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Question Number 138821 by mnjuly1970 last updated on 18/Apr/21

....... nice .. .. .. calculus...... find the value of: Θ=Σ_(n=−∞ ) ^∞ (1/(6n^2 +5n+1))=?

$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:.......\:{nice}\:..\:..\:..\:{calculus}...... \\ $$$${find}\:{the}\:{value}\:{of}: \\ $$$$\:\:\:\:\:\:\:\:\:\Theta=\underset{{n}=−\infty\:} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{6}{n}^{\mathrm{2}} +\mathrm{5}{n}+\mathrm{1}}=? \\ $$$$ \\ $$

Commented by Kamel last updated on 19/Apr/21

You can use digamma function or residue theorem.

$${You}\:{can}\:{use}\:{digamma}\:{function}\:{or}\:{residue}\:{theorem}. \\ $$

Answered by Kamel last updated on 19/Apr/21

Θ=Σ_(n=−∞) ^(+∞) (1/(6n^2 +5n+1))=(1/6)Σ_(n=−∞) ^(+∞) (1/((n+(1/2))(n+(1/3)))) =lim_(N→+∞) Σ_(n=0) ^N ((1/(n+(1/3)))−(1^ /(n+(1/2)))+(1/(−n+(1/3)))−(1/(−n+(1/2))))−1 =lim_(N→+∞) Σ_(n=0) ^N {3((1/(3n+1))−(1/(3n−1)))−2((1/(2n+1))−(1/(2n−1)))}−1 =lim_(N→+∞) Σ_(n=0) ^(N−1) 3((1/(3n+1))−(1/(3n+2)))=3Σ_(n=0) ^(+∞) (1/((3n+1)(3n+2))) Put: f(x)=Σ_(n=0) ^(+∞) (x^(3n+2) /((3n+1)(3n+2))), 0≤x<1. f′′(x)=Σ_(n=0) ^(+∞) x^(3n) =(1/(1−x^3 ))=(1/((1−x)(1+x+x^2 )))=(1/3)((1/(1−x))+((2x+1+3)/(2(x^2 +x+1)))) ∴ f′(x)+C=(1/3)(−Ln(1−x)+(1/2)Ln(x^2 +x+1)+(√3)Arctan((2/( (√3)))(x+(1/2))) f′(0)=0⇒C=(π/( 6(√3))) f(1)=−(1/3)∫_0 ^1 Ln(x)dx+(1/6)∫_0 ^1 Ln(x^2 +x+1)dx+(1/( (√3)))∫_0 ^1 Arctan((2/( (√3)))(x+(1/2)))dx−(π/( 6(√3))) I=∫Ln(x^2 +x+1)dx=xLn(x^2 +x+1)−∫((2(x^2 +x+1)−x−2)/(x^2 +x+1))dx =xLn(x^2 +x+1)−2x+∫((x+2)/(x^2 +x+1))dx =(x+(1/2))Ln(x^2 +x+1)−2x+(√3)Arctan((2/( (√3)))(x+(1/2))) ∫Arctan(x)dx=xArctan(x)−(1/2)Ln(1+x^2 ) So: f(1)=(1/3)+(1/6)((3/2)Ln(3)−2+((π(√3))/3)−((π(√3))/6))+(1/2)((π/( (√3)))−(π/(6(√3)))−(1/( 2))Ln(3))−(π/(6(√3))) =((π(√3))/(36))+((5π)/(12(√3)))−(π/( 6(√3)))=(π/(3(√3))), Θ=3f(1) ∴ Θ=(π/( (√3))) ⇒ 𝚺_(n=−∞) ^(+∞) (1/(6n^2 +5n+1))=((𝛑(√3))/3) BENAICHA Kamel

