Another question about Zeta function

**simplependulum** · May 27th 2009, 10:51 PM

Prove that $\sum_{j=1}^{n-1} 4^j \zeta(2j) \zeta(2n-2j) ~= \zeta(2n)( n + 2^{2n-1} )$

**NonCommAlg** · May 28th 2009, 07:21 AM

Originally Posted by simplependulum

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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