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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 \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:

  1. different versions of the problem, suitable for older/younger more advanced/less advanced students.
  2. Questions about the problem.

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)

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