Find the limit of the following expression if it converges:
$\displaystyle \sqrt {1 + \sqrt {2 + \sqrt {3 + ...}}}$
Also find a recurrence representation of the expression.

2. Originally Posted by mathwizard
Find the limit of the following expression if it converges:
$\displaystyle \sqrt {1 + \sqrt {2 + \sqrt {3 + ...}}}$
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 $\displaystyle k > 1$ and $\displaystyle a_n \geq 0, \ \forall n \geq 1.$ let $\displaystyle b_n=\sqrt[k]{a_1 + \sqrt[k]{a_2 + \ ... \ \sqrt[k]{a_{n-1} + \sqrt[k]{a_n}}}}.$ then the

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

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

let $\displaystyle b_n=\sqrt{1+\sqrt{2 + \ ... \ + \sqrt{n}}}, \ \ n \in \mathbb{N}.$ 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 $\displaystyle b_n < 2, \ \forall n \in \mathbb{N}.$ to prove this claim, we'll define a new sequence: let $\displaystyle a_1=2$ and

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

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

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

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

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