Prove-that-k-1-H-k-3-2-k-3-3-ln-3-2-pi-2-ln2-3-

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} \dots (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