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Math Help - An Alternating Infinite Series

  1. #1
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    An Alternating Infinite Series

    Challenge Problem: Find \displaystyle \sum_{k\ge 0}\frac{(-1)^k}{16k^2+16k+3}.

    Moderator approved CB
    Last edited by CaptainBlack; January 23rd 2011 at 10:36 AM.
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  2. #2
    MHF Contributor alexmahone's Avatar
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    Quote Originally Posted by TheCoffeeMachine View Post
    Challenge Problem: Find \displaystyle \sum_{k\ge 0}\frac{(-1)^k}{16k^2+16k+3}.
    \displaystyle 16k^2+16k+3=(4k+2)^2-1=(4k+3)(4k+1)

    \displaystyle \sum_{k\ge 0}\frac{(-1)^k}{16k^2+16k+3}=\sum_{k\ge 0}\frac{(-1)^k}{(4k+3)(4k+1)}

    \displaystyle = \sum_{k\ge 0}\frac{(-1)^k}{2}(\frac{1}{4k+1}-\frac{1}{4k+3})

    \displaystyle =\frac{1}{2}(\frac{1}{1}-\frac{1}{3})-\frac{1}{2}(\frac{1}{5}-\frac{1}{7})+...

    \displaystyle =\frac{1}{2}(\frac{1}{1}-\frac{1}{3}-\frac{1}{5}+\frac{1}{7}+...)

    I'm afraid I don't know how to proceed.
    Last edited by alexmahone; January 23rd 2011 at 07:27 AM. Reason: Stupid mistake
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  3. #3
    MHF Contributor chisigma's Avatar
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    Let suppose to have the periodic function defined in -\pi<x<\pi as...

    f(x)=\left\{\begin{array}{ll}1 ,\,\,|x|< \frac{\pi}{4}\\{}\\0 ,\,\, |x|>\frac{\pi}{4}\end{array}\right. (1)

    ... so that it can be written in Fourier series as...

    \displaystyle f(x)= \frac{a_{0}}{2} + \sum_{n=1}^{\infty} a_{n}\ \cos nx (2)

    The a_{n} can be computed in the usual way...

    \displaystyle \frac{a_{0}}{2} = \frac{1}{\pi}\ \int_{0}^{\frac{\pi}{4}} dx = \frac{1}{4}

    \displaystyle a_{n} = \frac{2}{\pi}\ \int_{0}^{\frac{\pi}{4}} \cos nx\ dx = \frac{2}{\pi\ n}\ \sin n\ \frac{\pi}{4} (3)

    Now we combine the results and obtain ...

    \displaystyle f(\frac{\pi}{4}) = \frac{1}{4} + \frac{2}{\pi}\ \sum_{n=1}^{\infty} \frac{\sin n\ \frac{\pi}{4}\ \cos\ n\frac{\pi}{4}}{n} = \frac{1}{2} \implies

    \displaystyle \implies \sum_{n=1}^{\infty} \frac{\sin n\ \frac{\pi}{4}\ \cos\ n\frac{\pi}{4}}{n} = \frac{1}{2}\ (1-\frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \frac{1}{9} - \frac{1}{11} + ...) =

    \displaystyle = \sum_{n=0}^{\infty} \frac{1}{16\ n^{2} + 16\ n +3} = \frac{\pi}{8} (4)

    Kind regards

    \chi \sigma
    Last edited by chisigma; January 24th 2011 at 04:08 AM. Reason: Corrected the error found by Coffemachine... sorry!...
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  4. #4
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    Quote Originally Posted by alexmahone View Post
    \displaystyle 16k^2+16k+3=(4k+2)^2-1=(4k+3)(4k+1)

    \displaystyle \sum_{k\ge 0}\frac{(-1)^k}{16k^2+16k+3}=\sum_{k\ge 0}\frac{(-1)^k}{(4k+3)(4k+1)}

    \displaystyle = \sum_{k\ge 0}\frac{(-1)^k}{2}(\frac{1}{4k+1}-\frac{1}{4k+3})

    \displaystyle =\frac{1}{2}(\frac{1}{1}-\frac{1}{3})-\frac{1}{2}(\frac{1}{5}-\frac{1}{7})+...

    \displaystyle =\frac{1}{2}(\frac{1}{1}-\frac{1}{3}-\frac{1}{5}+\frac{1}{7}+...)

    I'm afraid I don't know how to proceed.
    I didn't see anything wrong with what you had posted...
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  5. #5
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    Quote Originally Posted by alexmahone View Post
    I'm afraid I don't know how to proceed.
    I don't think it telescopes to any simple value or a familiar series...
    On the right track, though. I did exactly the same up to this step:

    \displaystyle \sum_{k\ge 0}\bigg(\frac{(-1)^k}{2(4k+1)}-\frac{(-1)^k}{2(4k+3)}\bigg).

