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A couple of things I need help proving.

  1. Feb 18, 2013 #1
    Basically, I was messing with my calculator,
    and i thought of this:


    where they go infinitely.
    I can't prove it, so I need help.

    another thing is 1/(a)+1/(a^2)+1/(a^3)..... = 1/(a-1)

    Perhaps the proof is something really obvious I missed?
  2. jcsd
  3. Feb 18, 2013 #2
    I'm not sure about the first off the top of my head, but I'll help with the second:

    What you have here is a geometric series: [itex] 1 + r + r^{2} + \cdots + r^{n} [/itex] with [itex] r = 1/a [/itex]

    Notice that you don't have the 1, so denoting the sum as [itex] S [/itex], and since [itex] (1 + r + r^{2} + \cdots + r^{n})(1-r) = 1 - r^{n+1} [/itex]

    [itex] S+1 = \lim_{n \rightarrow \infty} \frac{1 - r^{n+1}}{1-r} [/itex]

    This reduces to:

    [itex] S+1 = \frac{a}{a-1} \Rightarrow S = \frac{a}{a-1} - \frac{a-1}{a-1} = \frac{1}{a-1} [/itex]

    Edit: typo
  4. Feb 18, 2013 #3
    Actually I think the first one is deceptively simply. You know that [itex] (x^{m})^{n}= x^{mn}[/itex], so think of the square roots simply as [itex] \frac{1}{2} [/itex] exponents.

    For a finite example: [itex] [x^{x^{1/2}}]^{1/2} = x^{(1/2)x^{1/2}} = (x^{1/2})^{x^{1/2}}[/itex]

    Of course, this logic carries for an infinite number of exponents as well.
    Last edited: Feb 18, 2013
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