find the series Σ_(n=2) ^∞ (−1)^n [(1/(3n+1))+(1/(3n−2))]


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Question Number 110222 by mathdave last updated on 28/Aug/20

$f i n d t h e s e r i e s$ $\underset{n = 2}{\sum^{\infty}} {(- 1)}^{n} [\frac{1}{3 n + 1} + \frac{1}{3 n - 2}]$
	Answered by mnjuly1970 last updated on 27/Aug/20

	$f i n d t h e s e r i e s$ $\underset{n = 2}{\sum^{\infty}} {(- 1)}^{n} [\frac{1}{3 n + 1} + \frac{1}{3 n + 2}]$
	Answered by mathmax by abdo last updated on 28/Aug/20
	$S =Σ_(n=2) ^∞ (((−1)^n )/(3n+1)) +Σ_(n=2) ^∞ (((−1)^n )/(3n−2)) =A +B A =Σ_(n=0) ^(∞ ) (((−1)^n )/(3n+1)) −(1−(1/4)) =−(3/4)+Σ_(n=0) ^∞ (((−1)^n )/(3n+1)) let ϕ(x) =Σ_(n=0) ^∞ (−1)^n (x^(3n+1) /(3n+1)) (∣x∣≤1) ϕ^′ (x) =Σ_(n=0) ^∞ (−1)^n x^(3n) =(1/(1+x^3 )) ⇒ϕ(x) =∫_0 ^x (dx/(1+x^3 )) +c (c=ϕ(0)=0) F(x) =(1/((x+1)(x^2 −x+1))) =(a/(x+1)) +((bx +c)/(x^2 −x+1)) a=(1/3) ,lim_(x→+∞) xF(x) =0 =a+b ⇒b=−(1/3) F(0) =1 =a +c ⇒c=1−(1/3) =(2/3) ⇒F(x)=(1/(3(x+1)))+((−(1/3)x+(2/3))/(x^2 −x+1)) A =∫_0 ^1 (dx/(x^3 +1)) =(1/3)[ln∣x+1∣]_0 ^1 −(1/6)∫_0 ^1 ((2x−1−1)/(x^2 −x+1))dx +(2/3)∫_0 ^1 (dx/(x^2 −x+1)) =((ln2)/3) −(1/6)[ln(x^2 −x+1)]_0 ^1 +(5/6) ∫_0 ^1 (dx/(x^2 −x+1)) we have ∫_0 ^1 (dx/(x^2 −x+1)) =∫_0 ^1 (dx/((x−(1/2))^2 +(3/4))) =_(x−(1/2)=((√3)/2)t) (4/3) ∫_(−(1/(√3))) ^(1/(√3)) (1/(u^2 +1))((√3)/2)dt =(2/(√3))[arctanu]_(−(1/(√3))) ^(1/(√3)) =(2/(√3)){((2π)/6)} =((2π)/(3(√3))) ⇒ A =((ln2)/3)+(5/6).((2π)/(3(√3)))−(3/4) ⇒A =((ln(2))/3) +((5π)/(9(√3)))−(3/4) B =Σ_(n=2) ^∞ (((−1)^n )/(3n−2)) =_(n=p+1) Σ_(p=1) ^∞ (((−1)^(p+1) )/(3p+1)) =−Σ_(n=1) ^∞ (((−1)^n )/(3n+1)) =−(Σ_(n=0) ^∞ (((−1)^n )/(3n+1))−1)=1−Σ_(n=0) ^∞ (((−1)^n )/(3n+1)) =1−(((ln(2))/3)+((5π)/(9(√3)))) ⇒ S =A +B =((ln(2))/3)+((5π)/(9(√3)))−(3/4) +1−((ln2)/3)−((5π)/(9(√3))) ⇒ S =(1/4)$
	$S = \sum_{n = 2}^{\infty} \frac{{(- 1)}^{n}}{3 n + 1} + \sum_{n = 2}^{\infty} \frac{{(- 1)}^{n}}{3 n - 2} = A + B$ $A = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{3 n + 1} - (1 - \frac{1}{4}) = - \frac{3}{4} + \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{3 n + 1}$ $let φ (x) = \sum_{n = 0}^{\infty} {(- 1)}^{n} \frac{x^{3 n + 1}}{3 n + 1} (∣ x ∣⩽ 1)$ $φ^{^{'}} (x) = \sum_{n = 0}^{\infty} {(- 1)}^{n} x^{3 n} = \frac{1}{1 + x^{3}} \Rightarrow φ (x) = \int_{0}^{x} \frac{dx}{1 + x^{3}} + c (c = φ (0) = 0)$ $F (x) = \frac{1}{(x + 1) (x^{2} - x + 1)} = \frac{a}{x + 1} + \frac{bx + c}{x^{2} - x + 1}$ $a = \frac{1}{3}, \lim_{x \to + \infty} xF (x) = 0 = a + b \Rightarrow b = - \frac{1}{3}$ $F (0) = 1 = a + c \Rightarrow c = 1 - \frac{1}{3} = \frac{2}{3} \Rightarrow F (x) = \frac{1}{3 (x + 1)} + \frac{- \frac{1}{3} x + \frac{2}{3}}{x^{2} - x + 1}$ $A = \int_{0}^{1} \frac{dx}{x^{3} + 1} = \frac{1}{3} {[\ln ∣ x + 1 ∣]}_{0}^{1} - \frac{1}{6} \int_{0}^{1} \frac{2 x - 1 - 1}{x^{2} - x + 1} dx + \frac{2}{3} \int_{0}^{1} \frac{dx}{x^{2} - x + 1}$ $= \frac{\ln 2}{3} - \frac{1}{6} {[\ln (x^{2} - x + 1)]}_{0}^{1} + \frac{5}{6} \int_{0}^{1} \frac{dx}{x^{2} - x + 1} we have$ $\int_{0}^{1} \frac{dx}{x^{2} - x + 1} = \int_{0}^{1} \frac{dx}{{(x - \frac{1}{2})}^{2} + \frac{3}{4}} =_{x - \frac{1}{2} = \frac{\sqrt{3}}{2} t} \frac{4}{3} \int_{- \frac{1}{\sqrt{3}}}^{\frac{1}{\sqrt{3}}} \frac{1}{u^{2} + 1} \frac{\sqrt{3}}{2} dt$ $= \frac{2}{\sqrt{3}} {[arctanu]}_{- \frac{1}{\sqrt{3}}}^{\frac{1}{\sqrt{3}}} = \frac{2}{\sqrt{3}} {\frac{2 π}{6}} = \frac{2 π}{3 \sqrt{3}} \Rightarrow A = \frac{\ln 2}{3} + \frac{5}{6} . \frac{2 π}{3 \sqrt{3}} - \frac{3}{4}$ $\Rightarrow A = \frac{\ln (2)}{3} + \frac{5 π}{9 \sqrt{3}} - \frac{3}{4}$ $B = \sum_{n = 2}^{\infty} \frac{{(- 1)}^{n}}{3 n - 2} =_{n = p + 1} \sum_{p = 1}^{\infty} \frac{{(- 1)}^{p + 1}}{3 p + 1}$ $= - \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{3 n + 1} = - (\sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{3 n + 1} - 1) = 1 - \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{3 n + 1}$ $= 1 - (\frac{\ln (2)}{3} + \frac{5 π}{9 \sqrt{3}}) \Rightarrow$ $S = A + B = \frac{\ln (2)}{3} + \frac{5 π}{9 \sqrt{3}} - \frac{3}{4} + 1 - \frac{\ln 2}{3} - \frac{5 π}{9 \sqrt{3}} \Rightarrow S = \frac{1}{4}$
	Commented by mathdave last updated on 28/Aug/20

	$g o o d w o r k$
	Answered by mnjuly1970 last updated on 28/Aug/20

	Commented by mathdave last updated on 28/Aug/20

	$k e e p t h e s p i r i t u p$
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