# Sequences for infinitely nested radicals

##\large{\sqrt{2+\pi \sqrt{3+\pi\sqrt{4+\pi\sqrt{5+\dotsb}}}}}##
was discussed.

What sequence would you (fellow forum members) associate with that expression?

Define ##\{a_i\}## to be the sequence whose ##i## th term is a truncation of that expression after the ##i##th square root. So :
##a_1 = \sqrt{2}##
##a_2 = \sqrt{2 +\pi \sqrt{3 }}##
##a_3 = \sqrt{2 + \pi \sqrt{3 + \pi \sqrt{4}}}##.
etc.

I think of sequence ##\{a_i\}## and its limit as a safe and reliable context for discussing the meaning of the infinitely nested square roots, but others think of a different sequence.

This is my interpretation of the other sequence:
Define ##B = lim_{i \rightarrow \infty} a_i ##, assuming such a limit exists.
Define the sequence ##\{b_i\}## by ##b_1 = B## and the recurrence relation ##b_{n+1} = (b_n^2 - (n+1))/ \pi## .

So
## b_2 = ( B^2 - 2)/ \pi ## and we might denote it by ## \sqrt { 3 + \pi \sqrt{4 + ...}}##.
##b_3 = (b_2^2 - 3)/ \pi ## denoted by ## \sqrt{4 + \pi { \sqrt{5} + ...}}##.
etc.

There can be ambiguities in interpreting the "..." notation. (https://www.physicsforums.com/threads/ambiguities-of-the-notation.955191/#post-6055289 ) Are the sequences ##\{a_i\}## and ##\{b_i\}## the only reasonable seqeuences to use for assigning a value to the above expression of infinitely nested radicals?

.Scott
Homework Helper
Are the sequences ##\{a_i\}## and ##\{b_i\}## the only reasonable sequences to use for assigning a value to the above expression of infinitely nested radicals?
In the evaluation, we used ##c_i = \sqrt{(i+1)+\pi c_{i+1}}## and then restated it as ##c_{i+1} = {{c_i}^2 - (i+1)}/\pi##
In the first case (##c_i = \sqrt{(i+1)+\pi c_{i+1}}##), we are making a math statement only. In the second case, we are making both a math statement and a potential procedural statement, since we can define an initial ##a_1##.

What sequence would you (fellow forum members) associate with that expression?
I think the series that one associates with ##\sqrt{2+\pi\sqrt{3+\pi\sqrt{4+ ... }}}## is entirely dependent of one's purpose.

andrewkirk
Homework Helper
Gold Member
We can observe that the second sequence is no use in assigning a value to the sequence, because it has to have the value to start with. So only the first sequence is useful for that. I can't think of any other relevant sequences that do it.

I find it helps me get my head around sequences like this to write them out in prefix notation, like with an HP calculator but with the operators going in before the operands, and an operation happens only when operands have been entered that it can use. For that sequence, I get:

$$\sqrt\ +\ 2\ \times\ \pi\quad \sqrt\ +\ 3\ \times\ \pi\quad \sqrt\ +\ 4\ \times\ \pi\quad.....$$
None of these operators can ever operate, because they never get the required operands. So the sequence of ##a##s is obtained by setting ##a_j## equal to what happens when we delete the tail of the sequence from the ##j##th occurrence of ##\times## onwards by 0. THat then allows all prior operands to operate and it collapses to a number.

I don't think there's any way to reverse this order because - to put it very loosely - we'd have to start at infinity and work our way back.