Math
posted by Hannah .
a sequence {an} is given by a1 = (3)^1/2 , an+1 = (3+an)^1/2. Show that {an} is monotonic and bounded, and calculate the limit if it exists

A(n) = √(3 + √(3 + √(3 + ...))..) to infinity
let x = √(3 + √(3 + √(3 + ...))..)
square both sides
x^2 = 3 + √(3 + √(3 + √(3 + ...))..)
or
x^2 = 3 + x
x^2  x  3 = 0
by the formula
x = (1+√13)/2
= appr. 2.302775638... which I got by repeatedly doing the following on my calculator about 20 times
1. √3
2. =
3. +3
4. =
5. √
goto 2.