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Math Help - Infinite Series

  1. #1
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    Infinite Series

    A few infinite series that I thought would be fun.

    Find closed form expressions for:

    i) \sum_{r=0}^{\infty} (-1)^{r} \left(\frac{1}{4r+1} + \frac{1}{4r+3} \right)

    ii) \sum_{r=0}^{\infty} (-1)^{r} \frac{(4r)!}{2^{4r}(2r)!(2r+1)!}

    iii) \sum_{n=0}^{\infty} \frac{1}{2n+1} \left( \sum_{r=n+1}^{\infty} \frac{(-1)^{r-1}}{r} \right)

    A couple of them are based on other series I've seen, whereas the other I just happened to stumble across when contemplating something else. Hopefully they aren't too common and haven't been posted before.

    Obviously some calculus is going to be required.
    Last edited by mr fantastic; April 30th 2010 at 06:10 PM. Reason: Restored deleted question.
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  2. #2
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    Quote Originally Posted by Dystopia View Post

    A few infinite series that I thought would be fun.

    Find closed form expressions for:

    i) \sum_{r=0}^{\infty} (-1)^{r} \left(\frac{1}{4r+1} + \frac{1}{4r+3} \right)
    Sum equals \sum\limits_{r\,=\,0}^{\infty }{(-1)^{r}\left( \int_{0}^{1}{x^{4r}\,dx}+\int_{0}^{1}{x^{4r+2}\,dx  } \right)}.

    From here \sum\limits_{r\,=\,0}^{\infty }{(-1)^{r}\left\{ \int_{0}^{1}{x^{4r}\left( 1+x^{2} \right)\,dx} \right\}}, hence \int_{0}^{1}{\left\{ \left( 1+x^{2} \right)\sum\limits_{r\,=\,0}^{\infty }{\left( -x^{4} \right)^{r}} \right\}\,dx} and we happily get that the original sum equals \int_{0}^{1}{\frac{1+x^{2}}{1+x^{4}}\,dx}.

    There's your closed form.
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  3. #3
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    Quote Originally Posted by Krizalid View Post
    Sum equals \sum\limits_{r\,=\,0}^{\infty }{(-1)^{r}\left( \int_{0}^{1}{x^{4r}\,dx}+\int_{0}^{1}{x^{4r+2}\,dx  } \right)}.

    From here \sum\limits_{r\,=\,0}^{\infty }{(-1)^{r}\left\{ \int_{0}^{1}{x^{4r}\left( 1+x^{2} \right)\,dx} \right\}}, hence \int_{0}^{1}{\left\{ \left( 1+x^{2} \right)\sum\limits_{r\,=\,0}^{\infty }{\left( -x^{4} \right)^{r}} \right\}\,dx} and we happily get that the original sum equals \int_{0}^{1}{\frac{1+x^{2}}{1+x^{4}}\,dx}.

    There's your closed form.
    OK, how about an actual value. I don't blame you if you don't feel like expanding that using partial fractions, but there are nicer ways.
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  4. #4
    Super Member PaulRS's Avatar
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    Quote Originally Posted by Dystopia View Post
    OK, how about an actual value. I don't blame you if you don't feel like expanding that using partial fractions, but there are nicer ways.
    <br />
\int_0^1 {\frac{{1 + x^2 }}<br />
{{1 + x^4 }}dx} \underbrace  = _{u = \tfrac{1}<br />
{x}} - \int_{ + \infty }^1 {\left( {\tfrac{1}<br />
{{u^2 }}} \right)\frac{{1 + \tfrac{1}<br />
{{u^2 }}}}<br />
{{1 + \tfrac{1}<br />
{{u^4 }}}}du}  = \int_1^{ + \infty } {\frac{{u^2  + 1}}<br />
{{u^4  + 1}}du} <br />

    Thus: <br />
I = \int_0^1 {\frac{{1 + x^2 }}<br />
{{1 + x^4 }}dx}  \Rightarrow 2 \cdot I = \int_0^{ + \infty } {\frac{{1 + x^2 }}<br />
{{1 + x^4 }}dx} <br />

    By the linearity of the Integral: <br />
2 \cdot I = \int_0^{ + \infty } {\frac{{dx}}<br />
{{1 + x^4 }}}  + \int_0^{ + \infty } {\frac{{x^2 }}<br />
{{1 + x^4 }}dx} <br />

    Now note that: <br />
\int_0^{ + \infty } {\frac{{x^2 }}<br />
{{1 + x^4 }}dx} \underbrace  = _{u = \tfrac{1}<br />
{x}}\int_0^{ + \infty } {\frac{{du}}<br />
{{1 + u^4 }}} <br />

    Thus: <br />
I = \int_0^{ + \infty } {\frac{{dx}}<br />
{{1 + x^4 }}}  = \tfrac{\pi }<br />
{{2 \cdot \sqrt 2 }}<br />

    Where we used the formula: <br />
\int_0^{ + \infty } {\frac{{dx}}<br />
{{1 + x^n }}}  = \tfrac{{\tfrac{\pi }<br />
{n}}}<br />
{{\sin \left( {\tfrac{\pi }<br />
{n}} \right)}}<br />
with <br />
n \in \mathbb{N},n \geqslant 2<br />
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  5. #5
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    Quote Originally Posted by Dystopia View Post
    ii) \sum_{r=0}^{\infty} (-1)^{r} \frac{(4r)!}{2^{4r}(2r)!(2r+1)!}
    using this simple fact that \frac{1}{2^{4r}}\binom{4r}{2r}=\frac{2}{\pi}\int_0  ^{\frac{\pi}{2}} \sin^{4r} t \ dt, we will get:

