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Limit (infinity)

  1. Apr 8, 2009 #1
    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

    lim[tex]\sum[/tex](2n-1)/2^n
     
  2. jcsd
  3. Apr 8, 2009 #2

    Gib Z

    User Avatar
    Homework Helper

    Well, you did the first step correctly, in finding the general term. Now, if we split the numerator, the second term is just a geometric series, whilst the first term, [itex] n/2^{n-1}[/itex] is not quite as easy. To do this, we must use the result

    [tex]\sum_{n=0}^{\infty} nr^{n-1} = \frac{1}{(r-1)^2} [/tex]. Play around with it, see what you can substitute to get it to fit what you have.
     
  4. Apr 8, 2009 #3
    i took a bit of a different route, getting there but im stuck. maybe you can help me with my way before i try yours

    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??
     
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