1

If $S_0=1$ and $S_n=1-e^{-S_{n-1}}$ then prove that $0\le S_n\le1$ and $S_n$ converges.

My Solution: $S_0=1, \ S_n=1-e^{-S_{n-1}} \implies S_1=1-{1\over e}<S_0\\ Now, S_{n+1}-S_n=1-e^{-S_{n}}-(1-e^{-S_{n-1}})=e^{-S_{n-1}}-e^{-S_{n}}<0 \ \\ [Since, S_1<S_0 \implies e^{S_1}<e^{S_0} \implies e^{-S_0}-e^{-S_1}<0]\\ Hence, S_{n+1}<S_n\\ \text{Therefore, it's monotonic decreasing for all }n\\\text{How can I show this is of positive terms?} \\\text{Please help, Thanks in advance. }$

Chris
  • 748

1 Answers1

0

$e^{-x} \leq 1$ if $x \geq 0$. Hence, $S_n \geq 0$ implies $S_{n+1}=1-e^{-S_n} \geq 0$. By induction $S_n \geq 0$ for all $n$.

Kavi Rama Murthy
  • 311,013
  • How can we we say from this that $S_n>0$? As I have proved it's monotonically decreasing So it can include negative terms too. I need help on proof. – Chris Apr 21 '21 at 08:42
  • @Chris prove by induction. We know that $S_0=1\geq 0$. I proved that $S_n \geq 0$ implies $S_{n+1} \geq 0$. – Kavi Rama Murthy Apr 21 '21 at 08:44