I am trying to find out if

lim. n→∞ [√(2-√(3-√(4-√(5-√(6...-√n)))]

Converges. I'm fairly sure that it does, but I don't have any way of proving it. I'm also trying to figure out if that value can be written succinctly in terms of fundamental constants like π or e. Can anybody here point me in the right direction for solving this?

EDIT: To clarify this is an infinitely nested radical.

I get the strangest feeling that you're asking the wrong place.

Quote from Autistic Biscuit

I get the strangest feeling that you're asking the wrong place.

There are some intelligent individuals here, believe it or not. Unfortunately I hate math, so I'm no help here.

Jeff or Exilement can probably answer this.

Quote from Leokill

There are some intelligent individuals here, believe it or not. Unfortunately I hate math, so I'm no help here.

Jeff or Exilement can probably answer this.

I don't know about Jeff, but Exilement's a drummer isn't he? Good luck with that

Quote from Autistic Biscuit

I don't know about Jeff, but Exilement's a drummer isn't he? Good luck with that

Exilement is old. That increases the chances.

Jeff is a programmer, which has a lot to do with math. I'd be disappointed if he wouldn't be able to solve this pretty easily.

Grrr I knew about this, I like maths xD I'll try time things and let you know if I get the proof you need

Quote from Leokill

Jeff is a programmer, which has a lot to do with math. I'd be disappointed if he wouldn't be able to solve this pretty easily.

Nested radicals aren't exactly the easiest things to work with when the terms are different from each other. And there's not really any reason why you'd have to use them in programming.

Quote from Fusion

Nested radicals aren't exactly the easiest things to work with when the terms are different from each other. And there's not really any reason why you'd have to use them in programming.

Alright, but depending on where he has graduated, he probably has a pretty wide understanding of mathematics. If I'd go in to programming, I'd take all the math I can handle.

The economist in me likes graphs. Draw it out and you'll get:



I'm as surprised as you. My instinct was for convergence.

However, √n<√n+1 for all natural values of n. So the number you're adding (or subtracting, depending on how many negatives are before it so far) keeps on getting bigger as you go on.

That graph doesn't make sense at all for this; you should always have positive values because √(n+1) is always less than n for all n≥2. I think what you've done is put in something like √(2) - (√3) instead of √(2-√3); where the 2 - √3 is all under a radical, not two separate terms.

do you mean

lim. [√(2-√(3-√(4-√(5-√(6...-√n)))] = 0
n→∞

?
you never specified what you're trying to converge to.


i suspect you will need to define this as a function before you try to do this. i found this, which appears to be similar and may help in tackling this



however, your exclusion of √1 may complicate things a little

if you can't work this out i'll put in some more time too look at this later. i'm a little too busy at the moment

Quote from Scarecrow

do you mean

lim. [√(2-√(3-√(4-√(5-√(6...-√n)))] = 0
n→∞

No, I don't mean that because that statement clearly isn't true at all.

you never specified what you're trying to converge to.

That's because the point of convergence is the answer to my question. I don't *know* what it converges to and I'm trying to figure that out.

I already know how to do it with nested radicals of all of the same number; that's much easier. The whole reason why this is difficult is that they aren't all the same number.

Quote from Fusion

I think what you've done is put in something like √(2) - (√3) instead of √(2-√3); where the 2 - √3 is all under a radical, not two separate terms.

So I have. My bad.

Just by typing it in up to n=25 it looks like it's converging on 0.880984... so the bad news is it's probably not going to simplify to any nice irrational constant.

Not even a relation of them? It seems to me like there's always a simple way to write things like this.

i'm just gonna suggest you ask here
math.stackexchange.com

stack exchange is a great website and has always been reliable for me in the past