula $$x_{n+1} = - \ln(x_{1} +x_{2}+x_{3}+\dots+x_{n}) $$ Prove that the series $$\sum_{n=2}^{ \infty } x_{n}$$ is convergent and find the sum of it.

My attempt was to use the identity $$\ln(1+x)\lt x$$ somehow, but without any results. I've also determined that the elements of the sequence are positive and that $$x_{n+1} = - \ln(x_{1} +x_{2}+x_{3}+\dots+x_{n}) = \ln\left(\frac{1}{x_{1} +x_{2}+x_{3}+\dots+x_{n}}\right)$$

Let $s_n=\sum _{k=1}^n x_k$ for $n\geq 1$. Then $s_{n+1}=s_n-\log(s_n)$. So it comes down to showing that iterating $x\mapsto x-\log(x)$ converges when you start with some positive $x$. Now use that $x-\log(x) \geq 1$ with equality only for $x=1$ and that $1<x-\log(x)<x$ for $x>1$.

We have $e^{-x_{n+1}} = x_1+...+x_n \ge 1-x_{n+1}$. Hence $x_1+...+x_n + x_{n+1} \ge 1$, and consequently, $x_{n+1} \le 0$ for all $n>1$.

Hence the sequence $x_1+...+x_{n+1}$ is non-increasing and bounded below, so it converges. Since it converges, $x_n \to 0$, from which it follows that $x_1+...+x_n = e^{-x_{n+1}} \to 1$.