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How to prove this.

  1. Dec 2, 2005 #1
    how to prove that 1/2+1/4+1/8+............+1/2^n is always less than one?
  2. jcsd
  3. Dec 2, 2005 #2


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    The partial sum of the geometric series is:

    [tex]a+ar+ar^2+\cdot\cdot\cdot +ar^{n}=a\frac{1-r^{n+1}}{1-r}[/tex]
  4. Dec 2, 2005 #3


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    Proof by Contradiction
    1. Suppose the partial sum [tex]S_n[/tex] > 1.
    2. We know [tex]S_n<\lim_{n\rightarrow\infty}S_n=1[/tex].
    3. So [tex]1<S_n<1[/tex], a contradiction.
  5. Dec 4, 2005 #4
    [tex]S = a+ar+ar^2+\cdot\cdot\cdot +ar^{n}[/tex]

    [tex]rS = ar+ar^2 ar^3 +\cdot\cdot\cdot +ar^{n} + ar^{n+1}[/tex]

    [tex]rS - S = ar^{n+1} - a [/tex]

    [tex]S = a\frac{r^{n+1} - 1}{r-1}[/tex]

    For the general case it's

    [tex]ar^{c} + \cdot\cdot\cdot +ar^{n}=a\frac{r^{n+1} - r^{c}}{r-1}[/tex]

    Where n>c

    c is lowest power of the sum and n the highest.

    In your case, the formula would be

    [tex]\frac{2^{-1+1} - 2^{n}}{1}[/tex]

    [tex]1 - 2^{n}[/tex]

    2^n is always a positive integrer, thus the sum is alwasy inferior to 1.
    Last edited: Dec 4, 2005
  6. Dec 4, 2005 #5


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    You could do a geometric proof. Start with a unit square. Cut it into two equal parts and remove one of them. Repeat the process on the remaining half and so on. Your series represents the sum of the areas of the pieces you have removed and can never exceed 1 with a finite number of repetitions since some portion of the original square remains.
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