Prove that: Σ_(k=1) ^∞ (((H_k )^3 )/2^k ) = ((3ζ(3) + ln^3 2 + π^2 ln2)/3)


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Question Number 155386 by mathdanisur last updated on 29/Sep/21

$Prove that :$ $\underset{k = 1}{\sum^{\infty}} \frac{{(H_{k})}^{3}}{2^{k}} = \frac{3 ζ (3) + \ln^{3} 2 + π^{2} \ln 2}{3}$
	Answered by Kamel last updated on 30/Sep/21

	$Prove that :$ $\underset{k = 1}{\sum^{\infty}} \frac{{(H_{k})}^{3}}{2^{k}} = \frac{3 ζ (3) + \ln^{3} 2 + π^{2} \ln 2}{3}$ $Ω = \underset{n = 1}{\sum^{+ \infty}} \frac{{(H_{n})}^{3}}{2^{n}} = \underset{n = 1}{\sum^{+ \infty}} \frac{{(H_{n + 1} - \frac{1}{n + 1})}^{3}}{2^{n}}$ $= 2 \underset{n = 2}{\sum^{+ \infty}} \frac{{(H_{n})}^{3}}{2^{n}} - 6 \underset{n = 2}{\sum^{+ \infty}} \frac{{(H_{n})}^{2}}{n 2^{n}} + 6 \underset{n = 2}{\sum^{+ \infty}} \frac{H_{n}}{n^{2} 2^{n}} - 2 \underset{n = 2}{\sum^{+ \infty}} \frac{1}{n^{3} 2^{n}}$ $= 2 \underset{n = 1}{\sum^{+ \infty}} \frac{{(H_{n})}^{3}}{2^{n}} - 6 \underset{n = 1}{\sum^{+ \infty}} \frac{{(H_{n})}^{2}}{n 2^{n}} + 6 \underset{n = 1}{\sum^{+ \infty}} \frac{H_{n}}{n^{2} 2^{n}} - 2 \underset{n = 1}{\sum^{+ \infty}} \frac{1}{n^{3} 2^{n}}$ $∴ Ω = 6 \underset{n = 1}{\sum^{+ \infty}} \frac{{(H_{n})}^{2}}{n 2^{n}} - 6 \underset{n = 1}{\sum^{+ \infty}} \frac{H_{n}}{n^{2} 2^{n}} + 2 \underset{n = 1}{\sum^{+ \infty}} \frac{1}{n^{3} 2^{n}}$ $\underset{n = 1}{\sum^{+ \infty}} \frac{{(H_{n})}^{2}}{n 2^{n}} = \underset{n = 1}{\sum^{+ \infty}} \frac{1}{n 2^{n}} ({(H_{n + 1})}^{2} - \frac{2 H_{n + 1}}{n + 1} + \frac{1}{{(n + 1)}^{2}})$ $Ω_{1} = \underset{n = 1}{\sum^{+ \infty}} \frac{{(H_{n})}^{2}}{n 2^{n}}, f (x) = \underset{n = 1}{\sum^{+ \infty}} \frac{1}{n} H_{n}^{2} x^{n}$ $f^{'} (x) = \frac{1}{x} \underset{n = 1}{\sum^{+ \infty}} (H_{n + 1}^{2} - \frac{2 H_{n + 1}}{n + 1} + \frac{1}{{(n + 1)}^{2}}) x^{n - 1}$ $= \frac{1}{x} (f^{'} (x) - \frac{2}{x} ({L i}_{2} (x) + \frac{1}{2} {L n}^{2} (1 - x)) + \frac{{L i}_{2} (x)}{x}$ $∴ Ω_{1} = f (\frac{1}{2}) = \int_{0}^{\frac{1}{2}} (\frac{{L n}^{2} (1 - x)}{x (1 - x)} + \frac{{L i}_{2} (x)}{x} + \frac{{L i}_{2} (x)}{1 - x}) d x$ $= {L i}_{3} (\frac{1}{2}) + L n (2) {L i}_{2} (\frac{1}{2}) + \frac{{L n}^{3} (2)}{3} = \frac{7 ζ (3)}{8} . . . (1)$ $Ω_{2} = \underset{n = 1}{\sum^{+ \infty}} \frac{H_{n}}{n^{2} 2^{n}} = \int_{0}^{\frac{1}{2}} \frac{H_{n}}{n} x^{n - 1} d x = \int_{0}^{\frac{1}{2}} \frac{{L i}_{2} (x)}{x} d x + \frac{1}{2} \int_{0}^{\frac{1}{2}} \frac{{L n}^{2} (1 - x)}{x} d x$ $= - \frac{1}{2} \int_{0}^{\frac{1}{2}} \underset{n = 0}{\sum^{+ \infty}} x^{n} {L n}^{2} (x) d x + ζ (3) = ζ (3) - \frac{{L n}^{3} (2)}{2} - L n (2) {L i}_{2} (\frac{1}{2})$ $= ζ (3) - \frac{π^{2}}{12} L n (2)$ $Ω = 6 Ω_{1} - 6 Ω_{2} + 2 {L i}_{3} (\frac{1}{2}) = ζ (3) + \frac{{L n}^{3} (2) + π^{2} L n (2)}{3}$ $∴ \underset{n = 1}{\sum^{+ \infty}} \frac{{(H_{n})}^{3}}{2^{n}} = \frac{3 ζ (3) + π^{2} L n (2) + {L n}^{3} (2)}{3}$ $\underline{N o t e} : {L i}_{3} (\frac{1}{2}) = \int_{0}^{\frac{1}{2}} \frac{{L i}_{2} (x)}{x} d x = \frac{7 ζ (3)}{8} + \frac{{L n}^{3} (2)}{6} - \frac{π^{2}}{12} L n (2)$ $K A M E L B E N A I C H A$
	Commented by Tawa11 last updated on 01/Oct/21

	$Great sir$
	Commented by mathdanisur last updated on 01/Oct/21

	$thank you S er$
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