Sum of logarithmic series

Question

Let $x_1>0$ we define sequence $(x_n)$ with formula $x_{n+1}=-\ln(x_1+x_2+\cdots+x_n)$ Find sum of the series $\sum_{n=1}^\infty x_n$. How to deal which such summation with logarithms?

Answer 1

We note that the argument of the $\ln$ is just a partial sum of the series. We know that if the series converges, the terms must converge to zero, so the argument of the $\ln$ must go to to $1$, so the sum must go to $1$.

To establish convergence, we are given $x_1 =\sum_{n=1}^1 x_n \gt 1$ as the base case of an induction. Assume $\sum_{n=1}^k x_n \gt 1$ Then $x_{k+1} \lt 0$ so the sums are decreasing and $\sum_{n=1}^{k+1} x_n =\sum_{n=1}^k x_n-\ln (\sum_{n=1}^k x_n)\gt \sum_{n=1}^k x_n-( \sum_{n=1}^k x_n-1)=1$ where we used that $\log (1+y) \lt y$. As the sum is decreasing and bounded below, it converges.