\"Pile of roots.\" Consider the sequence (an) = ( , , , )(so a1 = and the subseq

Question

"Pile of roots." Consider the sequence (an) = ( , , , )(so a1 = and the subsequent elements are defined by the recursive formula an + 1 = ) Prove that this sequence converges. Hint: start by proving that it is bounded. To do that simplify the first few terms of the sequence bringing them into the form an = 2[some power dependent on n] You should see a familiar pattern in the exponents. Use that information to provide a bound on an and proceed from there. Find the limit of this sequence. Hint: Call this limit L. Take the limit n rightarrow infinity on both sides of the equation an + 1 = and then solve the resulting equation for L.

Explanation / Answer

I avoid looking at patterns in exponents, as to me this is against the spirit of the problem.

You could easily prove that each term in the sequence is of the form 2 ^ (1 - 1/(2^n)), the term in parentheses converges to 1, so the power converges to 2.

Thus, I go a different direction.

a) We shall show that it is bounded above (by 2) and monotone increasing. This implies that it converges.

We may define this series by an+1 = 2an and a0 = 1

a1 = 2, so a1 >  a0 and  a1 < 2.

Assume true for n = k.

Then, for n = k+1, ak+1 = 2ak

ak < 2, so ak*ak < 2*ak < 2 * 2

As square roots are a monotone transformation,

(ak*ak )< (2*ak )< (2 * 2)

ak < 2*ak < 2

Then, as  2*ak = ak+1

ak < ak+1 < 2

Thus, we have shown that this is a monotone increasing sequence bounded above by 2, so it has a limit.

b) As we have shown in a) that 2 is an upper bound, all we have to show is that it is the least upper bound.

We prove by contradiction. Let L be the limit less than 2. We know it is greater than 1, as 2, the first term, > 1, and it is monotone increasing.

Then, let = (2-L)/2

There exists an n such that ak > L - = L - (2-L)/2 = 3/2L - 1

Then,ak+1 2ak > 2(3/2L - 1) = 3L-2

We show that 3L-2 > L2 on (1,2), so 3L-2 > L2   = L, or  3L-2 > L

3L - 2 - L2 = -(L2 - 3 L + 2) = -(L - 2)(L - 1) For L on (1,2), L - 2 < 0 and L - 1 > 0, so  -(L - 2)(L - 1) > 0

Thus, ak+1 >   3L-2 > L, or  ak+1 > L.

Thus, as this is a monotone increasing sequence, the limit is greater than L. This is a contradiction, so

L < 2 is not the limit.

Thus, L = 2 is the limit.

(Note: we calculate the limit on problems like this by calculating 2x = x, so 2x = x2, and x = 2)

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