$$\Theta=\underset{{n}=−\infty} {\overset{+\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{6}{n}^{\mathrm{2}} +\mathrm{5}{n}+\mathrm{1}}=\frac{\mathrm{1}}{\mathrm{6}}\underset{{n}=−\infty} {\overset{+\infty} {\sum}}\frac{\mathrm{1}}{\left({n}+\frac{\mathrm{1}}{\mathrm{2}}\right)\left({n}+\frac{\mathrm{1}}{\mathrm{3}}\right)} \\ $$$$\:\:\:=\underset{{N}\rightarrow+\infty} {{lim}}\underset{{n}=\mathrm{0}} {\overset{{N}} {\sum}}\left(\frac{\mathrm{1}}{{n}+\frac{\mathrm{1}}{\mathrm{3}}}−\frac{\overset{} {\mathrm{1}}}{{n}+\frac{\mathrm{1}}{\mathrm{2}}}+\frac{\mathrm{1}}{−{n}+\frac{\mathrm{1}}{\mathrm{3}}}−\frac{\mathrm{1}}{−{n}+\frac{\mathrm{1}}{\mathrm{2}}}\right)−\mathrm{1} \\ $$$$\:\:\:=\underset{{N}\rightarrow+\infty} {{lim}}\underset{{n}=\mathrm{0}} {\overset{{N}} {\sum}}\left\{\mathrm{3}\left(\frac{\mathrm{1}}{\mathrm{3}{n}+\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{3}{n}−\mathrm{1}}\right)−\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{2}{n}−\mathrm{1}}\right)\right\}−\mathrm{1} \\ $$$$\:\:\:=\underset{{N}\rightarrow+\infty} {{lim}}\underset{{n}=\mathrm{0}} {\overset{{N}−\mathrm{1}} {\sum}}\mathrm{3}\left(\frac{\mathrm{1}}{\mathrm{3}{n}+\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{3}{n}+\mathrm{2}}\right)=\mathrm{3}\underset{{n}=\mathrm{0}} {\overset{+\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{3}{n}+\mathrm{1}\right)\left(\mathrm{3}{n}+\mathrm{2}\right)} \\ $$$$\:\:\:{Put}:\:{f}\left({x}\right)=\underset{{n}=\mathrm{0}} {\overset{+\infty} {\sum}}\frac{{x}^{\mathrm{3}{n}+\mathrm{2}} }{\left(\mathrm{3}{n}+\mathrm{1}\right)\left(\mathrm{3}{n}+\mathrm{2}\right)},\:\mathrm{0}\leqslant{x}<\mathrm{1}. \\ $$$${f}''\left({x}\right)=\underset{{n}=\mathrm{0}} {\overset{+\infty} {\sum}}{x}^{\mathrm{3}{n}} =\frac{\mathrm{1}}{\mathrm{1}−{x}^{\mathrm{3}} }=\frac{\mathrm{1}}{\left(\mathrm{1}−{x}\right)\left(\mathrm{1}+{x}+{x}^{\mathrm{2}} \right)}=\frac{\mathrm{1}}{\mathrm{3}}\left(\frac{\mathrm{1}}{\mathrm{1}−{x}}+\frac{\mathrm{2}{x}+\mathrm{1}+\mathrm{3}}{\mathrm{2}\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)}\right) \\ $$$$\:\therefore\:{f}'\left({x}\right)+{C}=\frac{\mathrm{1}}{\mathrm{3}}\left(−{Ln}\left(\mathrm{1}−{x}\right)+\frac{\mathrm{1}}{\mathrm{2}}{Ln}\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)+\sqrt{\mathrm{3}}{Arctan}\left(\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\left({x}+\frac{\mathrm{1}}{\mathrm{2}}\right)\right)\right. \\ $$$${f}'\left(\mathrm{0}\right)=\mathrm{0}\Rightarrow{C}=\frac{\pi}{\:\mathrm{6}\sqrt{\mathrm{3}}} \\ $$$${f}\left(\mathrm{1}\right)=−\frac{\mathrm{1}}{\mathrm{3}}\int_{\mathrm{0}} ^{\mathrm{1}} {Ln}\left({x}\right){dx}+\frac{\mathrm{1}}{\mathrm{6}}\int_{\mathrm{0}} ^{\mathrm{1}} {Ln}\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right){dx}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\int_{\mathrm{0}} ^{\mathrm{1}} {Arctan}\left(\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\left({x}+\frac{\mathrm{1}}{\mathrm{2}}\right)\right){dx}−\frac{\pi}{\:\mathrm{6}\sqrt{\mathrm{3}}} \\ $$$${I}=\int{Ln}\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right){dx}={xLn}\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)−\int\frac{\mathrm{2}\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)−{x}−\mathrm{2}}{{x}^{\mathrm{2}} +{x}+\mathrm{1}}{dx} \\ $$$$\:\:={xLn}\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)−\mathrm{2}{x}+\int\frac{{x}+\mathrm{2}}{{x}^{\mathrm{2}} +{x}+\mathrm{1}}{dx} \\ $$$$\:\:=\left({x}+\frac{\mathrm{1}}{\mathrm{2}}\right){Ln}\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)−\mathrm{2}{x}+\sqrt{\mathrm{3}}{Arctan}\left(\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\left({x}+\frac{\mathrm{1}}{\mathrm{2}}\right)\right) \\ $$$$\int{Arctan}\left({x}\right){dx}={xArctan}\left({x}\right)−\frac{\mathrm{1}}{\mathrm{2}}{Ln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right) \\ $$$${So}:\:{f}\left(\mathrm{1}\right)=\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{6}}\left(\frac{\mathrm{3}}{\mathrm{2}}{Ln}\left(\mathrm{3}\right)−\mathrm{2}+\frac{\pi\sqrt{\mathrm{3}}}{\mathrm{3}}−\frac{\pi\sqrt{\mathrm{3}}}{\mathrm{6}}\right)+\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\pi}{\:\sqrt{\mathrm{3}}}−\frac{\pi}{\mathrm{6}\sqrt{\mathrm{3}}}−\frac{\mathrm{1}}{\:\mathrm{2}}{Ln}\left(\mathrm{3}\right)\right)−\frac{\pi}{\mathrm{6}\sqrt{\mathrm{3}}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\pi\sqrt{\mathrm{3}}}{\mathrm{36}}+\frac{\mathrm{5}\pi}{\mathrm{12}\sqrt{\mathrm{3}}}−\frac{\pi}{\:\mathrm{6}\sqrt{\mathrm{3}}}=\frac{\pi}{\mathrm{3}\sqrt{\mathrm{3}}},\:\Theta=\mathrm{3}{f}\left(\mathrm{1}\right) \\ $$$$\therefore\:\:\:\:\:\:\:\Theta=\frac{\pi}{\:\sqrt{\mathrm{3}}}\:\:\:\:\:\:\:\Rightarrow\:\:\:\:\:\:\:\:\:\:\underset{\boldsymbol{{n}}=−\infty} {\overset{+\infty} {\boldsymbol{\sum}}}\frac{\mathrm{1}}{\mathrm{6}\boldsymbol{{n}}^{\mathrm{2}} +\mathrm{5}\boldsymbol{{n}}+\mathrm{1}}=\frac{\boldsymbol{\pi}\sqrt{\mathrm{3}}}{\mathrm{3}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{{BENAICHA}}\:\boldsymbol{{Kamel}}\:\:\:\:\:\:\: \\ $$$$ \\ $$

