calculate Σ_(n=0) ^∞ (((−1)^n )/(n^2 (3n+1)))


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Question Number 56330 by maxmathsup by imad last updated on 14/Mar/19

$c a l c u l a t e \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{n^{2} (3 n + 1)}$
Commented by maxmathsup by imad last updated on 24/Mar/19
$let decompose F(x)=(1/(x^2 (3x+1))) ⇒F(x)=(a/x) +(b/x^2 ) +(c/(3x+1)) b=lim_(x→0) x^2 F(x)=1 c =lim_(x→−(1/3)) (3x+1)F(x) =9 ⇒F(x)=(a/x) +(1/x^2 ) +(9/(3x+1)) F(−1) =−(1/2) =−a +1 −(9/2) ⇒−1 =−2a +2 −9 ⇒−1=−2a−7 ⇒ 2a =−6 ⇒a =−3 ⇒F(x)=−(3/x) +(1/x^2 ) +(9/(3x+1)) ⇒ Σ_(n=1) ^∞ (((−1)^n )/(n^2 (3n+1))) =−3 Σ_(n=1) ^∞ (((−1)^n )/n) +Σ_(n=1) ^∞ (((−1)^n )/n^2 ) +9 Σ_(n=1) ^∞ (((−1)^n )/(3n+1)) but Σ_(n=1) ^∞ (((−1)^n )/n) =−ln(2) let find Σ_(n=1) ^∞ (((−1)^n )/n^2 ) we have Σ_(n=1) ^∞ (((−1)^n )/n^2 ) =Σ_(n=1) ^∞ (1/(4n^2 )) −Σ_(n=0) ^∞ (1/((2n+1)^2 )) Σ_(n=1) ^∞ (1/n^2 ) =Σ_(n=1) ^∞ (1/(4n^2 )) +Σ_(n=0) ^∞ (1/((2n+1)^2 )) ⇒Σ_(n=0) ^∞ (1/((2n+1)^2 )) =(3/4) (π^2 /6) =(π^2 /8) ⇒ Σ_(n=1) ^∞ (((−1)^n )/n^2 ) =(π^2 /(24)) −(π^2 /8) =((π^2 −3π^2 )/(24)) =−(π^2 /(12)) let determine Σ_(n=1) ^∞ (((−1)^n )/(3n+1)) let S(x) =Σ_(n=1) ^∞ (−1)^n (x^(3n+1) /(3n+1)) with ∣x∣<1 ⇒(dS/dx)(x) =Σ_(n=1) ^∞ (−1)^n x^(3n) =Σ_(n=1) ^∞ (−x^3 )^n =(1/(1+x^3 )) −1 =((−x^3 )/(1+x^3 )) ⇒S(x) =∫_0 ^x ((−t^3 )/(1+t^3 )) dt +λ S(0)=0 ⇒λ =0 ⇒ S(x) =−∫_0 ^x ((t^3 +1−1)/(t^3 +1)) dt =−x + ∫_0 ^x (dt/(t^3 +1)) let decompose F(t) =(1/(t^3 +1)) =(1/((t+1)(t^2 −t+1))) F(t)=(a/(t+1)) +((bt +c)/(t^2 −t +1)) a =lim_(t→−1) (t+1)F(t)=(1/3) lim_(t→+∞) t F(t) =0 =a+b ⇒b =−(1/3) ⇒F(t) =(1/(3(t+1))) −(1/3) ((t−3c)/(t^2 −t +1)) F(o) =1 =(1/3) +c ⇒c =(2/3) ⇒F(t)=(1/(3(t+1))) −(1/3) ((t−2)/(t^2 −t+1)) ⇒ ∫ F(t)dt =(1/3)ln∣t+1∣ −(1/6) ∫ ((2t−1−3)/(t^2 −t+1)) dt =(1/3)ln∣t+1∣−(1/6)ln(t^2 −t +1) +(1/2) ∫ (dt/(t^2 −t +1)) but ∫ (dt/(t^2 −t +1)) =∫ (dt/((t−(1/2))^2 +(3/4))) =_(t+(1/2)=((√3)/2)u) ∫ (1/((3/4)(1+u^2 ))) ((√3)/2) du =(4/3) ((√3)/2) arctanu +c =(2/(√3)) arctan(((2t+1)/(√3))) ⇒ S(x)=−x +[(1/3)ln∣t+1∣−(1/6)ln(t^2 −t+1)+(1/(√3)) arctan(((2t+1)/(√3)))]_0 ^x =−x +{(1/3)ln∣x+1∣−(1/6)ln(x^2 −x+1)+(1/(√3)) arctan(((2x+1)/(√3)))−(1/(√3)) arctan((1/(√3)))} ⇒ Σ_(n=1) ^∞ (((−1)^n )/(3n+1)) =S(1) =−1 +(1/3)ln(2) +(1/(√3)) arctan((√3))−(1/(√3)) arctan((1/(√3))) =−1 +(1/3)ln(2) +(1/(√3)) arctan((√3))−(1/(√3))((π/2) −arctan((√3))) =−1 +((ln(2))/3) +(2/(√3)) arctan((√3)) −(π/(2(√3))) ⇒ Σ_(n=1) ^∞ (((−1)^n )/(n^2 (3n+1))) = 3ln(2) −(π^2 /(12)) −9 +3ln(2) +((18)/(√3)) arctan((√3)) −((9π)/(2(√3))) .$
