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 Post subject: Calculus (17) - Series
PostPosted: Fri, 2 Jan 2009 07:44:20 UTC 
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Evaluate $ \sum_{n=1}^{\infty} \frac{\zeta(2n)}{2^{n}}, where as usual $ \zeta(s) is the Riemann zeta function.


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PostPosted: Fri, 2 Jan 2009 09:19:51 UTC 
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commutative wrote:
Evaluate $ \sum_{n=1}^{\infty} \frac{\zeta(2n)}{2^{n}}, where as usual $ \zeta(s) is the Riemann zeta function.


Spoiler:
Via Bernoulli numbers?

It is well-known that
\zeta(2n)=(-1)^{n+1}\dfrac{B_{2n}(2\pi)^{2n}}{2(2n)!}\quad\forall n\in\mathbb{Z}_{>0}.
Taking the even part of the exponential generating function
\displaystyle \sum_{n=0}^\infty B_n\dfrac{x^n}{n!}=\dfrac{x}{e^x-1}
(radius of convergence=2*pi), we obtain
\displaystyle \sum_{n=1}^\infty B_{2n}\dfrac{x^{2n}}{(2n)!}=\dfrac{x/2}{\tanh(x/2)}-1
and hence
\displaystyle
\sum_{n=1}^\infty \dfrac{\zeta(2n)}{2^n}
=-\frac{1}{2}\sum_{n=1}^\infty B_{2n}\dfrac{(\pi\,\mathrm{i}\sqrt{2})^{2n}}{(2n)!}
=-\frac{1}{2}\left(\frac{\pi/\sqrt{2}}{\tan(\pi/\sqrt{2})}-1\right).

Err.. do you mean zeta(2n)/4^n? Then the answer is a nice number 1/2.


Last edited by outermeasure on Fri, 2 Jan 2009 09:59:54 UTC, edited 1 time in total.

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PostPosted: Fri, 2 Jan 2009 09:58:09 UTC 
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outermeasure wrote:

Spoiler:
Via Bernoulli numbers?

It is well-known that
\zeta(2n)=(-1)^{n+1}\dfrac{B_{2n}(2\pi)^{2n}}{2(2n)!}\quad\forall n\in\mathbb{Z}_{>0}.
Taking the even part of the exponential generating function
\displaystyle \sum_{n=0}^\infty B_n\dfrac{x^n}{n!}=\dfrac{x}{e^x-1}
(radius of convergence=2*pi), we obtain
\displaystyle \sum_{n=1}^\infty B_{2n}\dfrac{x^{2n}}{(2n)!}=\dfrac{x/2}{\tanh(x/2)}-1
and hence
\displaystyle
\sum_{n=1}^\infty \dfrac{\zeta(2n)}{2^n}
=-\frac{1}{2}\sum_{n=1}^\infty B_{2n}\dfrac{(\pi\,\mathrm{i})^{2n}}{(2n)!}
=-\frac{1}{2}(-1)=\frac{1}{2}.


that's nice but what you found is $ \sum_{n=1}^{\infty} \frac{\zeta(2n)}{4^{n}} not $ \sum_{n=1}^{\infty} \frac{\zeta(2n)}{2^{n}}. :)


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PostPosted: Fri, 2 Jan 2009 10:01:30 UTC 
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commutative wrote:
that's nice but what you found is $ \sum_{n=1}^{\infty} \frac{\zeta(2n)}{4^{n}} not $ \sum_{n=1}^{\infty} \frac{\zeta(2n)}{2^{n}}. :)


Err.. yes, was editing.... now corrected.


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PostPosted: Fri, 2 Jan 2009 10:13:48 UTC 
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outermeasure wrote:
Err.. yes, was editing.... now corrected.


correct! more generally, as you probabaly know, we have this well-known formula: $ \sum_{n=1}^{\infty} \zeta(2n)x^{2n}=\frac{1 - \pi x \cot(\pi x)}{2}, \ \ |x| < 1.


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