Given a sequence of positive real numbers

Question

Given a sequence of positive real numbers (a_n) such that a_n+1 = 2a_n + 3/a_n + 2 holds for all n. Consider the following statements:

(I) If a₁ is a rational number, then so are all a_n.

(II) If a₁ > √3, then a₂ > √3

(III) If a₁ > √3, then a₁ < a₂

(IV) If a₁₀₁ ≠ √3, then a₁₀₀ ≠ √3.

(V) If a_{1 <} √3, then a₂ < √3

How many of the above 5 statements are true?

Answer 1

Let's analyze the function

f(x) = 2x + 3/x + 2 on the domain of positive real numbers x.

We will need its derivative

f'(x) = 2 - 3/x²

First let's compute the minimum value of f(x).

That must occur where its derivative is 0, because it cannot occur at the bounds of its domain:

f(x) goes to infinity when x approaches the bounds of its domain, i.e, when x goes to 0 or x goes to infinity.

f'(x) = 0

2 - 3/x² = 0

x = √(3/2)

The minimum of f is

f(√(3/2)) = 2√(3/2) + 3/√(3/2) + 2 = 2√6 + 2 > √3

That implies that for any a_n, 2a_n + 3/a_n + 2 > √3. (1)

Secondly lets compare f(x) with the function g(x) = x.

For that consider under what conditions f(x) grows faster that g(x), i.e., where its derivative > 1.

f'(x) > 1

2 - 3/x² > 1

x > √3

Let's compare f(√3) with g(√3).

f(√3) = 2√3 + 3/√3 + 2 > √3

Since f(√3) > √3 and for all x > √3, f'(x) > g'(x), we get

for all x > √3, f(x) > x (2)

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(I) It is true, provided a₁ is the very first element of the sequence (i.e., there is no a₀),

because rational operations on rational numbers yield rational numbers.

In contrast, if the sequence starts with a₀, then the statement is false.

For example, a₁ = 8, a₀ = (3+√3)/2.

(II) It is true because of (1).

(III) It is true because of (2).

(IV) It is true because by (1) the premise is false, which makes the implication true.

(V) It is false, provided a₁ is the very first element of the sequence. (I.e., there is no a₀.)

Under that assumption the premise can be made true, while the conclusion is false by (1).

In contrast, if the sequence starts with a₀, then the statement is true:

By (1) the premise is false, hence the whole implication is true.