Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 183967 by SEKRET last updated on 01/Jan/23

Σ_(n=1) ^∞ (((−1)^(n+1) )/(3n−1)) = ?

$$\:\:\underset{\boldsymbol{\mathrm{n}}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\left(−\mathrm{1}\right)^{\boldsymbol{\mathrm{n}}+\mathrm{1}} }{\mathrm{3}\boldsymbol{\mathrm{n}}−\mathrm{1}}\:\:=\:\:\:? \\ $$

Answered by Ar Brandon last updated on 01/Jan/23

S=Σ_(n=1) ^∞ (((−1)^(n+1) )/(3n−1))=Σ_(n=1) ^∞ (((−1)^(3n−1) )/(3n−1)) S(t)=Σ_(n=1) ^∞ (t^(3n−1) /(3n−1)) ⇒S ′(t)=Σ_(n=1) ^∞ t^(3n−2) =(t/(1−t^3 )) ⇒S(t)=∫(t/(1−t^3 ))dt=−∫(t/((t−1)(t^2 +t+1)))dt ⇒S(t)=−∫((1/(3(t−1)))−((t−1)/(3(t^2 +t+1))))dt ⇒S(t)=−(1/3)ln∣t−1∣+(1/6)∫((2t+1)/(t^2 +t+1))dt−(1/2)∫(1/(t^2 +t+1))dt =−(1/3)ln∣t−1∣+(1/6)ln(t^2 +t+1)−(1/( (√3)))arctan(((2t+1)/( (√3))))+C S(0)=0=−(1/( (√3)))arctan((1/( (√3))))+C ⇒C=(π/(6(√3))) ⇒S(t)=−(1/( 3))ln∣t−1∣+(1/6)ln(t^2 +t+1)−(1/( (√3)))arctan(((2t+1)/( (√3))))+(π/(6(√3))) ⇒Σ_(n=1) ^∞ (((−1)^(n+1) )/(3n−1))=S(−1)=−((ln2)/3)−(1/( (√3)))arctan(−(1/( (√3))))+(π/(6(√3))) ⇒ determinant (((Σ_(n=1) ^∞ (((−1)^(n+1) )/(3n−1))=−((ln2)/3)+(π/(3(√3))))))

$${S}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} }{\mathrm{3}{n}−\mathrm{1}}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{\mathrm{3}{n}−\mathrm{1}} }{\mathrm{3}{n}−\mathrm{1}} \\ $$$${S}\left({t}\right)=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{t}^{\mathrm{3}{n}−\mathrm{1}} }{\mathrm{3}{n}−\mathrm{1}}\:\Rightarrow{S}\:'\left({t}\right)=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{t}^{\mathrm{3}{n}−\mathrm{2}} =\frac{{t}}{\mathrm{1}−{t}^{\mathrm{3}} } \\ $$$$\Rightarrow{S}\left({t}\right)=\int\frac{{t}}{\mathrm{1}−{t}^{\mathrm{3}} }{dt}=−\int\frac{{t}}{\left({t}−\mathrm{1}\right)\left({t}^{\mathrm{2}} +{t}+\mathrm{1}\right)}{dt} \\ $$$$\Rightarrow{S}\left({t}\right)=−\int\left(\frac{\mathrm{1}}{\mathrm{3}\left({t}−\mathrm{1}\right)}−\frac{{t}−\mathrm{1}}{\mathrm{3}\left({t}^{\mathrm{2}} +{t}+\mathrm{1}\right)}\right){dt} \\ $$$$\Rightarrow{S}\left({t}\right)=−\frac{\mathrm{1}}{\mathrm{3}}\mathrm{ln}\mid{t}−\mathrm{1}\mid+\frac{\mathrm{1}}{\mathrm{6}}\int\frac{\mathrm{2}{t}+\mathrm{1}}{{t}^{\mathrm{2}} +{t}+\mathrm{1}}{dt}−\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{1}}{{t}^{\mathrm{2}} +{t}+\mathrm{1}}{dt} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:=−\frac{\mathrm{1}}{\mathrm{3}}\mathrm{ln}\mid{t}−\mathrm{1}\mid+\frac{\mathrm{1}}{\mathrm{6}}\mathrm{ln}\left({t}^{\mathrm{2}} +{t}+\mathrm{1}\right)−\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\mathrm{arctan}\left(\frac{\mathrm{2}{t}+\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)+{C} \\ $$$${S}\left(\mathrm{0}\right)=\mathrm{0}=−\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\mathrm{arctan}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)+{C}\:\Rightarrow{C}=\frac{\pi}{\mathrm{6}\sqrt{\mathrm{3}}} \\ $$$$\Rightarrow{S}\left({t}\right)=−\frac{\mathrm{1}}{\:\mathrm{3}}\mathrm{ln}\mid{t}−\mathrm{1}\mid+\frac{\mathrm{1}}{\mathrm{6}}\mathrm{ln}\left({t}^{\mathrm{2}} +{t}+\mathrm{1}\right)−\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\mathrm{arctan}\left(\frac{\mathrm{2}{t}+\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)+\frac{\pi}{\mathrm{6}\sqrt{\mathrm{3}}} \\ $$$$\Rightarrow\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} }{\mathrm{3}{n}−\mathrm{1}}={S}\left(−\mathrm{1}\right)=−\frac{\mathrm{ln2}}{\mathrm{3}}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\mathrm{arctan}\left(−\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)+\frac{\pi}{\mathrm{6}\sqrt{\mathrm{3}}} \\ $$$$\Rightarrow\begin{array}{|c|}{\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} }{\mathrm{3}{n}−\mathrm{1}}=−\frac{\mathrm{ln2}}{\mathrm{3}}+\frac{\pi}{\mathrm{3}\sqrt{\mathrm{3}}}}\\\hline\end{array} \\ $$

