find-the-value-of-n-1-1-n-1-H-2n-n-where-H-n-1-1-2-1-3-1-n-

Question Number 197819 by mnjuly1970 last updated on 30/Sep/23

find the value of : 𝛗 = Σ_(n=1) ^∞ (( (−1)^(n−1) H_( 2n) )/n) = ? where,H_n =1+(1/2) +(1/3) +...+(1/n)

$$ \\ $$$$\:\:\:\:\:\:{find}\:{the}\:{value}\:{of}\:\:: \\ $$$$\:\:\:\:\:\:\boldsymbol{\phi}\:=\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\:\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \:{H}_{\:\mathrm{2}{n}} }{{n}}\:=\:? \\ $$$${where},{H}_{{n}} =\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{3}}\:+…+\frac{\mathrm{1}}{{n}} \\ $$

Answered by witcher3 last updated on 04/Oct/23

∫_0 ^1 −x^(n−1) ln(1−x)dx=(H_n /n) Σ(−1)^(n−1) (H_(2n) /n)=2Σ_(n≥1) ∫_0 ^1 (−1)^n x^(2n−2) ln(1−x) =2∫_0 ^1 −((ln(1−x))/((1+x^2 ))) =−2∫_0 ^(π/4) ln(cos(t)−sin(t))−ln(cos(t))dt =−2∫_0 ^(π/4) ln((√2)cos(t+(π/4)))−ln(cos(t))dt =−2ln((√2)).(π/4)−2∫_0 ^(π/4) ln(tg(t))dt =−((ln(2)π)/4)+2G

$$\int_{\mathrm{0}} ^{\mathrm{1}} −\mathrm{x}^{\mathrm{n}−\mathrm{1}} \mathrm{ln}\left(\mathrm{1}−\mathrm{x}\right)\mathrm{dx}=\frac{\mathrm{H}_{\mathrm{n}} }{\mathrm{n}} \\ $$$$\Sigma\left(−\mathrm{1}\right)^{\mathrm{n}−\mathrm{1}} \frac{\mathrm{H}_{\mathrm{2n}} }{\mathrm{n}}=\mathrm{2}\underset{\mathrm{n}\geqslant\mathrm{1}} {\sum}\int_{\mathrm{0}} ^{\mathrm{1}} \left(−\mathrm{1}\right)^{\mathrm{n}} \mathrm{x}^{\mathrm{2n}−\mathrm{2}} \mathrm{ln}\left(\mathrm{1}−\mathrm{x}\right) \\ $$$$=\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} −\frac{\mathrm{ln}\left(\mathrm{1}−\mathrm{x}\right)}{\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)} \\ $$$$=−\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \mathrm{ln}\left(\mathrm{cos}\left(\mathrm{t}\right)−\mathrm{sin}\left(\mathrm{t}\right)\right)−\mathrm{ln}\left(\mathrm{cos}\left(\mathrm{t}\right)\right)\mathrm{dt} \\ $$$$=−\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \mathrm{ln}\left(\sqrt{\mathrm{2}}\mathrm{cos}\left(\mathrm{t}+\frac{\pi}{\mathrm{4}}\right)\right)−\mathrm{ln}\left(\mathrm{cos}\left(\mathrm{t}\right)\right)\mathrm{dt} \\ $$$$=−\mathrm{2ln}\left(\sqrt{\mathrm{2}}\right).\frac{\pi}{\mathrm{4}}−\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \mathrm{ln}\left(\mathrm{tg}\left(\mathrm{t}\right)\right)\mathrm{dt} \\ $$$$=−\frac{\mathrm{ln}\left(\mathrm{2}\right)\pi}{\mathrm{4}}+\mathrm{2G} \\ $$$$ \\ $$

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