Here is what I am thinking:
An+1=an^2-1
L=L^2-1
0=L^2-L-1
L=1+\- sqrt 5/2 if An converges
Therefore if it does converge it does not
converge to zero
Is that enough?
Homework Statement
Suppose that a mathematically inclined child plays with a basket containing an infinite subset of integers (with some repetitions). If an integer k is present in the basket then there are initially |k| copies of it. The child pulls out the integers from the basket at...
a1 = real number > 0
a2 = (a1)^2 - 1 = (a1+1)(a1 - 1)
a3 = (a2)^2 - 1 = (a2 + 1)(a2-1)
a(n+1) = (an)^2 - 1 = (an+1)(an-1). This can only converge to 0 if and only if an equals either +1 or -1. However, that seems contradictory, doesn't it? How can something converge to 0 if it must converge...
a1 = real number > 0
a2 = (a1)^2 - 1 = (a1+1)(a1 - 1)
a3 = (a2)^2 - 1 = (a2 + 1)(a2-1)
a(n+1) = (an)^2 - 1 = (an+1)(an-1). This can only converge to 0 if and only if an equals either +1 or -1. However, that seems contradictory, doesn't it? How can something converge to 0 if it must converge...
Hi guys,
I'm new here at this forum, but I don't understand this problem.
Let a1 be a positive real number. Define a sequence an recursively by a(n+1) = (an)^2 - 1. Show that an does not converge to zero.
(Is there a1 such that the sequence an converges to some non-zero value?)
I'm...