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This is the place to offer methods of solutions for the puzzle below:

What is the value of $\frac{1}{1+\frac{1}{1 + \frac{1}{1 + \dotsb}}}$ ?

A month ago Dave Marain ran a discussion of $\sqrt{2 + \sqrt{2 + \sqrt{2 + \dotsb}}}$ over here.

I know this one has been done many many times before. But he generated a nice discussion.

There are also places for:

What makes these two problems so similar?

5 Comments leave one →
1. Edgardo permalink
April 3, 2007 pm30 9:48 pm 9:48 pm $\sqrt{2 + \sqrt{2 + \sqrt{2 + \dotsb}}}$

First examine what this ‘object’ is. It looks a like a sequence,
or better, the limit of the sequence.
The sequence can be described by: $a_1 = b \hspace{1cm}\mathrm{(initial value)}$ $a_2 = \sqrt{2+a_1}$ $a_3 = \sqrt{2+a_2} = \sqrt{2+ \sqrt{2+a_1}}$

In general: $a_{n+1}= \sqrt{2+ a_{n}}$

If the limit exists, then $a_{n}$ and $a_{n+1}$ have the same limit.
Thus, $a_{n}=a_{n+1}=B$, where $B$ is the limit of the sequence.

Pluggin this into the equation $a_{n+1}= \sqrt{2+ a_{n}}$
yields: $B = \sqrt{2 + B}$
This leads to a quadratic equation: $B^2-B-2=0$

Solving for B yields (B>0):
$B = 2$.

Of course, one has to prove the several steps.

2. Edgardo permalink
April 3, 2007 pm30 10:00 pm 10:00 pm

Correction for my post above:

If the limit exists, then $a_{n}$ and $a_{n+1}$ have the same limit. This can be written as $\rm{lim} a_{n} = B$ and $\rm{lim} a_{n+1} = B$
I forgot to write $\rm{lim}$ in my former post.

———–

Now, I think I know how this sequence was generated: $2 = \sqrt{2+2}$
Plugging this into the right hand side yields: $2 = \sqrt{2+ \sqrt{2+2}}$
and so on…

3. Edgardo permalink
April 3, 2007 pm30 10:20 pm 10:20 pm

I just figured out how such a sequence looks for 3 (this is fun): $3 = \sqrt{9}= \sqrt{6+3}$ $= \sqrt{ 6+ \sqrt{6+3} }$ $= \sqrt{ 6+ \sqrt{ 6 + \sqrt{ 6 + 3 }}}$
$latex = \sqrt{ 6 + \sqrt{ 6 + \sqrt{ 6 + \dotsb}}}$

4. Edgardo permalink
April 3, 2007 pm30 11:17 pm 11:17 pm $\frac{1}{1+\frac{1}{1 + \frac{1}{1 + \dotsb}}}$

This can be analyzed in a similar way.
The ‘object’ is the limit of a sequence: $a_{1} = b (\mathrm{initial value})$ $a_{2} = 1+ \frac{1}{a_{1}}$ $a_{3} = 1 + \frac{1}{a_{2}} = 1 + \frac{1}{1+ \frac{1}{a_{1}}}$

In general: $a_{n+1} = 1 + \frac{1}{a_{n}}$

If we assume that the sequence $a_{n}$ converges, then $a_{n}$ and $a_{n+1}$ have the same limit. This yields $B = 1 + \frac{1}{B}$, where B is the limit of the sequence.
It follows: $B^2-B-1 = 0$
Solving for B yields (B>0): $B = \frac{1}{2}(\sqrt{5}+1)$

How was this constructed? $\frac{1}{2} \left(\sqrt{5}+1 \right) = \frac{1}{2} \left(\sqrt{5}+1 \right) + 1 - 1$ $= 1+ \frac{1}{2} \left(\sqrt{5}+1 \right) -1$ $= 1+ \frac{\sqrt{5}-1}{2}$ $= 1+ \frac{5-1}{2(\sqrt{5}+1)}$ $= 1+ \frac{4}{2(\sqrt{5}+1)}$ $= 1+ \frac{2}{\sqrt{5}+1}$ $= 1+ \frac{1}{\frac{1}{2}(\sqrt{5}+1)}$

The tactic is obvious, in the first step a 1 is added because we want to have the 1 + 1/(1+/…) term. The remaining task is to manipulate the term after the 1 in such a way, that $\frac{1}{2} \left(\sqrt{5}+1 \right)$

appears again.

5. Edgardo permalink
April 3, 2007 pm30 11:26 pm 11:26 pm

I just noticed that I calculated the value of $1+\frac{1}{1+\frac{1}{1 + \frac{1}{1 + \dotsb}}}$
instead of $\frac{1}{1+\frac{1}{1 + \frac{1}{1 + \dotsb}}}$

Thus, the limit of $\frac{1}{1+\frac{1}{1 + \frac{1}{1 + \dotsb}}}$
is $\frac{1}{2}(\sqrt{5}-1)$