....... nice .. .. .. calculus...... find the value of: Θ=Σ


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

$. . . . . . . n i c e . . . . . . c a l c u l u s . . . . . .$ $f i n d t h e v a l u e o f :$ $Θ = \underset{n = - \infty}{\sum^{\infty}} \frac{1}{6 n^{2} + 5 n + 1} = ?$
Commented by Kamel last updated on 19/Apr/21

$Y o u c a n u s e d i g a m m a f u n c t i o n o r r e s i d u e t h e o r e m .$
	Answered by Kamel last updated on 19/Apr/21
	$Θ=Σ_(n=−∞) ^(+∞) (1/(6n^2 +5n+1))=(1/6)Σ_(n=−∞) ^(+∞) (1/((n+(1/2))(n+(1/3)))) =lim_(N→+∞) Σ_(n=0) ^N ((1/(n+(1/3)))−(1^ /(n+(1/2)))+(1/(−n+(1/3)))−(1/(−n+(1/2))))−1 =lim_(N→+∞) Σ_(n=0) ^N {3((1/(3n+1))−(1/(3n−1)))−2((1/(2n+1))−(1/(2n−1)))}−1 =lim_(N→+∞) Σ_(n=0) ^(N−1) 3((1/(3n+1))−(1/(3n+2)))=3Σ_(n=0) ^(+∞) (1/((3n+1)(3n+2))) Put: f(x)=Σ_(n=0) ^(+∞) (x^(3n+2) /((3n+1)(3n+2))), 0≤x<1. f′′(x)=Σ_(n=0) ^(+∞) x^(3n) =(1/(1−x^3 ))=(1/((1−x)(1+x+x^2 )))=(1/3)((1/(1−x))+((2x+1+3)/(2(x^2 +x+1)))) ∴ f′(x)+C=(1/3)(−Ln(1−x)+(1/2)Ln(x^2 +x+1)+(√3)Arctan((2/( (√3)))(x+(1/2))) f′(0)=0⇒C=(π/( 6(√3))) f(1)=−(1/3)∫_0 ^1 Ln(x)dx+(1/6)∫_0 ^1 Ln(x^2 +x+1)dx+(1/( (√3)))∫_0 ^1 Arctan((2/( (√3)))(x+(1/2)))dx−(π/( 6(√3))) I=∫Ln(x^2 +x+1)dx=xLn(x^2 +x+1)−∫((2(x^2 +x+1)−x−2)/(x^2 +x+1))dx =xLn(x^2 +x+1)−2x+∫((x+2)/(x^2 +x+1))dx =(x+(1/2))Ln(x^2 +x+1)−2x+(√3)Arctan((2/( (√3)))(x+(1/2))) ∫Arctan(x)dx=xArctan(x)−(1/2)Ln(1+x^2 ) So: f(1)=(1/3)+(1/6)((3/2)Ln(3)−2+((π(√3))/3)−((π(√3))/6))+(1/2)((π/( (√3)))−(π/(6(√3)))−(1/( 2))Ln(3))−(π/(6(√3))) =((π(√3))/(36))+((5π)/(12(√3)))−(π/( 6(√3)))=(π/(3(√3))), Θ=3f(1) ∴ Θ=(π/( (√3))) ⇒ 𝚺_(n=−∞) ^(+∞) (1/(6n^2 +5n+1))=((𝛑(√3))/3) BENAICHA Kamel$
	$Θ = \underset{n = - \infty}{\sum^{+ \infty}} \frac{1}{6 n^{2} + 5 n + 1} = \frac{1}{6} \underset{n = - \infty}{\sum^{+ \infty}} \frac{1}{(n + \frac{1}{2}) (n + \frac{1}{3})}$ $= \underset{N \to + \infty}{l i m} \underset{n = 0}{\sum^{N}} (\frac{1}{n + \frac{1}{3}} - \frac{\overset{}{1}}{n + \frac{1}{2}} + \frac{1}{- n + \frac{1}{3}} - \frac{1}{- n + \frac{1}{2}}) - 1$ $= \underset{N \to + \infty}{l i m} \underset{n = 0}{\sum^{N}} {3 (\frac{1}{3 n + 1} - \frac{1}{3 n - 1}) - 2 (\frac{1}{2 n + 1} - \frac{1}{2 n - 1})} - 1$ $= \underset{N \to + \infty}{l i m} \underset{n = 0}{\sum^{N - 1}} 3 (\frac{1}{3 n + 1} - \frac{1}{3 n + 2}) = 3 \underset{n = 0}{\sum^{+ \infty}} \frac{1}{(3 n + 1) (3 n + 2)}$ $P u t : f (x) = \underset{n = 0}{\sum^{+ \infty}} \frac{x^{3 n + 2}}{(3 n + 1) (3 n + 2)}, 0 ⩽ x < 1 .$ $f^{″} (x) = \underset{n = 0}{\sum^{+ \infty}} x^{3 n} = \frac{1}{1 - x^{3}} = \frac{1}{(1 - x) (1 + x + x^{2})} = \frac{1}{3} (\frac{1}{1 - x} + \frac{2 x + 1 + 3}{2 (x^{2} + x + 1)})$ $∴ f^{'} (x) + C = \frac{1}{3} (- L n (1 - x) + \frac{1}{2} L n (x^{2} + x + 1) + \sqrt{3} A r c t a n (\frac{2}{\sqrt{3}} (x + \frac{1}{2}))$ $f^{'} (0) = 0 \Rightarrow C = \frac{π}{6 \sqrt{3}}$ $f (1) = - \frac{1}{3} \int_{0}^{1} L n (x) d x + \frac{1}{6} \int_{0}^{1} L n (x^{2} + x + 1) d x + \frac{1}{\sqrt{3}} \int_{0}^{1} A r c t a n (\frac{2}{\sqrt{3}} (x + \frac{1}{2})) d x - \frac{π}{6 \sqrt{3}}$ $I = \int L n (x^{2} + x + 1) d x = x L n (x^{2} + x + 1) - \int \frac{2 (x^{2} + x + 1) - x - 2}{x^{2} + x + 1} d x$ $= x L n (x^{2} + x + 1) - 2 x + \int \frac{x + 2}{x^{2} + x + 1} d x$ $= (x + \frac{1}{2}) L n (x^{2} + x + 1) - 2 x + \sqrt{3} A r c t a n (\frac{2}{\sqrt{3}} (x + \frac{1}{2}))$ $\int A r c t a n (x) d x = x A r c t a n (x) - \frac{1}{2} L n (1 + x^{2})$ $S o : f (1) = \frac{1}{3} + \frac{1}{6} (\frac{3}{2} L n (3) - 2 + \frac{π \sqrt{3}}{3} - \frac{π \sqrt{3}}{6}) + \frac{1}{2} (\frac{π}{\sqrt{3}} - \frac{π}{6 \sqrt{3}} - \frac{1}{2} L n (3)) - \frac{π}{6 \sqrt{3}}$ $= \frac{π \sqrt{3}}{36} + \frac{5 π}{12 \sqrt{3}} - \frac{π}{6 \sqrt{3}} = \frac{π}{3 \sqrt{3}}, Θ = 3 f (1)$ $∴ Θ = \frac{π}{\sqrt{3}} \Rightarrow \underset{n = - \infty}{\sum^{+ \infty}} \frac{1}{6 n^{2} + 5 n + 1} = \frac{π \sqrt{3}}{3}$ $B E N A I C H A K a m e l$
	Commented by mnjuly1970 last updated on 19/Apr/21