Commented by Ar Brandon last updated on 01/Jan/23

★★Happy New Year friends ★★

$$\bigstar\bigstar\mathcal{H}{appy}\:\mathcal{N}{ew}\:\mathcal{Y}{ear}\:{friends}\:\bigstar\bigstar \\ $$

Commented by SEKRET last updated on 01/Jan/23

thanks sir

$$\:\:\boldsymbol{\mathrm{thanks}}\:\:\boldsymbol{\mathrm{sir}} \\ $$

Commented by SEKRET last updated on 01/Jan/23

thanks

$$\boldsymbol{\mathrm{thanks}}\: \\ $$

Answered by witcher3 last updated on 01/Jan/23

=−Σ_(n≥1) ∫_0 ^1 (−1)^(n+1) x^(3n−2) dx =−∫_0 ^1 (1/x^2 )Σ_(n≥1) (−x^3 )^n dx =−∫_0 ^1 ((−x)/(1+x^3 ))dx=−∫_0 ^1 (x/((x+1)(x^2 −x+1)))dx =∫_0 ^1 (((x+1)^2 −(x^2 −x+1))/((x+1)(x^2 −x+1)))dx ={∫_0 ^1 ((x+1)/(x^2 −x+1))−ln(2)} =∫_0 ^1 ((2x−1)/(2(x^2 −x+1)))dx+(3/2)∫_0 ^1 (dx/(x^2 −x+1))−ln(2) =−ln(2)+(3/2)∫_0 ^1 (dx/((x−(1/2))^2 +(3/4))) =−ln(2)+2∫_0 ^1 (dx/(1+(((2x)/( (√3)))−(1/( (√3))))^2 )) =−ln(2)+(√(3.))2.tan^(−1) ((1/( (√3)))) =−ln(2)+(π/( (√3)))

$$=−\underset{{n}\geqslant\mathrm{1}} {\sum}\int_{\mathrm{0}} ^{\mathrm{1}} \left(−\mathrm{1}\right)^{{n}+\mathrm{1}} {x}^{\mathrm{3}{n}−\mathrm{2}} {dx} \\ $$$$=−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{{x}^{\mathrm{2}} }\underset{{n}\geqslant\mathrm{1}} {\sum}\left(−{x}^{\mathrm{3}} \right)^{{n}} {dx} \\ $$$$=−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{−{x}}{\mathrm{1}+{x}^{\mathrm{3}} }{dx}=−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}}{\left({x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)}{dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\left({x}+\mathrm{1}\right)^{\mathrm{2}} −\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)}{\left({x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)}{dx} \\ $$$$=\left\{\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}+\mathrm{1}}{{x}^{\mathrm{2}} −{x}+\mathrm{1}}−{ln}\left(\mathrm{2}\right)\right\} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{2}{x}−\mathrm{1}}{\mathrm{2}\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)}{dx}+\frac{\mathrm{3}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{dx}}{{x}^{\mathrm{2}} −{x}+\mathrm{1}}−{ln}\left(\mathrm{2}\right) \\ $$$$=−{ln}\left(\mathrm{2}\right)+\frac{\mathrm{3}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{dx}}{\left({x}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{3}}{\mathrm{4}}} \\ $$$$=−{ln}\left(\mathrm{2}\right)+\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{dx}}{\mathrm{1}+\left(\frac{\mathrm{2}{x}}{\:\sqrt{\mathrm{3}}}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)^{\mathrm{2}} } \\ $$$$=−{ln}\left(\mathrm{2}\right)+\sqrt{\mathrm{3}.}\mathrm{2}.\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right) \\ $$$$=−{ln}\left(\mathrm{2}\right)+\frac{\pi}{\:\sqrt{\mathrm{3}}} \\ $$

Commented by Ar Brandon last updated on 01/Jan/23

(x+1)^2 −(x^2 −x+1)=3x

$$\left({x}+\mathrm{1}\right)^{\mathrm{2}} −\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)=\mathrm{3}{x} \\ $$

Commented by SEKRET last updated on 01/Jan/23

thanks sir

$$\boldsymbol{\mathrm{thanks}}\:\:\boldsymbol{\mathrm{sir}} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com