Question

Hello, I'm sorry for posting the same question again but I'm not sure if the tutors can see when I reply to an already answered question.

The sequence (a_n) is defined by a_1 = 1/2 and
a_n = a_(n - 1)^2 + a_(n - 1) for n >= 2.

Prove that
1\(a_1 + 1) + 1\(a_2 + 1) + ... + 1\(a_n + 1) < 2 for all n >= 1.

Here's my question regarding a hint that I was given when I last asked the question: How exactly could I show that a_(n+1) > 3/2 a_n? Please let me know what the first step is. Thank you!

Answer 1

a_1 = 1/2
a_2 = 3/4
a_3 = 21/16
Now, since x^2+x > x^2 and a_n > 1 for n > 2
and (x^2+x/x) = x + 1
a_4/a_3 > a_3+1 = 37/16 > 3/2
The a_n are always increasing, and since a_3 >3/2,
a_n > 5/2 for all n > 3
And, a_(n+1)/a_n = a(n) + 1 > 3/2

So, even though a_n is not greater than 3/2 for n=1,2,3 it is greater than 3/2 a_n for every n > 3

Answer 2

Thank you! I understand everything up until this line:

a_4/a_3 > a_3+1 = 37/16 > 3/2

Isn't a_4/a_3 = a_3+1?

Answer 3

Oh never mind, I completely understand the solution now. Thank you so much! This question really stumped me.