$a_{n+1}=\frac{a_n}{\sqrt{a_n^2+1}}$. 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!

Note  that $${ a }_{ 2 }=\frac { 1 }{ \sqrt { 2 }  } \\ { a }_{ 3 }=\frac { \frac { 1 }{ \sqrt { 2 }  }  }{ \sqrt { \frac { 3 }{ 2 }  }  } =\frac { 1 }{ \sqrt { 3 }  } \\ { a }_{ 4 }=\frac { \frac { 1 }{ \sqrt { 3 }  }  }{ \sqrt { \frac { 4 }{ 3 }  }  } =\frac { 1 }{ \sqrt { 4 }  }
$$

so $$\\ { a }_{ n }=\frac { 1 }{ \sqrt { n }  } \\ $$
