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Question Number 205774 by mnjuly1970 last updated on 30/Mar/24

−−−−−−− Ω = Σ_(n=0) ^∞ (( (−1)^( n) )/((−1)^( n) −n)) = ? −−−−−−−

$$ \\ $$$$\:\:\:\:\:\:−−−−−−− \\ $$$$\:\:\:\:\:\Omega\:=\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\:\left(−\mathrm{1}\right)^{\:{n}} }{\left(−\mathrm{1}\right)^{\:{n}} \:−{n}}\:=\:?\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:−−−−−−− \\ $$

Answered by Berbere last updated on 30/Mar/24

Ω=Σ_(n=0) ^∞ ((1/(1−2n))−(1/(−2n−2))) =Σ_(n≥0) ((1/(1−2n))+(1/(2(n+1))))=Σ_(n≥0) ((1/(2(n+1)))−(1/(2n−1))) =(3/2)−Σ_(n≥1) (1/(2n+1+1))−(1/(2n−2+1)) =(3/2)+Σ_(n≥1) (∫_0 ^1 x^(2n+1) −x^(2n−2) dx) S(N)=Σ_(n≥0) ∫_0 ^1 (x^3 /(1−x^2 ))−(1/(1−x^2 ))=(3/2)+∫_0 ^1 ((x^3 −1)/((1−x)(1+x)))dx =(3/2)−∫_0 ^1 ((x^2 +x+1)/(x+1))=(3/2)−∫_0 ^1 ((x(x+1)+1)/(x+1))dx =(3/2)−∫_0 ^1 x+(1/(1+x))dx=1−ln(2) Ω=1−ln(2)

$$\Omega=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{\mathrm{1}−\mathrm{2}{n}}−\frac{\mathrm{1}}{−\mathrm{2}{n}−\mathrm{2}}\right) \\ $$$$=\underset{{n}\geqslant\mathrm{0}} {\sum}\left(\frac{\mathrm{1}}{\mathrm{1}−\mathrm{2}{n}}+\frac{\mathrm{1}}{\mathrm{2}\left({n}+\mathrm{1}\right)}\right)=\underset{{n}\geqslant\mathrm{0}} {\sum}\left(\frac{\mathrm{1}}{\mathrm{2}\left({n}+\mathrm{1}\right)}−\frac{\mathrm{1}}{\mathrm{2}{n}−\mathrm{1}}\right) \\ $$$$=\frac{\mathrm{3}}{\mathrm{2}}−\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}+\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{2}{n}−\mathrm{2}+\mathrm{1}} \\ $$$$=\frac{\mathrm{3}}{\mathrm{2}}+\underset{{n}\geqslant\mathrm{1}} {\sum}\left(\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{\mathrm{2}{n}+\mathrm{1}} −{x}^{\mathrm{2}{n}−\mathrm{2}} {dx}\right) \\ $$$${S}\left({N}\right)=\underset{{n}\geqslant\mathrm{0}} {\sum}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{\mathrm{3}} }{\mathrm{1}−{x}^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{1}−{x}^{\mathrm{2}} }=\frac{\mathrm{3}}{\mathrm{2}}+\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{\mathrm{3}} −\mathrm{1}}{\left(\mathrm{1}−{x}\right)\left(\mathrm{1}+{x}\right)}{dx} \\ $$$$=\frac{\mathrm{3}}{\mathrm{2}}−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{\mathrm{2}} +{x}+\mathrm{1}}{{x}+\mathrm{1}}=\frac{\mathrm{3}}{\mathrm{2}}−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}\left({x}+\mathrm{1}\right)+\mathrm{1}}{{x}+\mathrm{1}}{dx} \\ $$$$=\frac{\mathrm{3}}{\mathrm{2}}−\int_{\mathrm{0}} ^{\mathrm{1}} {x}+\frac{\mathrm{1}}{\mathrm{1}+{x}}{dx}=\mathrm{1}−{ln}\left(\mathrm{2}\right) \\ $$$$\Omega=\mathrm{1}−{ln}\left(\mathrm{2}\right) \\ $$$$ \\ $$

Commented by mnjuly1970 last updated on 30/Mar/24

so good solution

$${so}\:{good}\:{solution} \\ $$

Answered by MM42 last updated on 30/Mar/24

Ω=((1/1))+(((−1)/(−1−1)))+((1/(1−2)))+(((−1)/(−1−3)))+((1/(1−4)))+(((−1)/(−1−5)))+((1/(1−6)))+(((−1)/(−1−7)))... 1+(1/2)−1+(1/4)−(1/3)+(1/6)−(1/5)+(1/8)−... =1−(1−(1/2)+(1/3)−(1/4)+(1/5)−(1/6)+..) =1−ln2 ✓

$$\Omega=\left(\frac{\mathrm{1}}{\mathrm{1}}\right)+\left(\frac{−\mathrm{1}}{−\mathrm{1}−\mathrm{1}}\right)+\left(\frac{\mathrm{1}}{\mathrm{1}−\mathrm{2}}\right)+\left(\frac{−\mathrm{1}}{−\mathrm{1}−\mathrm{3}}\right)+\left(\frac{\mathrm{1}}{\mathrm{1}−\mathrm{4}}\right)+\left(\frac{−\mathrm{1}}{−\mathrm{1}−\mathrm{5}}\right)+\left(\frac{\mathrm{1}}{\mathrm{1}−\mathrm{6}}\right)+\left(\frac{−\mathrm{1}}{\left.−\mathrm{1}−\mathrm{7}\right)}...\right. \\ $$$$\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{6}}−\frac{\mathrm{1}}{\mathrm{5}}+\frac{\mathrm{1}}{\mathrm{8}}−... \\ $$$$=\mathrm{1}−\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{5}}−\frac{\mathrm{1}}{\mathrm{6}}+..\right) \\ $$$$=\mathrm{1}−{ln}\mathrm{2}\:\checkmark \\ $$$$ \\ $$

Commented by mnjuly1970 last updated on 30/Mar/24

thanks alot

$${thanks}\:{alot} \\ $$

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