We have a sequence $a_1=\sqrt{6}\\a_2=\sqrt{6+\sqrt{6}}\\a_3=\sqrt{6+\sqrt{6+\sqrt{6}}}\\...$a)Find a recursion formula for $a_{n+1}$
b)Find a limit
Attempt: a) Tried finding the recursion formula: $a_{n+1}=\sqrt{6+a_n}$ I am not not sure about it because the problem does not say where n starts. So if n starts at zero $a_0$ is not defined. Or is it implied that n starts at 1 since the sequence starts with $a_1$.
b)Right away I assume that $a_{n}$ converges to some L which I will find. $n+1$ goes to infinity when n goes to infinity. So I asumme that $a_{n+1}$ converges to the same L. As a result I obtain the following: $L=\sqrt{6+L}\\-L^2+L+6=0$ Solving for L i get $L=3,-2$. What would be an argument that L converges to 3, but not to 2. Thanks.