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!
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
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?
Oh never mind, I completely understand the solution now. Thank you so much! This question really stumped me.
To prove that a_(n+1) > 3/2 a_n, you can start by solving for a_(n+1) and a_n, and then compare their values.
First, let's calculate a_(n+1):
a_(n+1) = a_n^2 + a_n
Next, let's calculate a_n:
a_n = a_(n-1)^2 + a_(n-1)
Now, substitute the value of a_n in the expression for a_(n+1):
a_(n+1) = (a_(n-1)^2 + a_(n-1))^2 + (a_(n-1)^2 + a_(n-1))
Simplifying this expression, we get:
a_(n+1) = a_(n-1)^4 + 2a_(n-1)^3 + a_(n-1)^2 + a_(n-1)
To compare a_(n+1) and a_n, we can subtract a_n from a_(n+1) and simplify the expression:
a_(n+1) - a_n = (a_(n-1)^4 + 2a_(n-1)^3 + a_(n-1)^2 + a_(n-1)) - (a_(n-1)^2 + a_(n-1))
Simplifying further, we get:
a_(n+1) - a_n = a_(n-1)^4 + 2a_(n-1)^3
Now, we want to show that a_(n+1) - a_n > 3/2 a_n. So, let's substitute the value of a_n in the inequality:
a_(n+1) - a_n > 3/2 a_n
a_(n-1)^4 + 2a_(n-1)^3 > 3/2 (a_(n-1)^2 + a_(n-1))
Next, divide both sides of the inequality by a_(n-1)^2 + a_(n-1) (which is greater than 0):
(a_(n-1)^2 + 2a_(n-1)) > 3/2
Now, if we can show that the left side of the inequality is greater than 3/2, then we have proven a_(n+1) > 3/2 a_n.
This is the approach to show that a_(n+1) > 3/2 a_n. To proceed further and complete your proof, you may need to use mathematical induction or other techniques. Let me know if you need assistance with the rest of the proof.