Advanced-Calculus-find-the-value-of-the-infinite-series-n-1-1-n-1-H-2n-2n-1-

Question Number 140885 by mnjuly1970 last updated on 13/May/21

\dots \dots . A d v a n c e d :::::::::: ★ ★ ★ :::::::::: C a l c u l u s \dots \dots .

f i n d t h e v a l u e o f t h e i n f i n i t e s e r i e s ::

Θ := \underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n - 1} H_{2 n}}{2 n - 1} = ? ? ?

\dots \dots . M . N . j u l y . 1970 \dots \dots . .

Answered by mindispower last updated on 15/May/21

H_{2 n} = H_{2 n - 1} + \frac{1}{2 n}

H_{n} = - n \int_{0}^{1} x^{n - 1} l n (1 - x) d x \dots

\Leftrightarrow Θ = \sum_{n ⩾ 1} {(- 1)}^{n - 1} . (- \int_{0}^{1} x^{2 n - 2} l n (1 - x) d x + \frac{1}{(2 n) (2 n - 1)})

= - \int_{0}^{1} \sum_{n ⩾ 1} {(- x^{2})}^{n - 1} l n (1 - x) d x + \sum_{n ⩾ 1} \frac{{(- 1)}^{n - 1}}{2 n - 1} + \sum_{n ⩾ 1} \frac{{(- 1)}^{n}}{2 n}

= \int_{0}^{1} \frac{- l n (1 - x)}{1 + x^{2}} d x + \sum_{n ⩾ 0} \frac{{(- 1)}^{n}}{2 n + 1} - \sum_{n ⩾ 0} \frac{{(- 1)}^{n}}{n + 1}

= \frac{π}{4} - l n (2) - \int_{0}^{1} \frac{l n (1 - x)}{1 + x^{2}} d \underset{= A}{x}

x = t g (t) \Leftrightarrow A = - \int_{0}^{\frac{π}{4}} l n (\frac{c o s (t) - s i n (t)}{c o s (t)}) d t

A = - \int_{0}^{\frac{π}{4}} l n (c o s (\frac{π}{4} + t)) d t - \int_{0}^{\frac{π}{4}} l n (c o s (t)) d t

= - \int_{0}^{\frac{π}{4}} l n (s i n (t)) d t - \int_{0}^{\frac{π}{4}} l n (c o s) t)) d t

= - \int_{0}^{\frac{π}{4}} l n (\frac{s i n (2 t)}{2}) d t = - \frac{1}{2} \int_{0}^{\frac{π}{2}} l n (s i n (t)) d t + \frac{π}{4} l n (2) .