Unraveling the Nested Radical: A Hint for Solving the Infinite Series Equation

[Daggy]

Problem --

Homework Statement

Give me a hint

Homework Equations

\sqrt{1 + \sqrt{1 + 2\sqrt{1 + 3\sqrt{1 + 4\sqrt{1 + \cdots}}}}} = ?

The Attempt at a Solution

Can someone help me get in the right direction, and give a little hint?

 

[Dick]

Cute. VERRY cute. Contemplate the following series of equations:
2=sqrt(1+1*3)
3=sqrt(1+2*4)
4=sqrt(1+3*5)
5=sqrt(1+4*6)
...

 

[Daggy]

seems likely to fit for 2

Then

<br /> <br /> 2 = \sqrt{1 + \sqrt{1 + 2\sqrt{1 + 3\sqrt{...}}}}<br />

Then n must be such that

<br /> <br /> n = \sqrt{n + ((n-1)(n +1))}<br /> <br />

Which fits for all n.. Is this correct resoning? Can someone give me a more formal explanation if it is (or isn't)..

 

[Dick]

2 doesn't solve n=sqrt(n+(n-1)(n+1)). But yes, I would say the whole thing is equal to 2. If you want to get formal, then you have to define what the whole iterated square root means to begin with. How to truncate it and define it as a convergent series?

 

[Daggy]

actually ment

n = \sqrt{1 + (n + 1)(n - 1)}
[/tex]
an so on. But that doesn't get me any further. Don't really know how to handle the rootexpressions. However, it seems to have som similarities with the root form of the golden ratio
<br /> \phi^2 = \phi + 1 <br />

<br /> \phi = \sqrt{1 + \sqrt{1 + \sqrt{1 + \sqrt{1 + \cdots}}}}<br /> <br />

 

[Dick]

I don't think you are seeing it. Substitute 3=sqrt(1+2*4) for the '3' in 2=sqrt(1+1*3). Now substitute 4=sqrt(1+3*5) into that. Etc etc.

 

[Daggy]

Ok, I see know. Thanks a lot

 

[Daggy]

But inserting it just gives av pinpoint of the solution. tried using mathematical induction to prove it formula, but i did'nt seem correct. are there better methods to prove this, eventually define this as a series in a way?