<s>[INST] $x_1=1,x_n=x_{n+1}+\ln (1+x_{n+1})$, prove $x_n\leq\frac{1}{2^{n-2}}$. $x_1=1,x_n=x_{n+1}+\ln (1+x_{n+1})$, prove


$x_n\leq\frac{1}{2^{n-2}}$.


I proved $0<x_{n+1}<x_n$ by contradiction, and I also get $x_n\geq\frac{1}{2^{n-1}}$ by induction. I tried $\ln(1+x)\geq x-\frac{1}{2}x^2$, but it did not work. Could anyone help me? Thanks!
I tried a new way
$\forall \epsilon>0, \ln(1+x)>(1-\epsilon)x$, then $x_n>(2-\epsilon)x_{n+1}$, thus $x_{n+1}<\frac{1}{(2-\epsilon)2^{n-2}}\to \frac{1}{2^{n-1}}$.
Is it right?
 [/INST] well, I think I misunderstand this question, sorry.
as you said
$x_{n+1}\ge ln(1+x_{n+1})\geq x-\frac{1}{2}x^2$
we can get $x_n\in(\frac{1}{2^{n-1}}, 2-\sqrt{4-2x_{n-1}})$
Thus we need to show $2-\sqrt{4-2x_{n-1}}\le \frac{1}{2^{n-2}}$
in fact, the above $2-\sqrt{4-2x_{n-1}}$ is determined by $ 2-\sqrt{2\sqrt{2\sqrt{2...}}}$ which you can check with $\frac{1}{2^{n-2}} $
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