find-the-sum-of-n-1-1-n-2n-1-3-n-

Question Number 90570 by Tony Lin last updated on 24/Apr/20

f i n d t h e s u m o f \underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n}}{(2 n + 1) 3^{n}}

Commented by mathmax by abdo last updated on 24/Apr/20

S = \sqrt{3} \sum_{n = 1}^{\infty} \frac{1}{2 n + 1} {(- 1)}^{n} {(\frac{1}{\sqrt{3}})}^{2 n + 1} = \sqrt{3} f (\frac{1}{\sqrt{3}}) w i t h

f (x) = \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{2 n + 1} x^{2 n + 1} = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{2 n + 1} x^{2 n + 1} - x (∣ x ∣< 1)

w e h a v e f^{^{'}} (x) = \sum_{n = 0}^{\infty} {(- 1)}^{n} x^{2 n} - 1 = \sum_{n = 0}^{\infty} {(- x^{2})}^{n} - 1 = \frac{1}{1 + x^{2}} - 1 \Rightarrow

f (x) = \int \frac{d x}{1 + x^{2}} - x + c = a r c t a n (x) - x + c

f (0) = 0 = c \Rightarrow f (x) = a r c t a n (x) - x \Rightarrow

S = \sqrt{3} {a r c t a n (\frac{1}{\sqrt{3}}) - \frac{1}{\sqrt{3}}} = \sqrt{3} a r c t a n (\frac{1}{\sqrt{3}}) - 1

S = \frac{π \sqrt{3}}{6} - 1

Answered by mr W last updated on 24/Apr/20

1 - x + x^{2} - x^{3} + \dots = \underset{n = 0}{\sum^{\infty}} {(- x)}^{n} = \frac{1}{1 + x} f o r ∣ x ∣< 1

r e p l a c e x w i t h x^{2}

\underset{n = 0}{\sum^{\infty}} {(- 1)}^{n} x^{2 n} = \frac{1}{1 + x^{2}}

\underset{n = 0}{\sum^{\infty}} [{(- 1)}^{n} \int_{0}^{x} x^{2 n} d x] = \int_{0}^{x} \frac{d x}{1 + x^{2}}

\underset{n = 0}{\sum^{\infty}} [{(- 1)}^{n} \frac{x^{2 n + 1}}{2 n + 1}] = \tan^{- 1} x

\underset{n = 0}{\sum^{\infty}} [{(- 1)}^{n} \frac{x^{2 n}}{2 n + 1}] = \frac{\tan^{- 1} x}{x}

r e p l a c e x w i t h \sqrt{x}

\underset{n = 0}{\sum^{\infty}} [{(- 1)}^{n} \frac{x^{n}}{2 n + 1}] = \frac{\tan^{- 1} \sqrt{x}}{\sqrt{x}}

r e p l a c e x w i t h \frac{1}{3}

\underset{n = 0}{\sum^{\infty}} \frac{{(- 1)}^{n}}{(2 n + 1) 3^{n}} = \sqrt{3} \tan^{- 1} \frac{1}{\sqrt{3}}

\Rightarrow \underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n}}{(2 n + 1) 3^{n}} = \sqrt{3} \tan^{- 1} \frac{1}{\sqrt{3}} - 1

Commented by Tony Lin last updated on 25/Apr/20

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