<s>[INST] 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.
 [/INST] 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 $\lim\limits_{r\to\infty}$.
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