Results 1 to 7 of 7

Math Help - series

  1. #1
    Super Member Random Variable's Avatar
    Joined
    May 2009
    Posts
    959
    Thanks
    3

    series

     \displaystyle \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n} \ \zeta(2n)
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor chisigma's Avatar
    Joined
    Mar 2009
    From
    near Piacenza (Italy)
    Posts
    2,162
    Thanks
    5
    If we define the 'factorial function' as...

    \displaystyle z!= \int_{0}^{\infty} t^{z}\ e^{-t}\ dt (1)

    ... with a little of patience we 'discover' that is...

    \displaystyle \ln \frac{1}{z!}= \gamma\ z - \sum_{n=2}^{\infty} (-1)^{n} \zeta(n)\ \frac{z^{n}}{n} (2)

    ... where \gamma is the 'Euler's constant'. To be continued in a succesive post...

    Kind regards

    \chi \sigma
    Last edited by chisigma; January 30th 2011 at 02:47 PM. Reason: see Random Variable post...
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member Random Variable's Avatar
    Joined
    May 2009
    Posts
    959
    Thanks
    3
    The answer is not zero.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor chisigma's Avatar
    Joined
    Mar 2009
    From
    near Piacenza (Italy)
    Posts
    2,162
    Thanks
    5
    Quote Originally Posted by chisigma View Post
    If we define the 'factorial function' as...

    \displaystyle z!= \int_{0}^{\infty} t^{z}\ e^{-t}\ dt (1)

    ... with a little of patience we 'discover' that is...

    \displaystyle \ln \frac{1}{z!}= \gamma\ z - \sum_{n=2}^{\infty} (-1)^{n} \zeta(n)\ \frac{z^{n}}{n} (2)

    ... where \gamma is the 'Euler's constant'...
    ... and that is...

    \displaystyle \varphi(z)= \frac{1}{z!}\ \frac{1}{(-z)!}= \frac{\sin \pi z}{\pi z} = sinc (z) (3)

    Combining (2) and (3) we obtain...

    \displaystyle \ln sinc(z) = -\sum_{n=1}^{\infty} \frac{\zeta(2n)}{n}\ z^{2n} (4)

    ... so that we have...

    \displaystyle \sum_{n=1}^{\infty} (-1)^{n-1} \frac{\zeta(2n)}{n} = \ln sinc(i) = \ln \frac{\sinh \pi}{\pi} (5)

    Kind regards

    \chi \sigma
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member Random Variable's Avatar
    Joined
    May 2009
    Posts
    959
    Thanks
    3
    The solution I had in mine is totally different.

     \displaystyle  \int^{\infty}_{0} \frac{x^{a}}{e^{x}-1} = \zeta (a+1) \Gamma(a+1), \ a>0

    so  \displaystyle \sum^{\infty}_{n=1} \frac{(-1)^{n-1}}{n} \zeta (2n) = \sum^{\infty}_{n=1} \frac{(-1)^{n-1}}{n} \frac{1}{\Gamma (2n)} \int^{\infty}_{0} \frac{x^{2n-1}}{e^{x}-1} \ dx

    Using the fact that  \displaystyle 1 - \cos x = \sum^{\infty}_{n=1} \frac{(-1)^{n-1}}{(2n-1)!} \frac{x^{2n}}{2n} (Integrate both sides of the Maclaurin series for \sin x and don't forget about the constant ).


     \displaystyle \sum^{\infty}_{n=1} \frac{(-1)^{n-1}}{n} \zeta (2n) = 2 \int^{\infty}_{0} \frac{ 1- \cos x}{x(e^{x}-1)} \ dx

     \displaystyle = 2 \int^{\infty}_{0} \frac{1-\cos x}{x} \sum_{n=1}^{\infty} e^{-nx} \ dx

     \displaystyle = 2 \sum^{\infty}_{n=1} \int^{\infty}_{0} \int^{\infty}_{0} (1-\cos x) \ e^{-(t+n)x} \ dt \ dx

    The integrand is always positive, so you don't have to worry about something potentially going haywire when you switch the order of integration.

     \displaystyle = \2 \sum^{\infty}_{n=1} \int^{\infty}_{0} \big( \mathcal{L} \{1\} (t+n) - \mathcal{L} \{\cos x\} (t+n) \big) \ dt

    \displaystyle =  2 \sum_{n=1}^{\infty} \int^{\infty}_{0} \Big( \frac{1}{t+n} - \frac{t+n}{(t+n)^{2}+1} \Big) \ dt

     \displaystyle  = \sum_{n=1}^{\infty} \ln \Big( \frac{(t+n)^{2}}{(t+n)^{2}+1} \Big) \Big|^{\infty}_{0}

     \displaystyle = \sum^{\infty}_{n=1} \ln \Big(1 + \frac{1}{n^{2}} \Big)


     \displaystyle \ln \Big(\frac{ \sinh \pi a}{\pi a} \Big) = \sum^{\infty}_{n=1}  \ln \Big( 1 + \frac{a^{2}}{n^{2}} \Big)

    so \displaystyle  \sum^{\infty}_{n=1} \frac{(-1)^{n-1}}{n} \zeta (2n) = \ln \Big( \frac{\sinh \pi}{\pi} \Big)
    Last edited by Random Variable; January 30th 2011 at 04:51 PM.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor chiph588@'s Avatar
    Joined
    Sep 2008
    From
    Champaign, Illinois
    Posts
    1,163
    Quote Originally Posted by Random Variable View Post
     \displaystyle \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n} \ \zeta(2n)
    We know  \displaystyle \frac{\sin\pi x}{\pi x} = \prod_{n=1}^\infty \left(1-\frac{x^2}{n^2}\right) \implies \log(\pi x) - \log(\sin\pi x) = \sum_{n=1}^\infty \frac{\zeta(2n)}{n}x^{2n} , for  |x|<1 .

    Thus  \displaystyle \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n}\zeta(2n) = \log(\sin\pi i)-\log(\pi i) = \log(i\sinh\pi)-\log(\pi i) = \log\left(\frac{\sinh\pi}\pi\right) .

    Notice we don't need to consider a which branch we choose here, since things cancel!
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21
    Quote Originally Posted by chiph588@ View Post
    We know  \displaystyle \frac{\sin\pi x}{\pi x} = \prod_{n=1}^\infty \left(1-\frac{x^2}{n^2}\right) \implies \log(\pi x) - \log(\sin\pi x) = \sum_{n=1}^\infty \frac{\zeta(2n)}{n}x^{2n} , for  |x|<1 .

    Thus  \displaystyle \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n}\zeta(2n) = \log(\sin\pi i)-\log(\pi i) = \log(i\sinh\pi)-\log(\pi i) = \log\left(\frac{\sinh\pi}\pi\right) .

    Notice we don't need to consider a which branch we choose here, since things cancel!
    It's actually true for -1\leqslant x<1 which is what enabled you to do this. The proof is slightly less clear than the one you did. You need to consider exactly what you said and then apply Abel's Limit Theorem.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 5
    Last Post: October 3rd 2011, 02:12 AM
  2. Replies: 3
    Last Post: September 29th 2010, 07:11 AM
  3. Replies: 0
    Last Post: January 26th 2010, 09:06 AM
  4. Replies: 1
    Last Post: May 5th 2008, 10:44 PM
  5. Replies: 11
    Last Post: April 1st 2008, 01:06 PM

Search Tags


/mathhelpforum @mathhelpforum