advanced-calculus-prove-that-n-1-H-n-n-4-3-5-2-3-where-H-n-1-1-2-1-3-1-n-

Question Number 126726 by mnjuly1970 last updated on 23/Dec/20

\dots a d v a n c e d c a l c u l u s \dots

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) \dots .

w h e r e :: H_{n} = 1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{n}

\dots \dots \dots .

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 \dots

Commented by mindispower last updated on 25/Dec/20

a l w a y s p l e a s u r