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Math Help - Series terms

  1. #1
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    Series terms

    The goal is to show how many terms of the series 1/n*2^n are needed to approximate ln 2 with 6 digit accuracy.

    So I set
    1/(2^(n+1)*(n+1)<10^-6

    And now this means

    2^n*(n+1)>500,000

    I know the answer is supposed to be 17, but the brain fade is in solving this inequality (unless I'm doing the whole problem wrong). I know it should be easy, but I'm at a little loss here.

    Thanks.
    Last edited by mr fantastic; December 8th 2009 at 11:52 AM. Reason: Changed post title, rt --> de
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  2. #2
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    You don't need to solve it do you ?
    It isn't as if you need to know the value of n to say 3 decimal places.
    You're looking for an integer value for n, so a simple crude trial and error substitution will do.
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  3. #3
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    Quote Originally Posted by zhupolongjoe View Post
    The goal is to show how many terms of the series 1/n*2^n are needed to approximate ln 2 with 6 digit accuracy.
    So I set
    1/(2^(n+1)*(n+1)<10^-6
    And now this means
    2^n*(n+1)>500,000
    I know the answer is supposed to be 17, but the brain fart is in solving this inequality (unless I'm doing the whole problem wrong). I know it should be easy, but I'm at a little loss here.
    Thanks.

     10^{-6} = 0.000001
    6 digit accuracy requires rounding, so half of the above is 0.0000005

     \dfrac{1}{n*2^n} < 0.0000005

    invert
     n*2^n > 2000000

     \ln(n) + n \ln(2) > \ln(2000000)

    \dfrac{ \ln(2000000) - l(n)}{\ln(2)} < n

    ln(2000000)=14.5087
    ln(2) = 0.6931

    \dfrac{ 14.5087 - log(n)}{ 0.6931} < n

    Here is "a simple crude trial and error substitution" method.

    n=1 : ln(n)= 0.00 :: \dfrac{ 14.5087 - 0.00}{ 0.6931} = 20.93

    n=20.93 : ln(20.93)=3.04 :: \dfrac{ 14.5087 - 3.04}{ 0.6931} = 16.55

    n=16.55 : ln(16.55)=2.81 :: \dfrac{ 14.5087 - 2.81}{ 0.6931} = 16.88

    n=16.88 : ln(16.88)=2.83 :: \dfrac{ 14.5087 - 2.83}{ 0.6931} = 16.85

    n=16.85 : ln(16.85)=2.824 :: \dfrac{ 14.5087 - 2.824}{ 0.6931} = 16.858

    The required value of n (rounded to whole number) is 17.

     \dfrac{1}{16*2^{16}}= 0.000000953 > 0.0000005 :  \dfrac{1}{17*2^{17}}= 0.000000448 < 0.0000005

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