My solution during the contest: We want to prove by induction: \[ 0<a_n<\frac{1}{\sqrt{2020n}} \]for $n\geq1$. Since $a_0>0$, we have $a_1=\frac{a_0}{\sqrt{1+2020a_0^2}}>0$ and $a_1=\frac{a_0}{\sqrt{1+2020a_0^2}}<\frac{a_0}{\sqrt{2020a_0^2}}=\frac{1}{\sqrt{2020}}$. If $0<a_n<\frac{1}{\sqrt{2020n}}$, we have $a_{n+1}=\frac{a_n}{\sqrt{1+2020a_n^2}}>0$ and \[ a_{n+1}=\frac{a_n}{\sqrt{1+2020a_n^2}}=\frac{1}{\sqrt{\frac{1}{a_n^2}+2020}}<\frac{1}{\sqrt{\frac{1}{\left(\frac{1}{\sqrt{2020n}}\right)^2}+2020}}=\frac{1}{\sqrt{2020(n+1)}} \] Thus $a_{2020}<\frac{1}{\sqrt{2020\cdot2020}}=\frac{1}{2020}$.

A much simpler solution: Note that \[ \frac{1}{a_n^2} = \frac{1}{a_{n-1}^2}+2020\implies \frac{1}{a_n^2}-\frac{1}{a_{n-1}^2} = 2020. \]Taking a telescoping sum, we thus obtain \[ \frac{1}{a_n^2} = \frac{1}{a_1^2}+2020(n-1). \]Now, \[ a_1 = \frac{a_0}{1+2020\cdot a_0^2}<\frac{1}{\sqrt{2020}}\implies \frac{1}{a_1^2}>2020. \]Hence, \[ \frac{1}{a_n^2}>2020n \implies a_n <\frac{1}{\sqrt{2020 n}}. \]Hence the result.

The problem can be restated as $(\tfrac{1}{a_n})^2-(\tfrac{1}{a_{n-1}})^2=2020$. So we get $$\frac{1}{a_{2020}^2} > \frac{1}{a_{2020}^2}-\frac{1}{a_0^2}=(2020)^2$$So $\tfrac{1}{2020}>a_{2020}$ as desired. $\square$

Tintarn wrote: Let $a_0>0$ be a real number, and let $$a_n=\frac{a_{n-1}}{\sqrt{1+2020\cdot a_{n-1}^2}}, \quad \textrm{for } n=1,2,\ldots ,2020.$$Show that $a_{2020}<\frac1{2020}$. Easy.... $(\frac{1}{a_n})^2=b_n\implies b_n=b_{n-1}+2020\implies b_{2020}=b_0+2020^2> 2020^2$ $2020^2< b_{2020}=(\frac{1}{a_{2020}})^2\implies a_{2020}< \frac{1}{2020}$ General Problem:Let $a_0,k,t>0$ be a real number, and let $$a_n=\frac{a_{n-1}}{\sqrt{k+t\cdot a_{n-1}^2}}, \quad \textrm{for } n=1,2,\ldots $$Show that $a_{n}< \frac{1}{\sqrt{t\cdot (k^{n-1}+k^{n-2}+...+1)}}$

Does $a_n$ this sequence convergence ? What is an equivalent of $a_n$ ?

Moubinool wrote: Does $a_n$ this sequence convergence ? What is an equivalent of $a_n$ ? You mean in the original problem? All of the solutions show $a_n<\frac{1}{\sqrt{2020n}}$ so the sequence clearly converges to $0$. In fact, solution #3 shows precisely that $a_n=\frac{1}{\sqrt{2020n+\frac{1}{a_0^2}}}$ so the sequence decreases like $a_n \sim \frac{1}{\sqrt{2020n}}$.

Ok Tintarn I did not read carefully #3

We prove the following claim: Claim: For any $n\geq 1,$ we have that $a_n^2=\frac{a_0^2}{1+2020na_0^2}$. Pf: We use induction. The base case is trivial. If we square both sides of the equation in the problem, we get that $a_n^2=\frac{a_{n-1}^2}{1+2020a_{n-1}^2}$. Assume that $a_n^2=\frac{a_0^2}{1+2020na_0^2}$. Then $a_{n+1}^2=\frac{a_0^2}{1+2020(n+1)a_0^2}$ by calculating. Therefore, the claim is proven. This easily finishes the problem as $a_{2020}^2<\frac{1}{2020^2}.$

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eazy_math wrote: i guess the error is for problem 14 I'm not sure I understand. I don't see any LaTeX error in the statement of Problem 14. If you find one, you can tell me and I will try to fix it immediately.

Note that $a_n^2 = \frac{a_{n-1}^2}{1+2020a_{n-1}^2} \implies \frac{1}{a_n^2} = 2020 + \frac{1}{a_{n-1}^2}$. $\frac{1}{a_{2020}^2} = \frac{1}{a_0^2}+2020^2 \implies a_{2020} = \frac{1}{\sqrt{\frac{1}{a_0^2}+2020^2}} < \frac{1}{2020}$.