...advanced calculus... prove that : Σ_(n=1 ) ^∞ (H_n /n^4 ) =^? 3ζ(5)−ζ(2)(3) .... where:: H_(n ) =1+(1/2)+(1/3) +...+(1/n) ..........


Question and Answers Forum

All Questions Topic List
Differentiation Questions
Previous in All Question Next in All Question
Previous in Differentiation Next in Differentiation
Question Number 126726 by mnjuly1970 last updated on 23/Dec/20

$. . . a d v a n c e d c a l c u l u s . . .$ $p r o v e t h a t :$ $\underset{n = 1}{\sum^{\infty}} \frac{H_{n}}{n^{4}} \overset{?}{=} 3 ζ (5) - ζ (2) (3) . . . .$ $w h e r e :: H_{n} = 1 + \frac{1}{2} + \frac{1}{3} + . . . + \frac{1}{n}$ $. . . . . . . . . .$
	Answered by mindispower last updated on 24/Dec/20

	$\int_{0}^{1} x^{n - 1} {l n}^{3} (x) d x = - \int_{0}^{\infty} e^{- n t} t^{3} d t = - \frac{1}{n^{4}} \int_{0}^{\infty} t^{3} e^{- t} d t$ $= - \frac{1}{n^{4}} . Γ (4) = - \frac{6}{n^{4}} \Rightarrow \frac{1}{n^{4}} = - \frac{1}{6} \int_{0}^{1} x^{n - 1} {l n}^{3} (x) d x$ $\sum_{n ⩾ 1} H_{n} x^{n} = - \frac{l n (1 - x)}{1 - x}$ $Σ \frac{H_{n}}{n^{4}} = Σ H_{n} . - \frac{1}{6} \int_{0}^{1} x^{n - 1} {l n}^{3} (x) d x$ $= - \frac{1}{6} \int_{0}^{1} {l n}^{3} (x) \sum_{n ⩾ 1} H_{n} x^{n - 1} d x$ $S = \frac{1}{6} \int_{0}^{1} \frac{{l n}^{3} (x) l n (1 - x)}{x (1 - x)} d x$ $β (a, b) = \int_{0}^{1} x^{a - 1} {(1 - x)}^{b - 1} d x$ $S = \frac{1}{6} . \lim_{a \to 0^{+}} . \lim_{b \to 0^{+}} . (\frac{\partial^{3}}{\partial a^{3}} . \frac{\partial}{\partial b^{}} β (a, b))$ $= (1 + \frac{4}{2}) ζ (5) - \frac{1}{2} \underset{k = 1}{\sum^{2}} ζ (k + 1) ζ (4 - k)$ $= 3 ζ (5) - \frac{1}{2} ζ (2) ζ 3) - \frac{1}{2} ζ (3) ζ (2) = 3 ζ (5) - ζ (2) ζ (3)$
	Commented by mnjuly1970 last updated on 24/Dec/20

	$v e r y n i c e a s a l w a y s s i r m i n d s . . .$
	Commented by mindispower last updated on 25/Dec/20

	$a l w a y s p l e a s u r$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum