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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 $$x_{n+1} = - \ln(x_{1} +x_{2}+x_{3}+\dots+x_{n}) = \ln\left(\frac{1}{x_{1} +x_{2}+x_{3}+\dots+x_{n}}\right)$$

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So we have $x_{n+1}=-\ln(x_1-\ln x_1-\ln(x_1-\ln x_1)-\ln(x_1-\ln x_1-\ln(x_1-\ln x_1))-\dots$ –  abiessu Jan 8 at 20:38
The elements of the sequence are not necessarily positive, unless $0\le x_1\lt 1$. If $x_1\gt1$, then $x_2=-\ln(x_1)$ is negative. –  Barry Cipra Jan 8 at 20:41
One simple observation: If the series converges, then $x_n\to0$. It follows from the assumed identity that the sum of the series is $e^0=1$. –  Harald Hanche-Olsen Jan 8 at 20:48

2 Answers 2

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 $x>1$.

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Ah, of course! You nailed it. –  Harald Hanche-Olsen Jan 8 at 20:51
In this case, the partial sum is more than the sum of its parts... –  robjohn Jan 8 at 21:52
Immeasurable comment. –  copper.hat Jan 8 at 21:56

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 $x_1+...+x_n = e^{-x_{n+1}} \to 1$.

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To atone for my earlier contractive sins... –  copper.hat Jan 8 at 21:42

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