Puzzle: one over one plus … – methods of solution
April 2, 2007 pm30 6:59 pm
This is the place to offer methods of solutions for the puzzle below:
What is the value of ?
A month ago Dave Marain ran a discussion of over here.
I know this one has been done many many times before. But he generated a nice discussion.
There are also places for:
- different versions of the problem, suitable for older/younger more advanced/less advanced students.
- Questions about the problem.
What makes these two problems so similar?
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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:
In general:

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



yields:
This leads to a quadratic equation:
Solving for B yields (B>0):
$B = 2$.
Of course, one has to prove the several steps.
Correction for my post above:
If the limit exists, then
and
have the same limit. This can be written as
and 
in my former post.
I forgot to write
———–
Now, I think I know how this sequence was generated:
Plugging this into the right hand side yields:
and so on…
I just figured out how such a sequence looks for 3 (this is fun):
$ latex = \sqrt{ 6 + \sqrt{ 6 + \sqrt{ 6 + \dotsb}}}$
This can be analyzed in a similar way.
The ‘object’ is the limit of a sequence:
In general:

If we assume that the sequence
converges, then
and
have the same limit. This yields
It follows:
Solving for B yields (B>0):
How was this constructed?
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

appears again.
I just noticed that I calculated the value of


instead of
Thus, the limit of


is