• Jun 15th 2008, 09:23 PM
mathwizard
Find the limit of the following expression if it converges:
$\sqrt {1 + \sqrt {2 + \sqrt {3 + ...}}}$
Also find a recurrence representation of the expression.
• Jun 16th 2008, 12:13 PM
NonCommAlg
Quote:

Originally Posted by mathwizard
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 $b_n=c_1 < a_1=2. \ \ \ \ \square$