$l e t d e c o m p o s e F (x) = \frac{1}{x^{2} (3 x + 1)} \Rightarrow F (x) = \frac{a}{x} + \frac{b}{x^{2}} + \frac{c}{3 x + 1}$ $b = {l i m}_{x \to 0} x^{2} F (x) = 1$ $c = {l i m}_{x \to - \frac{1}{3}} (3 x + 1) F (x) = 9 \Rightarrow F (x) = \frac{a}{x} + \frac{1}{x^{2}} + \frac{9}{3 x + 1}$ $F (- 1) = - \frac{1}{2} = - a + 1 - \frac{9}{2} \Rightarrow - 1 = - 2 a + 2 - 9 \Rightarrow - 1 = - 2 a - 7 \Rightarrow$ $2 a = - 6 \Rightarrow a = - 3 \Rightarrow F (x) = - \frac{3}{x} + \frac{1}{x^{2}} + \frac{9}{3 x + 1} \Rightarrow$ $\sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n^{2} (3 n + 1)} = - 3 \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n} + \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n^{2}} + 9 \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{3 n + 1}$ $b u t \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n} = - l n (2) l e t f i n d \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n^{2}}$ $w e h a v e \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n^{2}} = \sum_{n = 1}^{\infty} \frac{1}{4 n^{2}} - \sum_{n = 0}^{\infty} \frac{1}{{(2 n + 1)}^{2}}$ $\sum_{n = 1}^{\infty} \frac{1}{n^{2}} = \sum_{n = 1}^{\infty} \frac{1}{4 n^{2}} + \sum_{n = 0}^{\infty} \frac{1}{{(2 n + 1)}^{2}} \Rightarrow \sum_{n = 0}^{\infty} \frac{1}{{(2 n + 1)}^{2}} = \frac{3}{4} \frac{π^{2}}{6} = \frac{π^{2}}{8} \Rightarrow$ $\sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n^{2}} = \frac{π^{2}}{24} - \frac{π^{2}}{8} = \frac{π^{2} - 3 π^{2}}{24} = - \frac{π^{2}}{12} l e t d e t e r m i n e \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{3 n + 1}$ $l e t S (x) = \sum_{n = 1}^{\infty} {(- 1)}^{n} \frac{x^{3 n + 1}}{3 n + 1} w i t h ∣ x ∣< 1 \Rightarrow \frac{d S}{d x} (x) = \sum_{n = 1}^{\infty} {(- 1)}^{n} x^{3 n}$ $= \sum_{n = 1}^{\infty} {(- x^{3})}^{n} = \frac{1}{1 + x^{3}} - 1 = \frac{- x^{3}}{1 + x^{3}} \Rightarrow S (x) = \int_{0}^{x} \frac{- t^{3}}{1 + t^{3}} d t + λ$ $S (0) = 