calculate-n-1-1-n-3n-1-

Question Number 41349 by math khazana by abdo last updated on 06/Aug/18

c a l c u l a t e \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{3 n - 1}

Commented by maxmathsup by imad last updated on 06/Aug/18

l e t S = \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{3 n - 1} c h a n g e m e n t o f i n d i c e n = j + 1 g i v e

S = \sum_{j = 0}^{\infty} \frac{{(- 1)}^{j + 1}}{3 j + 2} = - \sum_{j = 0}^{\infty} \frac{{(- 1)}^{j}}{3 j + 2} \Rightarrow S = - \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{3 n + 2} l e t

w (x) = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{3 n + 2} x^{3 n + 2} w i t h ∣ x ∣< 1 w e h a v e S = w (1) a n d

w^{^{'}} (x) = \sum_{n = 0}^{\infty} {(- 1)}^{n} x^{3 n + 1} = x \sum_{n = 0}^{\infty} {(- x^{3})}^{n} = \frac{x}{1 + x^{3}} \Rightarrow

w (x) = \int \frac{x d x}{1 + x^{3}} + c l e t d e c o m p s e F (x) = \frac{x}{1 + x^{3}} = \frac{x}{(x + 1) (x^{2} - x + 1)}

F (x) = \frac{a}{x + 1} + \frac{b x + c}{x^{2} - x + 1}

a = {l i m}_{x \to - 1} (x + 1) F (x) = \frac{- 1}{3}

{l i m}_{x \to + \infty} x F (x) = 0 = a + b \Rightarrow b = \frac{1}{3} \Rightarrow F (x) = - \frac{1}{3 (x + 1)} + \frac{1}{3} \frac{x + 3 c}{x^{2} - x + 1}

F (0) = 0 = - \frac{1}{3} + c \Rightarrow c = \frac{1}{3} \Rightarrow F (x) = \frac{- 1}{3 (x + 1)} + \frac{1}{3} \frac{x + 1}{x^{2} - x + 1} \Rightarrow

\int F (x) d x = - \frac{1}{3} l n ∣ x + 1 ∣ + \frac{1}{6} \int \frac{2 x - 1 + 3}{x^{2} - x + 1} d x

= - \frac{1}{3} l n ∣ x + 1 ∣ + \frac{1}{6} l n (x^{2} - x + 1) + \frac{1}{2} \int \frac{d x}{{(x - \frac{1}{2})}^{2} + \frac{3}{4}} b u t

\int \frac{d x}{{(x - \frac{1}{2})}^{2} + \frac{3}{4}} =_{x - \frac{1}{2} = \frac{\sqrt{3}}{2} t} \frac{4}{3} \int \frac{1}{1 + t^{2}} \frac{\sqrt{3}}{2} d t

= \frac{2}{\sqrt{3}} a r c t a n (\frac{2 x - 1}{\sqrt{3}}) \Rightarrow w (x) = - \frac{1}{3} l n ∣ x + 1 ∣ + \frac{1}{6} l n (x^{2} - x + 1) + \frac{2}{\sqrt{3}} a r c t a n (\frac{2 x - 1}{\sqrt{3}}) + c

w (0) = 0 = - \frac{2}{\sqrt{3}} a r c t a n (\frac{1}{\sqrt{3}}) + c \Rightarrow c = \frac{2}{\sqrt{3}} a r c t a n (\frac{1}{\sqrt{3}}) = \frac{2}{\sqrt{3}} \frac{π}{6} = \frac{π}{3 \sqrt{3}} \Rightarrow

w (x) = - \frac{1}{3} l n ∣ x + 1 ∣ + \frac{1}{6} l n (x^{2} - x + 1) + \frac{2}{\sqrt{3}} a r c t a n (\frac{2 x - 1)}{\sqrt{3}}) + \frac{π}{3 \sqrt{3}}

S = w (1) = - \frac{1}{3} l n (2) + \frac{2}{\sqrt{3}} a r c t a n (\frac{1}{\sqrt{3}}) + \frac{π}{3 \sqrt{3}} = - \frac{l n (2)}{3} + \frac{2}{\sqrt{3}} \frac{π}{6} + \frac{π}{3 \sqrt{3}}

S = \frac{2 π}{3 \sqrt{3}} - \frac{l n (2)}{3} .

Commented by math khazana by abdo last updated on 07/Aug/18

S = - w (1) \Rightarrow S = \frac{l n (2)}{3} - \frac{2 π}{3 \sqrt{3}} .