Study convergence of a recurent sequence

Question

How to study convergence of $(X_n)$, with $X_{n+1}={\sqrt{2X_n+3}}$ and $x_1=\sqrt3$ ?

Answer 1

Hint:\begin{align}x_{n+1}-3 &= \sqrt{2x_n+3}-3 \\ &=\frac{2x_n+3-9}{\sqrt{2x_n+3}+3} \\ &=\frac{2}{\sqrt{2x_n+3}+3}(x_n-3)\end{align}

Answer 2

Clearly we have $x_n\geq \sqrt{3}$. Let see what is with monotonocy of the sequence $$x_{n+1}-x_{n} = \sqrt{2x_n+3}-x_n = {2x_n+3-x_n^2 \over \sqrt{2x_n+3}+x_n }$$ We used here $$\sqrt{a}-\sqrt{b} ={a-b \over \sqrt{a}+\sqrt{b} }$$ Since numerator is $>0$ we see that all depends of denumerator: $$-x_n^2+2x_n+3= -(x_n-3)(x_n+1)$$

Now since $x_n+1>0$ all depend on relation of $x_n$ with $3$ for all $n$. Now if $x_n<3$ then $$2x_n+3\leq 9\implies x_{n+1}< 3$$ Since $x_1< 3$ we see that $x_n< 3$ for all $n$ and so $x_{n+1}\geq x_n$ for all $n$.

So our sequence is increasing and bounded in $[\sqrt{3},3)$ so it is convergent.

Answer 3

Define $e_n=a_n-3$ therefore $$e_{n+1}=a_{n+1}-3=\sqrt{2(e_n+3)+3}-3$$therefore $$a_{n+1}=\sqrt{2e_n+9}-3={2e_n\over 3+\sqrt{2e_n+9}}$$and we have $$|e_{n+1}|\le {2\over 3}|e_n|$$ which means that $e_n\to 0$ and $a_n\to 3$