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Math Help - Another question about Zeta function

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    Another question about Zeta function

    Prove that  \sum_{j=1}^{n-1} 4^j \zeta(2j) \zeta(2n-2j) ~=  \zeta(2n)( n + 2^{2n-1} )
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    Quote Originally Posted by simplependulum View Post

    Prove that  \sum_{j=1}^{n-1} 4^j \zeta(2j) \zeta(2n-2j) ~= \zeta(2n)( n + 2^{2n-1} )
    differentiating both sides of \pi x \cot (\pi x) = 1 - 2\sum_{j=1}^{\infty} \zeta(2j)x^{2j}, which we'll call (1), will give us: \pi^2x^2(\cot^2(\pi x) - 1)=1-2 \pi^2 x^2 + 2 \sum_{j=1}^{\infty}(2j-1)\zeta(2j)x^{2j}. let's call this (2).

    next in (1) change x to 2x to get: 2\pi x \cot(2 \pi x) = 1- 2 \sum_{j=1}^{\infty} 4^j \zeta(2j)x^{2j}, which with (1) gives us: 2\pi^2x^2\cot(\pi x) \cot(2 \pi x) = \left(1 - 2\sum_{j=1}^{\infty} \zeta(2j)x^{2j} \right) \left(1- 2 \sum_{j=1}^{\infty} 4^j \zeta(2j)x^{2j} \right).

    call this result (3). but we have the trig identity 2 \pi^2 x^2 \cot(\pi x) \cot(2 \pi x)=\pi^2x^2(\cot^2(\pi x) - 1). thus by (2) and (3) we have:

    1-2 \pi^2 x^2 + 2 \sum_{j=1}^{\infty}(2j-1)\zeta(2j)x^{2j}=\left(1 - 2\sum_{j=1}^{\infty} \zeta(2j)x^{2j} \right) \left(1- 2 \sum_{j=1}^{\infty} 4^j \zeta(2j)x^{2j} \right). call this (4). finally equating the coefficients of x^{2n}, \ n \geq 2, in both sides of (4) gives us:

    (2n-1) \zeta(2n)=-4^n\zeta(2n) - \zeta(2n) + 2 \sum_{j=1}^{n-1} 4^j \zeta(2j) \zeta(2n-2j) and the result follows.
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