Finding the Limit of a Series: [a1]=√12, [an+1]=√(12+an)

[jaqueh]

Homework Statement

What does the series converge to?
[a₁]=√12
[a_n+1]=√(12+a_n)

Homework Equations

Let L = the limit it approaches

The Attempt at a Solution

I don't know if i did this correctly but I made
L = √(12+√(12+√(12+...)))
then L² = 12+√(12+√(12+...))
then L^∞ = 12^∞ + 12^∞-1...12
thus L = ∞√(12^∞ + 12^∞-1...12)

 

[Dick]

If it converges then a_n approaches L as n->infinity. So does a_(n+1). Put that into a_n=sqrt(12+a_(n+1)).

 

[jaqueh]

If it converges then a_n approaches L as n->infinity. So does a_(n+1). Put that into a_n=sqrt(12+a_(n+1)).

ok i get that a_n+1 converges now because it is in the a_n series, but what is the limit that it approaches?

 

[Dick]

ok i get that a_n+1 converges now because it is in the a_n series, but what is the limit that it approaches?

It approaches the same limit as a_n, call it L. Solve for L!