• Support PF! Buy your school textbooks, materials and every day products Here!

Proving an Increasing Sequence (a question about the answer)

  • Thread starter student34
  • Start date
  • #1
346
3

Homework Statement



Question from my professor: "Consider the sequence {a(n)} (from n=1 to ∞) defined inductively by a(1) = 0, and a(n+1) = √(a(n) + 2) for n ≥ 1. Prove that {a(n)} (from n=1 to ∞) is increasing".

Here's the first part of the answer from my professor: "Consider a(n+1)^2 − (a(n))^2 = a(n) + 2 − (a(n))^2 = −(a(n)^2 − a(n) −2) = −(a(n) − 2)(a(n) + 1). Note that a(n) ≥ 0 for all n. For a(n) ∈ [−1,2] we see that −(a(n) −2)(a(n) + 1) ≥ 0. The initial element a(1) = 0 belongs to the interval [0,2]".

How does he know that a(n) ≥ 0 for all n? Is this a given of some sort?
 
Last edited:

Answers and Replies

  • #2
CompuChip
Science Advisor
Homework Helper
4,302
47
All a(n) are defined as the square root of their predecessor, and the square root is always non-negative.

If you want to be completely rigorous, you can prove it inductively: assume that a(k) ≥ 0 for all k ≤ n. Then in particular a(n) ≥ 0, therefore a(n) + 2 ≥ 2 (> 0), so a(n + 1) is the square root of a non-negative number which is itself non-negative. And the base step -- a(1) ≥ 0 -- is trivial.
 
  • #3
346
3
All a(n) are defined as the square root of their predecessor, and the square root is always non-negative.

If you want to be completely rigorous, you can prove it inductively: assume that a(k) ≥ 0 for all k ≤ n. Then in particular a(n) ≥ 0, therefore a(n) + 2 ≥ 2 (> 0), so a(n + 1) is the square root of a non-negative number which is itself non-negative. And the base step -- a(1) ≥ 0 -- is trivial.
Why did you start off with this, a(k) ≥ 0 for all k ≤ n? I generally understand induction, but this use of k is unfamiliar to me.
 
  • #4
CompuChip
Science Advisor
Homework Helper
4,302
47
Actually "assume that a(n) ≥ 0 for some n" is enough.
 
  • #5
346
3
Actually "assume that a(n) ≥ 0 for some n" is enough.
Thank-you very much for helping me
 

Related Threads on Proving an Increasing Sequence (a question about the answer)

  • Last Post
Replies
2
Views
1K
Replies
7
Views
2K
  • Last Post
Replies
0
Views
843
  • Last Post
Replies
7
Views
746
  • Last Post
Replies
3
Views
701
  • Last Post
Replies
5
Views
1K
  • Last Post
Replies
11
Views
2K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
3
Views
1K
Replies
0
Views
2K
Top