Proving the convergence of limit and series Let be the sequence $a_{n+1}=\ln{(1+a_n)}, n\ge1, a_1=1$.
Show that $\lim_{n\to\infty}{a_n}=0$ and the series $\sum_{n=1}^{\infty}{a_n^2}$ converges.
My try:
I assumed that $a_n$ is monotonic (decreasing) and lower bounded by $0$ (intuitively). I denoted $L=\lim_{n\to\infty}{a_n}$, so $L=\ln(1+L)$ which gives $L=0$. If there exists mistakes, please tell me.
How do I prove that the series converges?
$$\sum_{n=1}^{\infty}{a_n^2}=\sum_{n=1}^{\infty}{\ln^2(1+\ln^2(1+\dots+\ln^2(1+1)))}$$
Thank you!
 A: By simple induction we see that $a_n\ge0$ forall $n$ and since
$$\log(1+x)\le x,\quad \forall x\ge0$$
then the sequence $(a_n)$ is decreasing and bounded below so it's convergent to $\ell$ with
$$\ell=\log(1+\ell)$$
hence $\ell=0$.
Now we have
$$a_{n+1}=\log(1+a_n)=a_n-\frac{a_n^2}{2}+o(a_n^2)$$
hence
$$\sum_{k=n}^\infty a_{k+1}-a_k=-a_n\sim-\frac{1}{2}\sum_{k=n}^\infty a_k^2$$
so the series is convergent.
A: Step I. We show first that $\lim_{n\to\infty}a_n=0$.
For every $x>0$, we have that 
$$
0<\log(1+x)<x,
$$
and thus
$$
a_1>a_2>\cdots>a_n>a_{n+1}>\cdots >0,
$$
and thus $a_n$ converges, with $a_n\to x\ge 0$. But $a_{n+1}=\log(1+a_n)\to \log(1+x)$.
Thus $\log(1+x)=x$, which implies that $x=0$. Hence $a_n\to 0$.
Step II. We show now that $\lim_{n\to\infty}n\,a_n=2$.
We have
$$
\frac{1}{a_{n+1}}-\frac{1}{a_n}=\frac{a_n-a_{n+1}}{a_na_{n+1}}=\frac{a_n}{a_{n+1}}\cdot
\frac{1-\frac{\log(1+a_n)}{a_n}}{a_{n}}\to \frac{1}{2},
$$
since standard methods (i.e., l'Hôpital's rule) provide that 
$$
\lim_{x\to 0}\frac{1-\frac{\log(1+x)}{x}}{x}=\frac{1}{2},
$$
and also
$$
\lim_{n\to\infty} \frac{a_n}{a_{n+1}}=\lim_{n\to\infty} \frac{a_n}{\log (1+a_{n})}
=\lim_{x\to 0}\frac{x}{\log(1+x)}=1.
$$ 
Using Stolz–Cesàro theorem
$$
\lim_{n\to\infty}\frac{1}{na_n}=\lim_{n\to\infty}\frac{\frac{1}{a_n}}{n}=
\lim_{n\to\infty}\frac{\frac{1}{a_{n+1}}-\frac{1}{a_n}}{(n+1)-n}=
\lim_{n\to\infty}\left(\frac{1}{a_{n+1}}-\frac{1}{a_n}\right)=\frac{1}{2}.
$$
This implies that the sequence $\{na_n\}$ is bounded, say by $M>0$, and hence 
$$
a_n^2\le \frac{M^2}{n^2},
$$
which implies that $\sum_{n=1}^\infty a_n^2<\infty$.
A: A standard inequality is
$$
x\ne0\implies1+x\lt e^x\tag{1}
$$
Thus,
$$
x_{n+1}=\log(1+x_n)\lt x_n\tag{2}
$$
Furthermore,
$$
x_n\gt0\implies x_{n+1}=\log(1+x_n)\gt0\tag{3}
$$
Thus, $x_n$ is a decreasing sequence bounded below, therefore $\lim\limits_{n\to\infty}x_n$ exists.  By continuity,
$$
\lim_{n\to\infty}x_n=\lim_{n\to\infty}x_{n+1}=\log\left(1+\lim_{n\to\infty}x_n\right)\tag{4}
$$
The contrapositive of $(1)$ says
$$
1+x\ge e^x\implies x=0\tag{5}
$$
$(4)$ and $(5)$ say that
$$
\lim_{n\to\infty}x_n=0\tag{6}
$$
