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A Sequence of Square Roots, Nested — and Bounded from Above

Date: 08/07/2012 at 02:17:31
From: raden
Subject: prove : sqrt(2(sqrt(3sqrt(4...)))) < 3

Prove

   sqrt(2(sqrt(3sqrt(4...)))) < 3



Date: 08/11/2012 at 18:44:34
From: Doctor Vogler
Subject: Re: prove : sqrt(2(sqrt(3sqrt(4...)))) < 3

Hi Raden,

Thanks for writing to Dr. Math.

Possibly the hardest part of this question is defining exactly what you
mean by the expression

   sqrt(2(sqrt(3sqrt(4...))))

This looks like the limit of a sequence. So I'm going to assume that
you mean the limit of the sequence

   1
   sqrt(2)
   sqrt(2sqrt(3))
   sqrt(2sqrt(3sqrt(4)))
   sqrt(2sqrt(3sqrt(4sqrt(5))))
   ...

I will write this sequence as

   f(1, 1)
   f(1, 2)
   f(1, 3)
   f(1, 4)
   f(1, 5)
   ...

Here,

   f(m, n) = m*sqrt(f(m + 1, n)) when m <= n, and
   f(n, n) = n, or f(n + 1, n) = 1

It is not immediately obvious that this sequence has a limit, but it turns
out that it does. I won't prove that here, since you asked about the
inequality. (You could prove it by showing that f(1, n) is a monotonically
increasing bounded sequence.)

Given that the limit exists, to prove that the limit is less than 3, it
will suffice to prove that this is true for all positive integers n:

   f(1, n) < 2.9

(Note that it would NOT be sufficient to prove that f(1, n) < 3 for all n,
since that wouldn't rule out the possibility that f(1, n) increases
monotonically to a limit of exactly 3. And you want to prove that the
limit is less than 3.)

Due to the definition of f(m, n), to prove that f(1, n) < 29/10, we need
to prove that ...

   sqrt(f(2, n)) < 29/10

... or ...

   f(2, n) < 841/100

This is equivalent to ...

   2*sqrt(f(3, n)) < 841/100

... or ...

   f(3, n) < 707281/40000,

... and so on. In fact, those numbers will start growing very quickly --
much more quickly than we need. Looking at the values of f(m, n) for large
n shows that it is actually very close to (m + 1)^2. So let's start by
showing that for all integers 1 <= m <= n,

   f(m, n) < (m + 1)^2

This is clearly true for m = n. So now we use induction to show that it's
true for smaller m.

The inductive hypothesis gives us

     f(m + 1, n) < (m + 2)^2

The definition of f(m, n) allows us to change the left side to

   [f(m, n)/m]^2 < (m + 2)^2

Therefore,

       f(m, n)/m < m + 2
         f(m, n) < m^2 + 2m

This implies that

         f(m, n) < m^2 + 2m + 1
         f(m, n) < (m + 1)^2

This completes the proof that for all integers 1 <= m <= n,

         f(m, n) < (m + 1)^2

Substituting m = 1 doesn't give us what we want, but substituting m = 3
gives us

         f(3, n) < 4^2 = 16 < 707281/40000

This is what we needed to prove that ...

         f(2, n) < 841/100,

... and ...

         f(1, n) < 29/10,

... which was our final goal.

If you have any questions about this or need more help, please write back
and show me what you have been able to do, and I will try to offer further
suggestions.

- Doctor Vogler, The Math Forum
  http://mathforum.org/dr.math/ 
Associated Topics:
High School Square & Cube Roots

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