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Math Help - series problem

  1. #1
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    series problem

    hello

    I need to find the sum:

    sigma 0-->infinity (-1/3)^n * 1/(2n+1)

    i looked at the power series : x^2n (1)
    the sum of this series is : 1/(1-x^2) (2)

    then by integration of (1) I got : x^2n+1 * 1/(2n+1) (3)
    by integration of (2) i got the sum of (3)

    I dont know what to do now

    help would be appreciated.
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  2. #2
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    I'm not sur but a primitive of 1/(1-x^2) is Argth(x) so by integration we have :
    \sum_{n=0}to{infty} (x^(2n+1))/(2n+1) = Argth(x) so
    x.\sum_{n=0}to{infty} (x^(2n))/(2n+1) = Argth(x)
    x.\sum_{n=0}to{infty} ((x^2)^n)/(2n+1) = Argth(x). Take x= i/sqrt(3) such that x^2= -1/3 and you have :
    i/sqrt(3).\sum_{n=0}to{infty} ((-1/3)^n)/(2n+1) = Argth(i/sqrt(3)) and finaly :
    \sum_{n=0}to{infty} ((-1/3)^n)/(2n+1) = Argth(i/sqrt(3))/(i/sqrt(3))

    I hope I didn't make mistake.
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by parallel
    hello

    I need to find the sum:

    sigma 0-->infinity (-1/3)^n * 1/(2n+1)

    i looked at the power series : x^2n (1)
    the sum of this series is : 1/(1-x^2) (2)

    then by integration of (1) I got : x^2n+1 * 1/(2n+1) (3)
    by integration of (2) i got the sum of (3)

    I dont know what to do now

    help would be appreciated.
    Consider the series expansion:

    <br />
\arctan(x)=\sum_0^{\infty}\frac{(-1)^n x^{2n+1}}{2n+1},\ \ |x|<1 <br />
,

    so:

    <br />
\arctan(\sqrt{x})=\sum_0^{\infty}\frac{(-1)^n x^{(2n+1)/2}}{2n+1}
    <br />
=x^{1/2}\sum_0^{\infty}\frac{(-x)^{n}}{2n+1},\ \ |x|<1 <br />
    and the rest should follow easily.

    RonL
    Last edited by CaptainBlack; August 18th 2006 at 02:15 PM. Reason: To correct error in series for arctan
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  4. #4
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    tize:

    the primitive of 1/1+x^2 is arctan(x) the primitive of 1/(1-x^2) is:
    1/2 * ln[ (1+x) / (1-x) ]
    but I understood the idea of what you were trying to do.

    but still,i can't get the result as it should be

    CaptainBlack

    do you mean that the sum is just arctan(1/sqrt(3))?

    i'm using an online calculator, the result should be 0.90~.
    and as I said before,I can't get the result,can you please help me with my solution too?

    i got that the sum is: i/sqrt(3) * 1/2 * ln[ (1+x) / (1-x) ] (x=i/sqrt(3))
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  5. #5
    Grand Panjandrum
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    Quote Originally Posted by parallel
    tize:

    the primitive of 1/1+x^2 is arctan(x) the primitive of 1/(1-x^2) is:
    1/2 * ln[ (1+x) / (1-x) ]
    but I understood the idea of what you were trying to do.

    but still,i can't get the result as it should be

    CaptainBlack

    do you mean that the sum is just arctan(1/sqrt(3))?

    i'm using an online calculator, the result should be 0.90~.
    and as I said before,I can't get the result,can you please help me with my solution too?



    i got that the sum is: i/sqrt(3) * 1/2 * ln[ (1+x) / (1-x) ] (x=i/sqrt(3))
    First there is an error in my handbook (Scham's) so rather than what I gave
    the relevant expansion for \arctan is:

    <br />
\arctan(x)=\sum_0^{\infty}\frac{(-1)^nx^{2n+1}}{2n+1},\ \ |x|<1<br />
,

    so

    <br />
\arctan(\sqrt{x})=\sum_0^{\infty}(-1)^n\frac{x^{(2n+1)/2}}{2n+1}<br />

    <br />
=x^{1/2}\sum_0^{\infty}\frac{(-x)^{n}}{2n+1},\ \ |x|<1<br />

    So:

    <br />
\sum_0^{\infty}\frac{(-x)^{n}}{2n+1}=\frac{\arctan(\sqrt{x})}{\sqrt{x}},<br />
\ \ |x|<1<br />
.

    Hence:

    <br />
\sum_0^{\infty}\frac{(-1/3)^{n}}{2n+1}=\frac{\arctan(\sqrt{1/3})}{\sqrt{1/3}}\approx 0.9069<br />
.

    RonL
    Last edited by CaptainBlack; August 18th 2006 at 02:30 PM.
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  6. #6
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    thanks for your help.
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