    Quote Originally Posted by chisigma View Post
    \displaystyle \implies \sum_{n=1}^{\infty} \frac{\sin n\ \frac{\pi}{4}\ \cos\ n\frac{\pi}{4}}{n} = \frac{1}{2}\ (1-\frac{1}{3} - \frac{1}{5} + \frac{1}{7} + \frac{1}{9} - \frac{1}{11} - \frac{1}{13} + ...) =

    \displaystyle = \sum_{n=0}^{\infty} \frac{(-1)^{n}}{16\ n^{2} + 16\ n +3} = \frac{\pi}{8} (4)
    I think what you've found is the sum of the series \displaystyle \sum_{k\ge 0}\frac{1}{16k^2+16k+3}, not \displaystyle \sum_{k\ge 0}\frac{(-1)^k}{16k^2+16k+3}.

    EDIT: I'm wondering whether \displaystyle \sum_{k\ge 0}\frac{(-1)^k}{16k^2+16k+3} can be found in the same way? It would be nice!
    Last edited by TheCoffeeMachine; January 23rd 2011 at 11:59 PM.
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  6. #6
    MHF Contributor alexmahone's Avatar
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    Quote Originally Posted by Prove It View Post
    I didn't see anything wrong with what you had posted...
    I had found the sum of the series \displaystyle \sum_{k\ge 0}\frac{1}{16k^2+16k+3}, not \displaystyle \sum_{k\ge 0}\frac{(-1)^k}{16k^2+16k+3}.
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  7. #7
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    First resolve the summand into partial fractions and we can see that it is equal to ( the result first obtained by alex )  \frac{1}{2} ( \frac{1}{1} - \frac{1}{3} - \frac{1}{5} + \frac{1}{7} + \frac{1}{9} - ... )

    or

     \frac{1}{2} \left[ \int_0^1 ( 1 - x^2(1+x^2)  + x^6(1+x^2) - x^{10}(1+x^2) + ... ) ~dx \right]

     = \frac{1}{2} \int_0^1 \left( 1 - \frac{x^2(1+x^2)}{1+x^4} \right) ~dx

     = \frac{1}{2} \int_0^1 \frac{1-x^2}{1+x^4}~dx

     = - \frac{1}{2} \int_0^1 \frac{1 - 1/x^2}{ ( x+1/x)^2 - 2 } ~dx

    Sub.  x + 1/x = t we have

     S = \frac{1}{2} \int_2^{\infty} \frac{dt}{t^2 - 2 } = \frac{1}{4\sqrt{2}} \left[ \ln[\frac{t - \sqrt{2} }{t + \sqrt{2}} ] \right]_2^{\infty} = \frac{1}{2\sqrt{2}} \ln(1+\sqrt{2} )
    Last edited by simplependulum; January 23rd 2011 at 11:30 PM.
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  8. #8
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    Quote Originally Posted by simplependulum View Post
    ...
    Correct! I did it in a similar fashion with minor differences:

    Writing the sum as an integral:

    \begin{aligned}S & = \sum_{k \ge 0}\frac{(-1)^k}{16k^2+16k+3} \\& = \sum_{k \ge 0}\frac{(-1)^k(4k+3)-(-1)^k(4k+1)}{2(4k+3)(4k+1)} \\& =  \sum_{k \ge 0}\frac{(-1)^k(4k+3)-(-1)^k(4k+1)}{2(4k+3)(4k+1)} \\& =  \sum_{k \ge 0} \bigg(\frac{(-1)^k}{2(4k+1)} -\frac{(-1)^k}{2(4k+3)}\bigg) \\&= \sum_{k \ge 0} \bigg(\frac{1}{2}\int_0^1(-1)^kx^{4k}~ dx -\frac{1}{2}\int_0^1(-1)^kx^{4k+2} ~ dx \bigg) \\& = \int_0^1 \bigg(\frac{1}{2}\sum_{k \ge 0}(-1)^kx^{4k} -\frac{1}{2}\sum_{k \ge 0}(-1)^kx^{4k+2}\bigg)\;dx \\& = \int_0^1 \bigg(\frac{1}{2+2x^4} -\frac{x^2}{2x+2x^4}\bigg)~ dx  = \int_0^1\frac{1-x^2}{2+2x^4}\;{dx} \end{aligned}

