find-the-value-of-n-0-1-n-3n-1-

Question Number 26565 by abdo imad last updated on 26/Dec/17

f i n d t h e v a l u e o f \sum_{n = 0}^{\propto} \frac{{(- 1)}^{n}}{3 n + 1}

Commented by abdo imad last updated on 29/Dec/17

l e t p u t s (x) = \sum_{n = 0}^{n =\propto} \frac{x^{3 n + 1}}{3 n + 1} w i t h / x / < 1

s (^{3} \sqrt{x}) = {\sum_{n = 0}}^{\propto} \frac{{(^{3} \sqrt{x})}^{3 n + 1}}{3 n + 1} =^{3} \sqrt{x} \sum_{n = 0}^{\propto} \frac{x^{n}}{3 n + 1} a n d f o r x \neq 0

\sum_{n = 0}^{\propto} \frac{x^{n}}{3 n + 1} =^{3} {(\sqrt{x})}^{- 1} s (^{3} \sqrt{x)} a n d \sum_{n = 0}^{\propto} \frac{{(- 1)}^{n}}{3 n + 1} = - s (- 1)

s^{,} (x) = \sum_{n = 0}^{\propto} x^{3 n} = \frac{1}{1 - x^{3}} \Rightarrow ? s (x) = λ - \int \frac{d x}{x^{3} - 1}

a f t e r t h a t w e d e c o m p o s e F (x) = \frac{1}{x^{3} - 1} = \frac{a}{x - 1} + \frac{b x + c}{x^{2} + x + 1}

w e f i n d F (x) = \frac{1}{3 (x - 1)} - \frac{x + 2}{3 (x^{2} + x + 1)} a n d

\int F (x) d x = \frac{1}{3} l n / x - 1 / - \frac{1}{6} l n / x^{2} + x + 1 / + \frac{\sqrt{3}}{3} a r c t a n (\frac{2}{\sqrt{3}} (x + \frac{1}{2}))

s (x) = λ - \int / F (x) d x \Rightarrow s (0) = 0 = λ + \frac{\sqrt{3}}{3} a r c t a n (\frac{1}{\sqrt{3}})

λ = - \frac{π \sqrt{3}}{18} a n d s (x) = - \frac{π \sqrt{3}}{18} - \frac{1}{3} l n / x - 1 / + \frac{1}{6} l n / x^{2} + x + 1 / + \frac{\sqrt{3}}{3} a r c t a n (\frac{2}{\sqrt{3}} (x + \frac{1}{2}))

f i n a l l y \sum_{n = 0}^{\propto} \frac{{(- 1)}^{n}}{3 n + 1} = - s (- 1) = \frac{π \sqrt{3}}{9} + \frac{1}{3} l n 2 .