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How are they getting this?

  1. Nov 6, 2004 #1
    This is in the series and convergence chapter

    infinit sum (2^(n)+1)/2^(n+1)

    lim as n goes to infinity of
    (2^(n)+1)/2^(n+1) = [tex]\frac{1+2^{-n}}{2}=1/2[/tex]

    couldn't get latex to work right for the first part.
     
    Last edited: Nov 6, 2004
  2. jcsd
  3. Nov 6, 2004 #2
    [tex]Lim_{n \inf} {(\frac{1}{2})}^n = 0 [/tex]
     
    Last edited: Nov 6, 2004
  4. Nov 6, 2004 #3

    arildno

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    You have:
    [tex]\frac{2^{n}+1}{2^{n+1}}=\frac{1}{2}\frac{2^{n}+1}{2^{n}}=\frac{1}{2}(1+2^{-n})[/tex]
     
    Last edited: Nov 6, 2004
  5. Nov 6, 2004 #4
    Thanks, my algebra still seems to be a little rusty.

    forgot that [tex]2^{n+1}=2^n * 2^1[/tex]

    makes sense now and so do a few other ones that have been giving me headaches this morning.
     
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