Ω= Σ_(n=1) ^∞ (( H_( n) )/(n(n+1))) = −−−−−− Ω = Σ_(n=1) ^∞ −(1/(n+1)) ∫_(0 ) ^( 1) x^( n−1) ln(1−x )dx = ∫_0 ^( 1) {−(1/x^2 )ln(1−x).Σ_(n=1) (x^( n+1) /(n+1))}dx = ∫_0 ^( 1) {((−ln(1−x))/x^( 2) )Σ_(n=2) ^∞ (x^( n) /n)}dx = ∫_0 ^( 1) ((−ln(1−x))/x^( 2) ) {−x +Σ_(n=1) ^∞ (x^( n) /n) }dx = −li_( 2) ( 1) +[ ∫_0 ^( 1) ((ln^( 2) ( 1−x ))/x^( 2) )dx=_(derived) ^(earlier) (π^( 2) /3) ] = −(π^( 2) /6) + (π^( 2) /3) = (( π^( 2) )/6) = ζ (2) ■ m.n


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Question Number 166916 by mnjuly1970 last updated on 02/Mar/22
$Ω= Σ_(n=1) ^∞ (( H_( n) )/(n(n+1))) = −−−−−− Ω = Σ_(n=1) ^∞ −(1/(n+1)) ∫_(0 ) ^( 1) x^( n−1) ln(1−x )dx = ∫_0 ^( 1) {−(1/x^2 )ln(1−x).Σ_(n=1) (x^( n+1) /(n+1))}dx = ∫_0 ^( 1) {((−ln(1−x))/x^( 2) )Σ_(n=2) ^∞ (x^( n) /n)}dx = ∫_0 ^( 1) ((−ln(1−x))/x^( 2) ) {−x +Σ_(n=1) ^∞ (x^( n) /n) }dx = −li_( 2) ( 1) +[ ∫_0 ^( 1) ((ln^( 2) ( 1−x ))/x^( 2) )dx=_(derived) ^(earlier) (π^( 2) /3) ] = −(π^( 2) /6) + (π^( 2) /3) = (( π^( 2) )/6) = ζ (2) ■ m.n$
$Ω = \underset{n = 1}{\sum^{\infty}} \frac{H_{n}}{n (n + 1)} =$ $- - - - - -$ $Ω = \underset{n = 1}{\sum^{\infty}} - \frac{1}{n + 1} \int_{0}^{1} x^{n - 1} l n (1 - x) d x$ $= \int_{0}^{1} {- \frac{1}{x^{2}} l n (1 - x) . \sum_{n = 1} \frac{x^{n + 1}}{n + 1}} d x$ $= \int_{0}^{1} {\frac{- l n (1 - x)}{x^{2}} \underset{n = 2}{\sum^{\infty}} \frac{x^{n}}{n}} d x$ $= \int_{0}^{1} \frac{- l n (1 - x)}{x^{2}} {- x + \underset{n = 1}{\sum^{\infty}} \frac{x^{n}}{n}} d x$ $= - {l i}_{2} (1) + [\int_{0}^{1} \frac{{l n}^{2} (1 - x)}{x^{2}} d x \underset{d e r i v e d}{\overset{e a r l i e r}{=}} \frac{π^{2}}{3}]$ $= - \frac{π^{2}}{6} + \frac{π^{2}}{3} = \frac{π^{2}}{6} = ζ (2) ◼ m . n$
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