find-the-series-n-2-1-n-1-3n-1-1-3n-2-

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 = \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

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