Find the limit of $a_1 = 1 $ , $ a_{n+1} = \frac{\sqrt{1+a_n^2}-1}{a_n}$ , $n \in \mathbb{N}$ $a_1 = 1 $ ,  $ a_{n+1} = \frac{\sqrt{1+a_n^2}-1}{a_n}$ , $n \in \mathbb{N}$
It is easy to prove that the limit exists: the boundary of that expression is $0$ and it is monotonically decreasing.
The problem is to actually find the limit (it is $0$) because if I take arithmetic of limits:
$ \lim_{x \to +\infty} a_{n+1} =  \frac{\sqrt{1+ \lim_{n \to +\infty} a_n^2+1}}{\lim_{n \to +\infty} a_n}$
$g = \frac{\sqrt{1+g^2+1}}{g}  \implies g^2 = {\sqrt{1+g^2}-1} \implies g^2 + 1 = {\sqrt{1+g^2}} \implies 0 = 0$
 A: If you have already proved that the limit exists and is greater than or equal to $0$, then assume by way of contradiction that
$$\lim_{n\to\infty}a_n=L>0$$
Then following your work (which is all valid since $L>0$) we get
$$L^2+1=\sqrt{L^2+1}$$
$$L^4+2L^2+1=L^2+1$$
$$L^4+L^2=0$$
$$L^2(L^2+1)=0$$
This has solutions $L=0$, $L=i$, and $L=-i$. Since $L$ is a positive real number, we have reached a contradiction. Thus, $L$ is a non-negative non-positive real number. Of course, the only such number is $L=0$. We conclude
$$\lim_{n\to\infty}a_n=0$$
A: The expression in the right hand side inspires me to use $$a_m=\tan(b_m),-\dfrac\pi2\le b_m<\dfrac\pi2$$
$m=1\implies b_1=?$
$$a_{n+1}=\tan(b_{n+1})=\cdots=\tan\dfrac{b_n}2=\cdots=\tan\dfrac{b_{n+1-r}}{2^r},0\le r\le n$$
Set $r=n$ and then $n\to\infty$
A: You can prove it directly like this
$$a_{n+1} = \frac{\sqrt{1+a_n^2}-1}{a_n} < \frac{\sqrt{1+a_n^2+\frac{1}{4}a_n^4}-1}{a_n} = \frac{1+\frac{1}{2}a_n^2-1}{a_n} = \frac 12 a_n$$
Therefore $a_n \to 0$, $n\to \infty$.
A: $$a_{n+1} = \frac{\sqrt{1+a_n^2}-1}{a_n}=\\ \frac{\sqrt{1+a_n^2}-1}{a_n}.\frac{\sqrt{1+a_n^2}+1}{\sqrt{1+a_n^2}+1}=\\\frac{1+a_n^2-1}{a_n(\sqrt{1+a_n^2}+1)}=\\
\frac{a_n}{(\sqrt{1+a_n^2}+1)}\leq \frac{a_n}{(\sqrt{1}+1)}=\frac{a_n}{2} $$so $$0<a_{n+1}\leq \frac 12a_n\leq\frac14 a_{n-1}\leq\frac18a_{n-2}\cdots<1$$