Commented by mnjuly1970 last updated on 19/Apr/21

thanks alot mr Kmel...

$${thanks}\:{alot}\:{mr}\:{Kmel}... \\ $$

Answered by Dwaipayan Shikari last updated on 19/Apr/21

Σ_(n=−∞) ^∞ (1/((an^2 +bn+c)))=(1/( (√(b^2 −4ac))))(ψ(((b+(√(b^2 −4ac)))/2))−ψ(((b−(√(b^2 −4ac)))/2)))+Λ Λ=(1/( (√(b^2 −4ac))))(ψ(1−(b/(2a))+((√(b^2 −4ac))/(2a)))−ψ(1−(b/(2a))−((√(b^2 −4ac))/(2a)))) Σ_(n=−∞) ^∞ (1/((6n^2 +5n+1)))=ψ((1/2))−ψ((1/3))+ψ((2/3))−ψ((1/2)) =πcot((π/3))=(π/( (√3)))

$$\underset{{n}=−\infty} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left({an}^{\mathrm{2}} +{bn}+{c}\right)}=\frac{\mathrm{1}}{\:\sqrt{{b}^{\mathrm{2}} −\mathrm{4}{ac}}}\left(\psi\left(\frac{{b}+\sqrt{{b}^{\mathrm{2}} −\mathrm{4}{ac}}}{\mathrm{2}}\right)−\psi\left(\frac{{b}−\sqrt{{b}^{\mathrm{2}} −\mathrm{4}{ac}}}{\mathrm{2}}\right)\right)+\Lambda \\ $$$$\Lambda=\frac{\mathrm{1}}{\:\sqrt{{b}^{\mathrm{2}} −\mathrm{4}{ac}}}\left(\psi\left(\mathrm{1}−\frac{{b}}{\mathrm{2}{a}}+\frac{\sqrt{{b}^{\mathrm{2}} −\mathrm{4}{ac}}}{\mathrm{2}{a}}\right)−\psi\left(\mathrm{1}−\frac{{b}}{\mathrm{2}{a}}−\frac{\sqrt{{b}^{\mathrm{2}} −\mathrm{4}{ac}}}{\mathrm{2}{a}}\right)\right) \\ $$$$\underset{{n}=−\infty} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{6}{n}^{\mathrm{2}} +\mathrm{5}{n}+\mathrm{1}\right)}=\psi\left(\frac{\mathrm{1}}{\mathrm{2}}\right)−\psi\left(\frac{\mathrm{1}}{\mathrm{3}}\right)+\psi\left(\frac{\mathrm{2}}{\mathrm{3}}\right)−\psi\left(\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$=\pi{cot}\left(\frac{\pi}{\mathrm{3}}\right)=\frac{\pi}{\:\sqrt{\mathrm{3}}} \\ $$

Commented by mnjuly1970 last updated on 19/Apr/21

thank you mr Payan...

$$\:{thank}\:{you}\:{mr}\:{Payan}... \\ $$

Answered by mnjuly1970 last updated on 19/Apr/21

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