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Math Help - A question about Riemann Zeta function

  1. #1
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    A question about Riemann Zeta function

    Prove that  \sum_{j=1}^{n} \frac{\zeta(2j)(-1)^{j+1}}{{\pi}^{2j}{(2n-2j+1)!}} ~= \frac{n}{{(2n+1)!}}
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  2. #2
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    Quote Originally Posted by simplependulum View Post

    Prove that  \sum_{j=1}^{n} \frac{\zeta(2j)(-1)^{j+1}}{{\pi}^{2j}{(2n-2j+1)!}} ~= \frac{n}{{(2n+1)!}}
    changing x to \frac{x}{\pi} in \frac{1 - \pi x \cot(\pi x)}{2}= \sum_{j=1}^{\infty} \zeta(2j)x^{2j} gives us: x \cot x=1 - 2\sum_{j=1}^{\infty} \frac{\zeta(2j)}{\pi^{2j}} x^{2j}. call this (1). we also have \frac{\sin x}{x} = \sum_{j=0}^{\infty} \frac{(-1)^j}{(2j+1)!}x^{2j}. call this (2). multiply (1) by (2) to get:

    \sum_{j=0}^n \frac{(-1)^j}{(2j)!} x^{2j}=\cos x = \left(1 - 2\sum_{j=1}^{\infty} \frac{\zeta(2j)}{\pi^{2j}} x^{2j} \right)\sum_{j=0}^{\infty} \frac{(-1)^j}{(2j+1)!}x^{2j}. call this (3). now equating the coefficients of x^{2n}, \ n \geq 1, in both sides of (3) gives us:

    \frac{(-1)^n}{(2n)!}=\frac{(-1)^n}{(2n+1)!} - 2 \sum_{j=1}^n \frac{(-1)^{n-j}\zeta(2j)}{\pi^{2j}(2n-2j + 1)!} and the result follows.
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  3. #3
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    Let's start with the Fourier Series (It may also be done in other ways, see (*) ): <br />
\sum\limits_{k = 1}^\infty  {\tfrac{{\sin \left( {k \cdot x} \right)}}<br />
{k}}  = \tfrac{{\pi  - x}}<br />
{2}{\text{ }}\forall x \in \left( {0,2 \cdot \pi } \right)<br />

    <br />
 \Rightarrow \sum\limits_{k = 1}^\infty  {\tfrac{{\int_0^x {\sin \left( {k \cdot t} \right)dt} }}<br />
{k}}  = \sum\limits_{k = 1}^\infty  {\tfrac{{1 - \cos \left( {k \cdot x} \right)}}<br />
{{k^2 }}}  = \tfrac{{2\pi  \cdot x - x^2 }}<br />
{{2 \cdot 2}}<br />
    <br />
 \Rightarrow \zeta \left( 2 \right) - \sum\limits_{k = 1}^\infty  {\tfrac{{\cos \left( {k \cdot x} \right)}}<br />
{{k^2 }}}  = \tfrac{{2\pi  \cdot x - x^2 }}<br />
{{2 \cdot 2}} \Rightarrow \zeta \left( 2 \right) \cdot x - \sum\limits_{k = 1}^\infty  {\tfrac{{\sin \left( {k \cdot x} \right)}}<br />
{{k^3 }}}  = \tfrac{{3\pi  \cdot x^2  - x^3 }}<br />
{{2 \cdot \left( {2 \cdot 3} \right)}}<br />
- integrating again-

    Repeating the process we get: <br />
\zeta \left( 2 \right) \cdot \tfrac{{x^3 }}<br />
{{3!}} - \zeta \left( 4 \right) \cdot x + \sum\limits_{k = 1}^\infty  {\tfrac{{\sin \left( {k \cdot x} \right)}}<br />
{{k^3 }}}  = \tfrac{{5\pi  \cdot x^4  - x^5 }}<br />
{{2 \cdot 5!}}<br />

    <br />
 \vdots <br /> <br />
    <br />
\sum\limits_{k = 1}^n {\tfrac{{\zeta \left( {2 \cdot k} \right) \cdot x^{2 \cdot \left( {n - k} \right) + 1} }}<br />
{{\left[ {2 \cdot \left( {n - k} \right) + 1} \right]!}} \cdot \left( { - 1} \right){}^{k + 1}}  + \left( { - 1} \right)^n  \cdot \sum\limits_{k = 1}^\infty  {\tfrac{{\sin \left( {k \cdot x} \right)}}<br />
{{k^{2n + 1} }}}  = \tfrac{{\left( {2n + 1} \right) \cdot \pi  \cdot x^{2n}  - x^{2n + 1} }}<br />
{{2 \cdot \left( {2n + 1} \right)!}}<br />
for 0<x<2\pi

    Now set: <br />
x = \pi  \Rightarrow \sum\limits_{k = 1}^n {\tfrac{{\zeta \left( {2 \cdot k} \right) \cdot \pi ^{2 \cdot \left( {n - k} \right) + 1} }}<br />
{{\left[ {2 \cdot \left( {n - k} \right) + 1} \right]!}} \cdot \left( { - 1} \right){}^{k + 1}}  = n \cdot \tfrac{{\pi ^{2n + 1} }}<br />
{{\left( {2n + 1} \right)!}}\square <br />

    (*) Here
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  4. #4
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    Excellent ! I couldn't imgaine that there is one more solution for this question .
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