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Irrational infinite sum

  1. Oct 31, 2009 #1


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    I'm curious to answer (or at least reasonably understand) what the answer to:


    might be, where a>0.

    It doesn't follow any ordinary pattern, such as an arithmetic or geometric progression. Also, if there is for any reason an easily derivable answer only for certain values of 'a', then that would also be interesting to hear.

    edit: This problem diverges as hamster143 has noticed.
    Instead, [tex]\sqrt{a}+\sqrt{a-\sqrt{a}}+\sqrt{a-\sqrt{a-\sqrt{a}}}+...[/tex] where a>1
    Last edited: Oct 31, 2009
  2. jcsd
  3. Oct 31, 2009 #2


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    Have you experimented with it at all?

    For example, looked at the first few terms for a chosen value of a?
  4. Oct 31, 2009 #3


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    Yes I basically stared at the problem for a=2 and came up wth absolutely nothing. This is where I realized it's not a geometric sum.

    The problem arose from when I was solving [tex]\sqrt{a+\sqrt{a+\sqrt{a}}}...[/tex] and thought if I could expand it to this new problem. However, I'm out of luck.
  5. Oct 31, 2009 #4
    Are you sure you got the expression right? As stated in your post, the sum diverges for all a except zero. Individual terms converge at [tex](1+\sqrt{1+4a})/2[/tex].
  6. Oct 31, 2009 #5


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    Oh it diverges, thanks for spotting that. A big wasted effort that was...

    I think I'll change the question to:

    [tex]\sqrt{a}+\sqrt{a-\sqrt{a}}+\sqrt{a-\sqrt{a-\sqrt{a}}}+...[/tex], a>1
  7. Oct 31, 2009 #6
    That still diverges, terms converge at [tex](-1+\sqrt{1+4a})/2[/tex].
  8. Oct 31, 2009 #7


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    How frustrating. I can't even construct a problem that converges with the simple criteria that it be an infinite sum and the nth term having n nested surds among it.

    I'll come back when I have a legitimate question.
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