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.
 A: 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}
$.
A: Hint: to finish $(a)$. You already proved that $a_{n+1}<\frac{a_n^2}{2}$, use it further, i.e.
$$0<a_{n+1}<\frac{a_n^2}{2}<\frac{1}{2}\cdot\left(\frac{a_{n-1}^2}{2}\right)^{2}=
\frac{a_{n-1}^4}{2^3}<\\
\frac{1}{2^3}\left(\frac{a_{n-2}^2}{2}\right)^4=
\frac{a_{n-2}^8}{2^7}<\\
\frac{1}{2^7}\left(\frac{a_{n-3}^2}{2}\right)^8=
\frac{a_{n-3}^{16}}{2^{15}}$$
The patterns reveals as
$$0<a_{n+1}<\frac{a_{n-r}^{2^{r+1}}}{2^{2^{r+1}-1}} \Rightarrow
0<a_{k+r+1}<\frac{a_{k}^{2^{r+1}}}{2^{2^{r+1}-1}} \tag{1}$$ 
Since you also proved $\lim\limits_{n\to \infty}a_n=0$, then $0<a_k<1$ from some $k_0$ onwards, thus
$$0<a_{k_0+r+1}<\frac{1}{2^{2^{r+1}-1}} \Rightarrow \\
0<(k_0+r+1)^{2019}a_{k_0+r+1} < \frac{(k_0+r+1)^{2019}}{2^{2^{r+1}-1}} =\\
\frac{r^{2019}}{2^{2^{r+1}-1}}\cdot \left(\frac{k_0}{r}+1+\frac{1}{r}\right)^{2019} $$
and take the $\lim\limits_{r\to\infty}$.