0 \Rightarrow λ = 0 \Rightarrow S (x) = - \int_{0}^{x} \frac{t^{3} + 1 - 1}{t^{3} + 1} d t$ $= - x + \int_{0}^{x} \frac{d t}{t^{3} + 1} l e t d e c o m p o s e F (t) = \frac{1}{t^{3} + 1} = \frac{1}{(t + 1) (t^{2} - t + 1)}$ $F (t) = \frac{a}{t + 1} + \frac{b t + c}{t^{2} - t + 1}$ $a = {l i m}_{t \to - 1} (t + 1) F (t) = \frac{1}{3}$ ${l i m}_{t \to + \infty} t F (t) = 0 = a + b \Rightarrow b = - \frac{1}{3} \Rightarrow F (t) = \frac{1}{3 (t + 1)} - \frac{1}{3} \frac{t - 3 c}{t^{2} - t + 1}$ $F (o) = 1 = \frac{1}{3} + c \Rightarrow c = \frac{2}{3} \Rightarrow F (t) = \frac{1}{3 (t + 1)} - \frac{1}{3} \frac{t - 2}{t^{2} - t + 1} \Rightarrow$ $\int F (t) d t = \frac{1}{3} l n ∣ t + 1 ∣ - \frac{1}{6} \int \frac{2 t - 1 - 3}{t^{2} - t + 1} d t$ $= \frac{1}{3} l n ∣ t + 1 ∣ - \frac{1}{6} l n (t^{2} - t + 1) + \frac{1}{2} \int \frac{d t}{t^{2} - t + 1} b u t$ $\int \frac{d t}{t^{2} - t + 1} = \int \frac{d t}{{(t - \frac{1}{2})}^{2} + \frac{3}{4}} =_{t + \frac{1}{2} = \frac{\sqrt{3}}{2} u} \int \frac{1}{\frac{3}{4} (1 + u^{2})} \frac{\sqrt{3}}{2} d u$ $= \frac{4}{3} \frac{\sqrt{3}}{2} a r c t a n u + c = \frac{2}{\sqrt{3}} a r c t a n (\frac{2 t + 1}{\sqrt{3}}) \Rightarrow$ $S (x) = - x + {[\frac{1}{3} l n ∣ t + 1 ∣ - \frac{1}{6} l n (t^{2} - t + 1) + \frac{1}{\sqrt{3}} a r c t a n (\frac{2 t + 1}{\sqrt{3}})]}_{0}^{x}$ $= - x + {\frac{1}{3} l n ∣ x + 1 ∣ - \frac{1}{6} l n (x^{2} - x + 1) + \frac{1}{\sqrt{3}} a r c t a n (\frac{2 x + 1}{\sqrt{3}}) - \frac{1}{\sqrt{3}} a r c t a n (\frac{1}{\sqrt{3}})}$ $\Rightarrow \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{3 n + 1} = S (1) = - 1 + \frac{1}{3} l n (2) + \frac{1}{\sqrt{3}} a r c t a n (\sqrt{3}) - \frac{1}{\sqrt{3}} a r c t a n (\frac{1}{\sqrt{3}})$ $= - 1 + \frac{1}{3} l n (2) + \frac{1}{\sqrt{3}} a r c t a n (\sqrt{3}) - \frac{1}{\sqrt{3}} (\frac{π}{2} - a r c t a n (\sqrt{3}))$ $= - 1 + \frac{l n (2)}{3} + \frac{2}{\sqrt{3}} a r c t a n (\sqrt{3}) - \frac{π}{2 \sqrt{3}} \Rightarrow$ $\sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n^{2} (3 n + 1)} = 3 l n (2) - \frac{π^{2}}{12} - 9 + 3 l n (2) + \frac{18}{\sqrt{3}} a r c t a n (\sqrt{3}) - \frac{9 π}{2 \sqrt{3}} .$
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