$a_{n+1}=\frac{a_n}{\sqrt{a_n^2+1}}$. Find $a_n$. Given $\{a_n\}$: $a_{n+1}=\frac{a_n}{\sqrt{a_n^2+1}}$, $a_1=1$.
Find $a_n$ like a function of $n$.
My trying.
Let $a_n=\tan\alpha_n$, where $\alpha_n\in\left(0,\frac{\pi}{2}\right)$.
Hence, $a_{n+1}=\sin\alpha_n$ and what is the rest?
Thank you!

Squaring gives,
$$a_{n+1}^2=\frac{a_n^2}{1+a_n^2}$$
Let $b_n=a_n^2$ then we have,
$$b_{n+1}=\frac{b_{n}}{1+b_{n}}$$
$$=\frac{1+b_n-1}{1+b_n}$$
$$=1-\frac{1}{1+b_n}$$
$$=1-\frac{b_{n+1}}{b_n}$$
$$=\frac{b_{n}-b_{n+1}}{b_{n}}$$
And hence,
$$\frac{b_{n}-b_{n+1}}{b_{n}b_{n+1}}=1$$
Thus,
$$\sum_{n=1}^{N-1} \frac{b_{n}-b_{n+1}}{b_{n}b_{n+1}}=N-1$$
$$\sum_{n=1}^{N-1} \left(\frac{1}{b_{n+1}}-\frac{1}{b_{n}} \right)=N-1$$
$$\frac{1}{b_{N}}-\frac{1}{b_1}=N-1$$
$$b_{N}=\frac{1}{N-1+1}$$
$$a_{N}=\sqrt{\frac{1}{N-1+1}}$$
$$a_{N}=\frac{1}{\sqrt{N}}$$
