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polygamma

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In a previous thread I showed how to express $\zeta'(-1)$ in terms of the Glaisher-Kinkelin constant.

http://mathhelpboards.com/challenge-questions-puzzles-28/euler-maclaurin-summation-formula-riemann-zeta-function-7702.html

This thread is about expressing $\zeta(3)$ (sometimes referred to as Apery's constant) in terms of a constant similar to the Glaisher-Kinkelin constant.

Specifically, $$\zeta(3) = 4 \pi^{2} \log B$$ where $$\log B = \lim_{n \to \infty} \left[ \sum_{k=1}^{n} k^{2} \log k - \left(\frac{n^{3}}{3} + \frac{n^{2}}{2} + \frac{n}{6} \right) \log n + \frac{n^{3}}{9} - \frac{n}{12} \right] $$

Use the Euler-Maclaurin summation formula (or perhaps summation by parts) to show that the constant $B$ exists.Then using the representation of the Riemann zeta function derived in the other thread,

$$ \zeta(s) = \lim_{n \to \infty} \left( \sum_{k=1}^{n} k^{-s} - \frac{n^{1-s}}{1-s} - \frac{n^{-s}}{2} + \frac{s n^{-s-1}}{12} \right) \ \ \big(\text{Re}(s) > -3 \big) $$

show that

$$ \zeta'(-2) = - \log B $$Finally use the functional equation of the Riemann zeta function to show that $$ \zeta(3) = 4 \pi^{2} \log B $$

http://mathhelpboards.com/challenge-questions-puzzles-28/euler-maclaurin-summation-formula-riemann-zeta-function-7702.html

This thread is about expressing $\zeta(3)$ (sometimes referred to as Apery's constant) in terms of a constant similar to the Glaisher-Kinkelin constant.

Specifically, $$\zeta(3) = 4 \pi^{2} \log B$$ where $$\log B = \lim_{n \to \infty} \left[ \sum_{k=1}^{n} k^{2} \log k - \left(\frac{n^{3}}{3} + \frac{n^{2}}{2} + \frac{n}{6} \right) \log n + \frac{n^{3}}{9} - \frac{n}{12} \right] $$

Use the Euler-Maclaurin summation formula (or perhaps summation by parts) to show that the constant $B$ exists.Then using the representation of the Riemann zeta function derived in the other thread,

$$ \zeta(s) = \lim_{n \to \infty} \left( \sum_{k=1}^{n} k^{-s} - \frac{n^{1-s}}{1-s} - \frac{n^{-s}}{2} + \frac{s n^{-s-1}}{12} \right) \ \ \big(\text{Re}(s) > -3 \big) $$

show that

$$ \zeta'(-2) = - \log B $$Finally use the functional equation of the Riemann zeta function to show that $$ \zeta(3) = 4 \pi^{2} \log B $$

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