Σ_(n=1) ^∞ (((−1)^(n+1) )/(3n−1)) = ?


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

$\underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n + 1}}{3 n - 1} = ?$
	Answered by Ar Brandon last updated on 01/Jan/23

	$S = \underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n + 1}}{3 n - 1} = \underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{3 n - 1}}{3 n - 1}$ $S (t) = \underset{n = 1}{\sum^{\infty}} \frac{t^{3 n - 1}}{3 n - 1} \Rightarrow S^{'} (t) = \underset{n = 1}{\sum^{\infty}} t^{3 n - 2} = \frac{t}{1 - t^{3}}$ $\Rightarrow S (t) = \int \frac{t}{1 - t^{3}} d t = - \int \frac{t}{(t - 1) (t^{2} + t + 1)} d t$ $\Rightarrow S (t) = - \int (\frac{1}{3 (t - 1)} - \frac{t - 1}{3 (t^{2} + t + 1)}) d t$ $\Rightarrow S (t) = - \frac{1}{3} \ln ∣ t - 1 ∣ + \frac{1}{6} \int \frac{2 t + 1}{t^{2} + t + 1} d t - \frac{1}{2} \int \frac{1}{t^{2} + t + 1} d t$ $= - \frac{1}{3} \ln ∣ t - 1 ∣ + \frac{1}{6} \ln (t^{2} + t + 1) - \frac{1}{\sqrt{3}} \arctan (\frac{2 t + 1}{\sqrt{3}}) + C$ $S (0) = 0 = - \frac{1}{\sqrt{3}} \arctan (\frac{1}{\sqrt{3}}) + C \Rightarrow C = \frac{π}{6 \sqrt{3}}$ $\Rightarrow S (t) = - \frac{1}{3} \ln ∣ t - 1 ∣ + \frac{1}{6} \ln (t^{2} + t + 1) - \frac{1}{\sqrt{3}} \arctan (\frac{2 t + 1}{\sqrt{3}}) + \frac{π}{6 \sqrt{3}}$ $\Rightarrow \underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n + 1}}{3 n - 1} = S (- 1) = - \frac{\ln 2}{3} - \frac{1}{\sqrt{3}} \arctan (- \frac{1}{\sqrt{3}}) + \frac{π}{6 \sqrt{3}}$ $\Rightarrow \begin{matrix} \underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n + 1}}{3 n - 1} = - \frac{\ln 2}{3} + \frac{π}{3 \sqrt{3}} \end{matrix}$
	Commented by Ar Brandon last updated on 01/Jan/23

	$★ ★ H a p p y N e w Y e a r f r i e n d s ★ ★$
	Commented by SEKRET last updated on 01/Jan/23

	$thanks sir$
	Commented by SEKRET last updated on 01/Jan/23

	$thanks$
	Answered by witcher3 last updated on 01/Jan/23
	$=−Σ_(n≥1) ∫_0 ^1 (−1)^(n+1) x^(3n−2) dx =−∫_0 ^1 (1/x^2 )Σ_(n≥1) (−x^3 )^n dx =−∫_0 ^1 ((−x)/(1+x^3 ))dx=−∫_0 ^1 (x/((x+1)(x^2 −x+1)))dx =∫_0 ^1 (((x+1)^2 −(x^2 −x+1))/((x+1)(x^2 −x+1)))dx ={∫_0 ^1 ((x+1)/(x^2 −x+1))−ln(2)} =∫_0 ^1 ((2x−1)/(2(x^2 −x+1)))dx+(3/2)∫_0 ^1 (dx/(x^2 −x+1))−ln(2) =−ln(2)+(3/2)∫_0 ^1 (dx/((x−(1/2))^2 +(3/4))) =−ln(2)+2∫_0 ^1 (dx/(1+(((2x)/( (√3)))−(1/( (√3))))^2 )) =−ln(2)+(√(3.))2.tan^(−1) ((1/( (√3)))) =−ln(2)+(π/( (√3)))$
	$= - \sum_{n ⩾ 1} \int_{0}^{1} {(- 1)}^{n + 1} x^{3 n - 2} d x$ $= - \int_{0}^{1} \frac{1}{x^{2}} \sum_{n ⩾ 1} {(- x^{3})}^{n} d x$ $= - \int_{0}^{1} \frac{- x}{1 + x^{3}} d x = - \int_{0}^{1} \frac{x}{(x + 1) (x^{2} - x + 1)} d x$ $= \int_{0}^{1} \frac{{(x + 1)}^{2} - (x^{2} - x + 1)}{(x + 1) (x^{2} - x + 1)} d x$ $= {\int_{0}^{1} \frac{x + 1}{x^{2} - x + 1} - l n (2)}$ $= \int_{0}^{1} \frac{2 x - 1}{2 (x^{2} - x + 1)} d x + \frac{3}{2} \int_{0}^{1} \frac{d x}{x^{2} - x + 1} - l n (2)$ $= - l n (2) + \frac{3}{2} \int_{0}^{1} \frac{d x}{{(x - \frac{1}{2})}^{2} + \frac{3}{4}}$ $= - l n (2) + 2 \int_{0}^{1} \frac{d x}{1 + {(\frac{2 x}{\sqrt{3}} - \frac{1}{\sqrt{3}})}^{2}}$ $= - l n (2) + \sqrt{3 .} 2 . \tan^{- 1} (\frac{1}{\sqrt{3}})$ $= - l n (2) + \frac{π}{\sqrt{3}}$
	Commented by Ar Brandon last updated on 01/Jan/23

	${(x + 1)}^{2} - (x^{2} - x + 1) = 3 x$
	Commented by SEKRET last updated on 01/Jan/23

	$thanks sir$
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