	$t h a n k s a l o t m r K m e l . . .$
	Answered by Dwaipayan Shikari last updated on 19/Apr/21

	$\underset{n = - \infty}{\sum^{\infty}} \frac{1}{({a n}^{2} + b n + c)} = \frac{1}{\sqrt{b^{2} - 4 a c}} (ψ (\frac{b + \sqrt{b^{2} - 4 a c}}{2}) - ψ (\frac{b - \sqrt{b^{2} - 4 a c}}{2})) + Λ$ $Λ = \frac{1}{\sqrt{b^{2} - 4 a c}} (ψ (1 - \frac{b}{2 a} + \frac{\sqrt{b^{2} - 4 a c}}{2 a}) - ψ (1 - \frac{b}{2 a} - \frac{\sqrt{b^{2} - 4 a c}}{2 a}))$ $\underset{n = - \infty}{\sum^{\infty}} \frac{1}{(6 n^{2} + 5 n + 1)} = ψ (\frac{1}{2}) - ψ (\frac{1}{3}) + ψ (\frac{2}{3}) - ψ (\frac{1}{2})$ $= π c o t (\frac{π}{3}) = \frac{π}{\sqrt{3}}$
	Commented by mnjuly1970 last updated on 19/Apr/21

	$t h a n k y o u m r P a y a n . . .$
	Answered by mnjuly1970 last updated on 19/Apr/21

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