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

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Question Number 140885 by mnjuly1970 last updated on 13/May/21
....... Advanced ::::::::::★★★:::::::::: Calculus....... find the value of the infinite series:: Θ := Σ_(n=1) ^∞ (((−1)^(n−1) H_( 2n) )/(2n−1)) = ??? .......M.N.july.1970........
$$\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:…….\:{Advanced}\:::::::::::\bigstar\bigstar\bigstar::::::::::\:{Calculus}……. \\ $$$$\:\:\:\:\:\:{find}\:{the}\:{value}\:{of}\:\:{the}\:{infinite}\:{series}:: \\ $$$$\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\Theta\::=\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \mathrm{H}_{\:\mathrm{2}{n}} }{\mathrm{2}{n}−\mathrm{1}}\:=\:??? \\ $$$$\:\:\:\:\:\:\:\:\:\:…….\mathscr{M}.\mathscr{N}.{july}.\mathrm{1970}…….. \\ $$
Answered by mindispower last updated on 15/May/21
H_(2n) =H_(2n−1) +(1/(2n)) H_n =−n∫_0 ^1 x^(n−1) ln(1−x)dx... ⇔Θ=Σ_(n≥1) (−1)^(n−1) .(−∫_0 ^1 x^(2n−2) ln(1−x)dx+(1/((2n)(2n−1)))) =−∫_0 ^1 Σ_(n≥1) (−x^2 )^(n−1) ln(1−x)dx+Σ_(n≥1) (((−1)^(n−1) )/(2n−1))+Σ_(n≥1) (((−1)^n )/(2n)) =∫_0 ^1 ((−ln(1−x))/(1+x^2 ))dx+Σ_(n≥0) (((−1)^n )/(2n+1))−Σ_(n≥0) (((−1)^n )/(n+1)) =(π/4)−ln(2)−∫_0 ^1 ((ln(1−x))/(1+x^2 ))dx_(=A) x=tg(t)⇔A=−∫_0 ^(π/4) ln(((cos(t)−sin(t))/(cos(t))))dt A=−∫_0 ^(π/4) ln(cos((π/4)+t))dt−∫_0 ^(π/4) ln(cos(t))dt =−∫_0 ^(π/4) ln(sin(t))dt−∫_0 ^(π/4) ln(cos)t))dt =−∫_0 ^(π/4) ln(((sin(2t))/2))dt =−(1/2)∫_0 ^(π/2) ln(sin(t))dt+(π/4)ln(2).
$${H}_{\mathrm{2}{n}} ={H}_{\mathrm{2}{n}−\mathrm{1}} +\frac{\mathrm{1}}{\mathrm{2}{n}} \\ $$$${H}_{{n}} =−{n}\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{n}−\mathrm{1}} {ln}\left(\mathrm{1}−{x}\right){dx}… \\ $$$$\Leftrightarrow\Theta=\underset{{n}\geqslant\mathrm{1}} {\sum}\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} .\left(−\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{\mathrm{2}{n}−\mathrm{2}} {ln}\left(\mathrm{1}−{x}\right){dx}+\frac{\mathrm{1}}{\left(\mathrm{2}{n}\right)\left(\mathrm{2}{n}−\mathrm{1}\right)}\right) \\ $$$$=−\int_{\mathrm{0}} ^{\mathrm{1}} \underset{{n}\geqslant\mathrm{1}} {\sum}\left(−{x}^{\mathrm{2}} \right)^{{n}−\mathrm{1}} {ln}\left(\mathrm{1}−{x}\right){dx}+\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{\mathrm{2}{n}−\mathrm{1}}+\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{2}{n}} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{−{ln}\left(\mathrm{1}−{x}\right)}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}+\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{2}{n}+\mathrm{1}}−\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}+\mathrm{1}} \\ $$$$=\frac{\pi}{\mathrm{4}}−{ln}\left(\mathrm{2}\right)−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left(\mathrm{1}−{x}\right)}{\mathrm{1}+{x}^{\mathrm{2}} }{d}\underset{={A}} {{x}} \\ $$$${x}={tg}\left({t}\right)\Leftrightarrow{A}=−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {ln}\left(\frac{{cos}\left({t}\right)−{sin}\left({t}\right)}{{cos}\left({t}\right)}\right){dt} \\ $$$${A}=−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {ln}\left({cos}\left(\frac{\pi}{\mathrm{4}}+{t}\right)\right){dt}−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {ln}\left({cos}\left({t}\right)\right){dt} \\ $$$$\left.=\left.−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {ln}\left({sin}\left({t}\right)\right){dt}−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {ln}\left({cos}\right){t}\right)\right){dt} \\ $$$$=−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {ln}\left(\frac{{sin}\left(\mathrm{2}{t}\right)}{\mathrm{2}}\right){dt}\:=−\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left({sin}\left({t}\right)\right){dt}+\frac{\pi}{\mathrm{4}}{ln}\left(\mathrm{2}\right). \\ $$$$ \\ $$

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