# Infinite series

1. Dec 7, 2011

### jaqueh

1. The problem statement, all variables and given/known data
What does the series converge to?
[a1]=√12
[an+1]=√(12+an)

2. Relevant equations
Let L = the limit it approaches

3. 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)

File size:
23.1 KB
Views:
48
2. Dec 7, 2011

### 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)).

3. Dec 7, 2011

### jaqueh

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

4. Dec 7, 2011

### Dick

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