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Thread: A question about Riemann Zeta function

  1. #1
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    A question about Riemann Zeta function

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

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

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

    $\displaystyle \frac{(-1)^n}{(2n)!}=\frac{(-1)^n}{(2n+1)!} - 2 \sum_{j=1}^n \frac{(-1)^{n-j}\zeta(2j)}{\pi^{2j}(2n-2j + 1)!}$ 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 $\displaystyle (*)$ ): $\displaystyle
    \sum\limits_{k = 1}^\infty {\tfrac{{\sin \left( {k \cdot x} \right)}}
    {k}} = \tfrac{{\pi - x}}
    {2}{\text{ }}\forall x \in \left( {0,2 \cdot \pi } \right)
    $

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

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

    $\displaystyle
    \vdots

    $
    $\displaystyle
    \sum\limits_{k = 1}^n {\tfrac{{\zeta \left( {2 \cdot k} \right) \cdot x^{2 \cdot \left( {n - k} \right) + 1} }}
    {{\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)}}
    {{k^{2n + 1} }}} = \tfrac{{\left( {2n + 1} \right) \cdot \pi \cdot x^{2n} - x^{2n + 1} }}
    {{2 \cdot \left( {2n + 1} \right)!}}
    $ for $\displaystyle 0<x<2\pi $

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

    $\displaystyle (*)$ 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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