Σ_(n=1) ^∞ (H_n /(2n+1))(π^2 +2H_n ^((2)) −8H_(2n) ^((2)) ) =((83)/4)ζ(4)−7log(2)ζ(3)−8log^2 (2)ζ(2)−(2/3)log^4 (2)−16 Li_4 ((1/2)) where H_n ^((m)) =1+(1/2^m )+.....+(1/n^m ) represents the nth generalized harmonic number of order m , ζ denotes the Riemann zeta function,and Li


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Question Number 88004 by M±th+et£s last updated on 07/Apr/20

$\underset{n = 1}{\sum^{\infty}} \frac{H_{n}}{2 n + 1} (π^{2} + 2 H_{n}^{(2)} - 8 H_{2 n}^{(2)})$ $= \frac{83}{4} ζ (4) - 7 l o g (2) ζ (3) - 8 {l o g}^{2} (2) ζ (2) - \frac{2}{3} {l o g}^{4} (2) - 16 {L i}_{4} (\frac{1}{2})$ $w h e r e H_{n}^{(m)} = 1 + \frac{1}{2^{m}} + . . . . . + \frac{1}{n^{m}} r e p r e s e n t s t h e n t h g e n e r a l i z e d$ $h a r m o n i c n u m b e r o f o r d e r m, ζ d e n o t e s t h e R i e m a n n$ $z e t a f u n c t i o n, a n d {L i}_{n} d e s i g n a t e s t h e p o l y l o g a r i t h m$
	Answered by mind is power last updated on 08/Apr/20

	$i w i l l T r y$
	Commented by M±th+et£s last updated on 08/Apr/20

	$t h a n k y o u s i r$
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