    I=\sum_{r=0}^{\infty}(-1)^r \frac{(4r)!}{2^{4r}(2r)!(2r+1)!}=\frac{2}{\pi}\sum  _{r=0}^{\infty} \int _0^1 \int_0^{\frac{\pi}{2}} (-x^2)^r\sin^{4r}t \ dt \ dx

    = \frac{2}{\pi} \int_0^1 \int_0^{\frac{\pi}{2}} \sum_{r=0}^{\infty} (-x^2 \sin^4 t)^r \ dt \ dx = \frac{2}{\pi} \int_0^{\frac{\pi}{2}} \int_0^1 \frac{1}{1 + x^2 \sin^4 t} \ dx \ dt

    = \frac{2}{\pi} \int_0^{\frac{\pi}{2}} \frac{\tan^{-1}(\sin^2 t)}{\sin^2 t} \ dt=\frac{4}{\pi} \int_0^{\frac{\pi}{2}} \frac{\cos^2 t}{1 + \sin^4 t} \ dt. now let \tan t = u. then:

    I=\frac{4}{\pi}\int_0^{\infty} \frac{1}{2u^4 + 2u^2 + 1} \ du=\frac{2}{\pi}\int_0^{\infty} \frac{\sqrt{2}+u^{-2}}{(\sqrt{2}u - u^{-1})^2 + 2\sqrt{2} + 2} \ du \ -

    \frac{2}{\pi} \int_0^{\infty} \frac{\sqrt{2}-u^{-2}}{(\sqrt{2} u + u^{-1})^2 -(2\sqrt{2}-2)} \ du=\sqrt{2\sqrt{2}-2}. \ \ \ \square


    iii) \sum_{n=0}^{\infty} \frac{1}{2n+1} \left( \sum_{r=n+1}^{\infty} \frac{(-1)^{r-1}}{r} \right)
    \sum_{n=0}^{\infty} \frac{1}{2n+1} \left(\sum_{r=n+1}^{\infty} \frac{(-1)^{r-1}}{r} \right)=\sum_{n=0}^{\infty} \frac{1}{2n+1} \sum_{r=n+1}^{\infty}\int_0^1 (-x)^{r-1} \ dx

    =\sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1} \int_0^1 \frac{x^n}{x+1} \ dx=2 \sum_{n=0}^{\infty} \frac{(-1)^n}{2n+1} \int_0^1 \frac{t^{2n+1}}{t^2 + 1} \ dt

    =2 \int_0^1 \frac{1}{t^2+1} \sum_{n=0}^{\infty} \frac{(-1)^n t^{2n+1}}{2n+1} \ dt=2 \int_0^1 \frac{\tan^{-1} t}{t^2 +1} \ dt = \frac{\pi^2}{16}. \ \ \ \square
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  6. #6
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    Quote Originally Posted by Dystopia View Post
    i) \sum_{r=0}^{\infty} (-1)^{r} \left(\frac{1}{4r+1} + \frac{1}{4r+3} \right)
    Note that,
    \sum_{n=-\infty}^{-1} (-1)^n \left( \frac{1}{4n+1}+\frac{1}{4n+3} \right) = \sum_{n=-\infty}^{-1}(-1)^{(1-n)-1}\left( \frac{-1}{4(1-n)+3}+\frac{-1}{4(1-n)+1} \right) = \sum_{n=0}^{\infty} (-1)^n \left( \frac{1}{4n+1}+\frac{1}{4n+3} \right)
    Therefore,
    \sum_{n=-\infty}^{\infty}(-1)^n \left( \frac1{4n+1}+\frac1{4n+3}\right) = 2\sum_{n=0}^{\infty}(-1)^n\left( \frac1{4n+1}+\frac1{4n+3}\right).

    Now define f(z) = \frac{1}{4z+1}+\frac{1}{4z+3} this function has poles at \xi = -1/4 and \mu = -3/4.

    If F(z) = \pi f(z) \csc \pi z then by complex analysis methods we have:
    \sum_{n=-\infty}^{\infty} (-1)^n f(n) = - \mbox{Res}(F,\xi) - \mbox{Res}(F,\mu) = \frac{\pi}{2\sqrt{2}}+\frac{\pi}{2\sqrt{2}} = \frac{\pi}{\sqrt{2}}.

    Thus, \sum_{n=0}^{\infty} (-1)^n \left( \frac{1}{4n+1}+\frac{1}{4n+3} \right) = \frac{1}{2}\cdot \frac{\pi}{\sqrt{2}}.
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  7. #7
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    Those answers are correct. The methods used are interesting. Here is how I did them:

    i) If \sin\theta > 0, then \sum_{r=0}^{\infty} \frac{\sin(2r+1)\theta}{2r+1} = \frac{\pi}{4}.

    Put \theta = \frac{\pi}{4}

    ii) By De Moivre's theorem, \cos\frac{x}{2} = \Re (\cos x + i\sin x)^{-\frac{1}{2}}

    \Rightarrow \cos\frac{x}{2}\sqrt{\cos x} = \sum_{r=0}^{\infty} \frac{(-1)^{r} (4r)!}{2^{4r}(2r)!^{2}} \tan^{2r}x \;\;\;\; (|\tan x| \leq 1)

    Multiply by \sec^{2}x and integrate with limits pi/4, 0.

    iii) Same as NCA.
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