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Math Help - infinite series Euler's constant

  1. #1
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    infinite series Euler's constant

    (a) If Xn=1-1/2+1/3-...-1/(2n), show that Xn -> log 2 as n -> oo, by using the definition of Euler's constant. Hint: Show that Xn=C2n-Cn + log2 in the notation

    (b) Prove that the partial sums of the series

    1+1/2-1/3+1/4+1/5-1/6++- ...

    approach + oo as n -> oo. Make use of Euler's constant

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  2. #2
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by mathsohard View Post
    (a) If Xn=1-1/2+1/3-...-1/(2n), show that Xn -> log 2 as n -> oo, by using the definition of Euler's constant. Hint: Show that Xn=C2n-Cn + log2 in the notation
    Using the notation...

    \displaystyle S_{n}= 1 + \frac{1}{2} + \frac{1}{3} + ... + \frac{1}{n}

    \displaystyle S_{2 n}= 1 + \frac{1}{2} + \frac{1}{3} + ... + \frac{1}{2n}

    \displaystyle X_{2n}= 1 - \frac{1}{2} + \frac{1}{3} + ... - \frac{1}{2n} (1)

    ... it is easy to verify that is...

    X_{2n}= S_{2n} - S_{n} (2)

    But by definition of Euler's constant is...

    \displaystyle \gamma= \lim_{n \rightarrow \infty} S_{2n} - \ln 2n = \lim_{n \rightarrow \infty} S_{2n} - \ln n - \ln 2 = \lim_{n \rightarrow \infty} S_{n} - \ln n  (3)

    ... so that combining (2) and (3) we obtain...

    \displaystyle \lim_{n \rightarrow \infty} X_{2n} =\lim_{n \rightarrow \infty} S_{2n} - S_{n} = \ln 2 (4)

    Kind regards

    \chi \sigma
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  3. #3
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    need some help on (b) as well
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  4. #4
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    b)
    It seems to me that

    <br />
X_{3n}=S_{3n}-\frac{2}{3}S_n<br />
    Last edited by zzzoak; February 18th 2011 at 11:23 PM.
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  5. #5
    MHF Contributor chisigma's Avatar
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    Perfect!!!... so that is...

    \displaystyle \lim_{n \rightarrow \infty} X_{3n} = \lim_{n \rightarrow \infty} (\gamma + \ln 3 + \ln n - \frac{2}{3}\ \gamma - \frac{2}{3}\ \ln n) = \lim_{n \rightarrow \infty} \frac{\gamma}{3} + \ln 3 + \frac{\ln n}{3}

    Kind regards

    \chi \sigma
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