    From there it was just a matter of finding the value of the integral:

    \begin{aligned} S & = \frac{1}{2}\int_0^1\frac{1-x^2}{1+x^4}\;{dx} = \frac{1}{2}\int_0^1\frac{1-x^2}{(x^2+1)^2-2x^2}{\;dx} \\& = \frac{1}{2}\int_0^1\frac{1-x^2}{(x^2+\sqrt{2}x+1)(x^2-\sqrt{2}x+1)} ~ {\;dx}\\& = \frac{1}{4\sqrt{2}}\int_0^1\bigg\{\frac{(x^2+\sqrt  {2}x+1)'}{(x^2+\sqrt{2}x+1)}-\frac{(x^2-\sqrt{2}x+1)'}{(x^2-\sqrt{2}x+1)}\bigg\} ~ dx \\& = \frac{1}{4\sqrt{2}} \biggl\| \ln\bigg(\frac{x^2+\sqrt{2}x+1}{x^2-\sqrt{2}x+1}\bigg) \biggl\|_{x = 0}^{x = 1} = \frac{1}{4\sqrt{2}}\ln\bigg(\frac{2+\sqrt{2}}{2-\sqrt{2}}\bigg).\end{aligned}
    Last edited by TheCoffeeMachine; January 24th 2011 at 12:47 AM.
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  9. #9
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    The sum that Alex and Chisigma calculated earlier can also be found in the same way:

    \begin{aligned}S & = \sum_{k \ge 0}\frac{1}{16k^2+16k+3} \\& = \sum_{k \ge 0}\frac{(4k+3)-(4k+1)}{2(4k+3)(4k+1)} \\& = \sum_{k \ge 0}\frac{(4k+3)-(4k+1)}{2(4k+3)(4k+1)} \\& = \sum_{k \ge 0} \bigg(\frac{1}{2(4k+1)} -\frac{1}{2(4k+3)}\bigg) \\&= \sum_{k \ge 0} \bigg(\frac{1}{2}\int_0^1x^{4k}~ dx -\frac{1}{2}\int_0^1 x^{4k+2} ~ dx \bigg) \\& = \int_0^1 \bigg(\frac{1}{2}\sum_{k \ge 0}x^{4k} -\frac{1}{2}\sum_{k \ge 0}x^{4k+2}\bigg)\;dx \\& = \int_0^1 \bigg(\frac{1}{2-2x^4} -\frac{x^2}{2x-2x^4}\bigg)~ dx = \int_0^1\frac{1-x^2}{2-2x^4}\;{dx} \end{aligned}

    And this integral is rather much simpler than the other one:

    \begin{aligned}& S  =  \int_0^1\frac{1-x^2}{2-2x^4}\;{dx}  = \frac{1}{2}\int_0^1 \frac{1-x^2}{(1-x^2)(1+x^2)}\;{dx} \\& = \frac{1}{2}\int_0^{1}\frac{1}{1+x^2}\;{dx} = \frac{1}{2}\biggl\| \arctan{x}\biggl\| _{x=0}^{x=1} = \frac{1}{2}\left(\frac{\pi}{4}\right) = \frac{\pi}{8}.\end{aligned}
    Last edited by TheCoffeeMachine; January 24th 2011 at 01:42 AM.
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  10. #10
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    It turns out that \displaystyle \sum_{k \ge 0}\frac{1}{16k^2+16k+3} is also a special case of the more general result:

    \displaystyle \sum_{k \ge 0}\frac{1}{k^2+k-\alpha} = \frac{\pi\tan\left(\frac{1}{2}\pi\sqrt{4\alpha+1}\  right)}{\sqrt{4\alpha+1}}

    which is very handy and punishes lots of sums, but I don't seem to be able to prove it!
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  11. #11
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    an alternate computation could be the following:

    \begin{aligned}<br />
   \sum\limits_{k=0}^{\infty }{\frac{{{(-1)}^{k}}}{(4k+3)(4k+1)}}&=\int_{0}^{1}{\int_{0}^{1  }{{{y}^{2}}\left( \sum\limits_{k=0}^{\infty }{{{\left( -{{x}^{4}}{{y}^{4}} \right)}^{k}}} \right)\,dx}\,dy} \\ <br />
 & =\int_{0}^{1}{\int_{0}^{1}{\frac{{{y}^{2}}}{1+{{x}  ^{4}}{{y}^{4}}}\,dx}\,dy} \\ <br />
 & =\int_{0}^{1}{\int_{0}^{y}{\frac{y}{1+{{t}^{4}}}\,  dt}\,dy} \\ <br />
 & =\int_{0}^{1}{\int_{t}^{1}{\frac{y}{1+{{t}^{4}}}\,  dy}\,dt} \\ <br />
 & =\frac{1}{2}\int_{0}^{1}{\frac{1-{{t}^{2}}}{1+{{t}^{4}}}\,dt}, \\ <br />
\end{aligned}

    the last integral is routine, and yields the expected result.
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