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Limit of a series

  1. Feb 14, 2006 #1
    With the equation [tex]\frac{(-1)^{x+1}}{2\cdot2^{x - 1}}[/tex] can I just use the argument that 2^(x-1) will reach infinity faster than (-1)^(n+ 1) so the limit as x -> inf is 0? Because I don't see what I can do the equation to make it more "obvious".
  2. jcsd
  3. Feb 14, 2006 #2
    By rewriting the equation to [tex]-\frac{(-1)^x}{2^x}[/tex], it might be easier to see the solution.
  4. Feb 14, 2006 #3


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    The numerator is bounded, it will always be either 1 or -1.
    The denumerator, as you say, will go to infinity when x tends to infinity making the fraction tend to 0 indeed.
  5. Feb 14, 2006 #4


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    By the way, [tex]\frac{(-1)^{x+1}}{2\cdot2^{x - 1}}[/tex] is not an equation! :)
  6. Feb 14, 2006 #5

    Oops again :tongue2:

    Anyway... thanks!
  7. Feb 14, 2006 #6


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    You're welcome :smile:
  8. Feb 15, 2006 #7


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    And since this is a series, rather than a sequence, you might want to note that, since
    [tex]\frac{(-1)^{x+1}}{2\cdot2^{x - 1}}= -\left(\frac{-1}{2}\right)^n[/tex]
    as you were told before, this is a geometric series.
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