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Math Help - Infinite nested radicals

  1. #1
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    Infinite nested radicals

    Find the limit of the following expression if it converges:
    \sqrt {1 + \sqrt {2 + \sqrt {3 + ...}}}
    Also find a recurrence representation of the expression.
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  2. #2
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    Quote Originally Posted by mathwizard View Post
    Find the limit of the following expression if it converges:
    \sqrt {1 + \sqrt {2 + \sqrt {3 + ...}}}
    Also find a recurrence representation of the expression.
    the limit is called Vijayaraghavan constant and it's almost 1.7579. see here for more details. what is fascinating about this constant

    is that no closed form (in terms of other known constants) is known for Vijayaraghaven constant!! anyway, there's a theorem that

    gives a necessary and sufficient condition for these kind of nested radicals to be convergent. here it is:

    Theorem (Vijayaraghavan-Herschfeld): let k > 1 and a_n \geq 0, \ \forall n \geq 1. let b_n=\sqrt[k]{a_1 + \sqrt[k]{a_2 + \ ... \ \sqrt[k]{a_{n-1} + \sqrt[k]{a_n}}}}. then the

    sequence \{b_n\}_{n \geq 1} is convergent if and only if the sequence \{a_n^{\frac{1}{k^n}} \}_{n \geq 1} is bounded.


    so by the above theorem, your nested radical is obviously convergent. here's a more interesting, yet elementary, proof:


    let b_n=\sqrt{1+\sqrt{2 + \ ... \ + \sqrt{n}}}, \ \ n \in \mathbb{N}. clearly the sequence is positive and increasing. so if we prove that it's bounded, then we're

    done! we'll do even better, i.e. we'll prove that b_n < 2, \ \forall n \in \mathbb{N}. to prove this claim, we'll define a new sequence: let a_1=2 and

    a_{n+1}=a_n^2 - n, \ \forall n \in \mathbb{N}. call this (1). an easy induction shows that a_n \geq n+1 > \sqrt{n}, \ \forall n. call this result (2). now fix an n and define

    c_k=\sqrt{k+\sqrt{k+1 + \ ... \ + \sqrt{n}}}. we'll prove that c_k < a_k, \ \forall k \leq n. the proof is by reverse induction:

    if k=n, then c_n=\sqrt{n} < a_n, by (2). so c_n < a_n, which is the induction base! (remember, we're using reverse induction!) now suppose

    the claim is true for k+1 and we'll prove it's also true for k: so we have that c_{k+1} < a_{k+1}. thus c_k^2 - k = c_{k+1} < a_{k+1}, and hence

    c_k^2 < a_{k+1} + k = a_k^2, by (1). so: c_k < a_k, which completes the induction. now put k=1 to get b_n=c_1 < a_1=2. \ \ \ \ \square
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