A sequence $(a_n)_{n\ge 1}$ such that $a_1>0$ and $a_{n+1}=a_n-\ln(1+a_n)$ Let $(a_n)_{n\ge 1}$ be a sequence  such that $a_1>0$ and $$a_{n+1}=a_n-\ln(1+a_n)$$
a) Prove that $a_{n+1}<\frac{a_n^2}{2}, \forall n\in \mathbb{N}$ and $\lim\limits_{n\to \infty} (n^{2019}a_n)=0$.
b) If $a_1<2$, prove that $a_{12} \in (0,10^{-300})$.
It is easy to see that $a_n >0$, $\forall n\in \mathbb{N}$. I considered the function $f:(0,\infty)\to \mathbb{R}$, $f(x)=x-\ln(1+x)-\frac{x^2}{2}$ and I could easily prove that this function is strictly decreasing, so it follows that $$x-\ln(1+x)<\frac{x^2}{2}, \forall x>0 \tag{*}$$
Now from $(*)$ we get that $a_{n+1}<\frac{a_n^2}{2}, \forall n\in \mathbb{N}$.
I couldn't make much further progress. I could show that $\lim\limits_{n\to \infty}a_n=0$, but then I got stuck.

From
$ a_{n+1}
<\frac{a_n^2}{2}
$
we get
$ a_{n+2}
<\frac{a_{n+1}^2}{2}
<\frac{(\frac{a_{n}^2}{2})^2}{2}
=\frac{a_{n}^4}{8}
=\frac{a_{n}^4}{2^3}
$
and
$ a_{n+3}
<\frac{a_{n+2}^2}{2}
<\frac{(\frac{a_{n}^4}{2^3})^2}{2}
=\frac{a_{n}^8}{2^7}
$.
By induction
we can show that
$ a_{n+m}
<\frac{a_{n}^{2^m}}{2^{2^{m}-1}}
$.
If this holds that
$ a_{n+m+1}
<\frac{a_{n+m}^2}{2}
<\frac{(\frac{a_{n}^{2^m}}{2^{2^{m}-1}})^2}{2}
=\frac{a_{n}^{2^{m+1}}}{2^{2^{m+1}-2+1}}
=\frac{a_{n}^{2^{m+1}}}{2^{2^{m+1}-1}}
$.
If $a_1 < 2$
then,
since
$x-\ln(1+x)$
is increasing,
$a_2
< 2-\ln(3)
< .902
$
If $m=10$ and $n=2$ then
$ a_{12}
<\frac{a_2^{2^{10}}}{2^{2^{10}-1}}
<\frac{.902^{1024}}{2^{1023}}
<1.501 × 10^{-354}
$.
