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Where have i gone wrong with this lim

  1. Apr 8, 2009 #1
    where am i going wrong with this lim?
    how do i find the limit for the following , where n->infinity

    1/2 + 3/4 + 5/8 + 7/16 +9/32 +......

    i see that the numerator starts at 1 and has jumps of +2, giving me all the odd numbers

    the denominator starts at 2 with jumps of *2 giving all the powers of 2

    so i have.... + (2n-1)/2^n

    but how do i find the sum of the series? where n=infinity

    ??

    if i had one fraction, (1+3+5+7+9....)/(2+4+8+16....) then i know i could use the equations for sum of a series, but how do i dela with each one as its own fraction.

    i am looking for the lim of the sum, not the lim of (2n-1)/2^n


    the sum S is

    S = 1/2 + 3/4 + 5/8 + 7/16 +9/32 +...... (2n-1)/2^n

    now i take double that and i get
    2S = 1 + 3/2 + 5/4 + 7/8 + 9/16 +......(4n-2)/2^n

    now if i subtract 2S-S i get
    2S = 1 + 3/2 + 5/4 + 7/8 + 9/16 +......(4n-2)/2^n
    S =//// 1/2 + 3/4 + 5/8 + 7/16 +9/32 +...... (2n-1)/2^n

    as you see, if i subtract S from 2S all the middle fractions have a matching one (eg, 3/2 -1/2 =2/2 5/4 -3/4=2/4 etc) only the 1 from 2S and the (2n-1)/2^n from S are left with the sequence (2/2 +2/4 +2/8....)

    2S – S = 1 + 2/2 + 2/4 + 2/8 + …2/2^n - (2n-1)/2^n

    S = 1 + 2*( 1/2 + 1/4 + 1/8 …) - (2n-1)/2^n

    now i know that 2*( 1/2 + 1/4 + 1/8 …) =2*1=2
    so

    S=1+2-(2n-1)/2^n
    =3-(2n-1)/2^n


    only that the answer is wrong, and the correct one is

    3-(2n+3)/2^n

    can you see where i have gone wrong??
     
  2. jcsd
  3. Apr 8, 2009 #2

    Tom Mattson

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    I don't understand why the sum isn't just 3. You clearly have:

    [tex]S=1+2\sum_{n=1}^{\infty}\left(\frac{1}{2}\right)^n[/tex]

    The sum of the geometric series above is known to be exactly 1, so S=1+2(1)=3, no?
     
  4. Apr 9, 2009 #3
    correct, but as i have shown before, when i subtract S from 2S i subtract the 'n+1' number of S from the 'n' number of 2S, leaving me with the 1st from 2S and the last from S, with the geometric series between, if you look at the answer '3-(2n+3)/2^n' it is correct for ANY n,
    n=1, S=1/2
    n=2 S=7/4.....

    how do i get to this?
     
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