a sequence {an} is given by a1 = (3)^1/2 , an+1 = (3+an)^1/2. Show that {an} is monotonic and bounded, and calculate the limit if it exists
A(n) = √(3 + √(3 + √(3 + ...))..) to infinity
let x = √(3 + √(3 + √(3 + ...))..)
square both sides
x^2 = 3 + √(3 + √(3 + √(3 + ...))..)
or
x^2 = 3 + x
x^2 - x - 3 = 0
by the formula
x = (1+√13)/2
= appr. 2.302775638... which I got by repeatedly doing the following on my calculator about 20 times
1. √3
2. =
3. +3
4. =
5. √
goto 2.
To show that the sequence {an} is monotonic and bounded, we need to show two things:
1) Monotonicity: We need to prove that the sequence is either increasing or decreasing, meaning that an+1 ≥ an or an+1 ≤ an for all n.
2) Boundedness: We need to prove that the sequence is bounded, meaning that there is a finite number M such that |an| ≤ M for all n.
Let's start by proving the monotonicity of the sequence {an}:
To prove that {an} is monotonic, we can use mathematical induction.
First, let's check the base case: a1 = (√3)^1/2 = √3. This is the first term of the sequence.
Now, let's assume that an ≥ an-1 for some value n.
Then, we can compare an+1 and an:
an+1 = (√(3 + an))^1/2
Using the assumption an ≥ an-1, we can substitute an-1 for an in the above expression:
an+1 = (√(3 + an-1))^1/2
Now, let's compare an+1 and an:
an+1 - an = (√(3 + an-1))^1/2 - (√(3 + an))^1/2
To simplify this expression, we can multiply both the numerator and denominator by the conjugate of (√(3 + an))^1/2, which is (√(3 + an))^1/2 + (√(3 + an-1))^1/2:
an+1 - an = (√(3 + an-1))^1/2 - (√(3 + an))^1/2 * (√(3 + an))^1/2 + (√(3 + an-1))^1/2
Using the difference of squares formula, we can simplify further:
an+1 - an = (√(3 + an-1) - √(3 + an)) * (√(3 + an-1)) + (√(3 + an))
Since both (√(3 + an-1)) and (√(3 + an)) are positive, we can conclude that the sign of (an+1 - an) depends on (√(3 + an-1) - √(3 + an)).
Now, by assuming an ≥ an-1, we can conclude that (√(3 + an-1) - √(3 + an)) ≤ 0, as (√(3 + an-1) - √(3 + an)) is always negative.
Therefore, we have proved by induction that an ≥ an-1 for all n, which means that {an} is a decreasing sequence.
Now let's prove the boundedness of the sequence {an}:
To prove that {an} is bounded, we need to find a finite number M such that |an| ≤ M for all n.
Let's consider the base case: a1 = √3.
For all n ≥ 1, we have an = (√(3 + an-1))^1/2. Since (√3)^1/2 > 0, it follows that an > 0 for all n.
Now, let's assume that an ≤ an-1 for some value n.
We can substitute an-1 for an in the expression:
an = (√(3 + an))^1/2.
Since (√3)^1/2 > 0, it follows that an > 0.
Therefore, |an| = an ≤ (√(3 + an))^1/2.
Simplifying the inequality, we have an^2 ≤ 3 + an.
Rearranging the terms, we get an^2 - an - 3 ≤ 0.
Now, we need to find the roots of the quadratic equation an^2 - an - 3 = 0.
Using the quadratic formula, we have:
an = (1 ± √(1 + 4(3)))/2 = (1 ± √13)/2.
Since an > 0, we have an = (1 - √13)/2 < 0.
Therefore, we can conclude that 0 < an < (√13 - 1)/2 for all n.
Thus, we have shown that the sequence {an} is bounded between 0 and (√13 - 1)/2.
Finally, let's calculate the limit of the sequence if it exists:
Since the sequence {an} is decreasing and bounded, we know that it must converge to a limit. Let's denote this limit as L.
Taking the limit of both sides of the recurrence relation an+1 = (√(3 + an))^1/2 as n approaches infinity, we have:
L = (√(3 + L))^1/2.
Squaring both sides of the equation, we get:
L^2 = 3 + L.
Simplifying further, we have the quadratic equation:
L^2 - L - 3 = 0.
Using the quadratic formula, we can solve for L:
L = (1 ± √(1 + 4(3)))/2 = (1 ± √13)/2.
Since the sequence {an} is decreasing, we have an ≥ (√(3 + an-1))^1/2 for all n.
Therefore, the limit of the sequence is L = (1 - √13)/2.
In summary, we have shown that the sequence {an} is monotonic (decreasing) and bounded between 0 and (√13 - 1)/2, and the limit of the sequence is (1 - √13)/2.