?prove :Σ_(n=1) ^∞ (((−1)^(n−1) H_(2n) )/(2n+1))=(π/8)ln(2)..


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Question Number 134016 by mnjuly1970 last updated on 26/Feb/21

$? p r o v e : \underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n - 1} H_{2 n}}{2 n + 1} = \frac{π}{8} l n (2) . .$
	Answered by mnjuly1970 last updated on 27/Feb/21
	$Σ_(n=1) ^∞ (((−1)^n H_(2n) )/(2n+1))=(1/2)Σ_(n=1) ^∞ (((i^n H_n )/(n+1)))+(1/2)Σ_(n=1) ^∞ ((((−1)^n i^n H_n )/(n+1))) =(1/2)Σ_(n=1) ^∞ {i^n H_n ∫_0 ^( 1) x^n dx}+(1/2)Σ_(n=) ^∞ {(−i)^n H_n ∫_0 ^( 1) x^n dx} =−(1/2)∫_0 ^( 1) ((ln(1−ix))/(1−ix))dx−(1/2)∫_0 ^( 1) ((ln(1+ix))/(1+ix))dx =(1/(4i))[ln^2 (1−xi)]_0 ^1 −(1/(4i))[ln^2 (1+xi)]_(0 ) ^1 =(1/(4i))(ln((√2) e^((−πi)/4) ))^2 −(1/(4i))(ln((√2) e^((πi)/4) ))^2 =(1/(4i))[(ln((√2) )−((iπ)/4))^2 −(ln(√(2))) +((iπ)/4))^2 ] =(1/(4i))(−4ln((√2) )(((iπ)/4)))=((−π)/8)ln(2) ∴ S=(π/8)ln(2) ...✓✓ ...m.n...$
	$\underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n} H_{2 n}}{2 n + 1} = \frac{1}{2} \underset{n = 1}{\sum^{\infty}} (\frac{i^{n} H_{n}}{n + 1}) + \frac{1}{2} \underset{n = 1}{\sum^{\infty}} (\frac{{(- 1)}^{n} i^{n} H_{n}}{n + 1})$ $= \frac{1}{2} \underset{n = 1}{\sum^{\infty}} {i^{n} H_{n} \int_{0}^{1} x^{n} d x} + \frac{1}{2} \underset{n =}{\sum^{\infty}} {{(- i)}^{n} H_{n} \int_{0}^{1} x^{n} d x}$ $= - \frac{1}{2} \int_{0}^{1} \frac{l n (1 - i x)}{1 - i x} d x - \frac{1}{2} \int_{0}^{1} \frac{l n (1 + i x)}{1 + i x} d x$ $= \frac{1}{4 i} {[{l n}^{2} (1 - x i)]}_{0}^{1} - \frac{1}{4 i} {[{l n}^{2} (1 + x i)]}_{0}^{1}$ $= \frac{1}{4 i} {(l n (\sqrt{2} e^{\frac{- π i}{4}}))}^{2} - \frac{1}{4 i} {(l n (\sqrt{2} e^{\frac{π i}{4}}))}^{2}$ $= \frac{1}{4 i} [{(l n (\sqrt{2}) - \frac{i π}{4})}^{2} - {(l n \sqrt{2)} + \frac{i π}{4})}^{2}]$ $= \frac{1}{4 i} (- 4 l n (\sqrt{2}) (\frac{i π}{4})) = \frac{- π}{8} l n (2)$ $∴ S = \frac{π}{8} l n (2) . . . ✓ ✓$ $. . . m . n . . .$
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