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Question Number 115405 by mathdave last updated on 25/Sep/20

f i n d t h e c l o s e f o r m o f

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

Answered by mathmax by abdo last updated on 25/Sep/20

S = \lim_{n \to + \infty} S_{n} with S_{n} = \sum_{k = 0}^{n} \frac{{(- 1)}^{k}}{(k + 1) (k + 2) (2 k + 1) (2 k + 3)}

let decompose F (x) = \frac{1}{(x + 1) (x + 2) (2 x + 1) (2 x + 3)}

F (x) = \frac{a}{x + 1} + \frac{b}{x + 2} + \frac{c}{2 x + 1} + \frac{d}{2 x + 3}

a = \frac{1}{(- 1)} = - 1, b = \frac{1}{(- 1) (- 1) (1)} = 1

c = \frac{1}{(\frac{1}{2}) (\frac{3}{2}) (2)} = \frac{2}{3}, d = \frac{1}{(\frac{- 1}{2}) (\frac{1}{2}) (- 2)} = 2 \Rightarrow

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

S_{n} = - \sum_{k = 0}^{n} \frac{{(- 1)}^{k}}{k + 1} + \sum_{k = 0}^{n} \frac{{(- 1)}^{k}}{k + 2} + \frac{2}{3} \sum_{k = 0}^{n} \frac{{(- 1)}^{k}}{2 k + 1} + 2 \sum_{k = 0}^{n} \frac{{(- 1)}^{k}}{2 k + 3}

we have - \sum_{k = 0}^{n} \frac{{(- 1)}^{k}}{k + 1} = - \sum_{k = 1}^{n + 1} \frac{{(- 1)}^{k - 1}}{k} = \sum_{k = 1}^{n + 1} \frac{{(- 1)}^{k}}{k} \to - \ln (2)

\sum_{k = 0}^{n} \frac{{(- 1)}^{k}}{k + 2} = \sum_{k = 2}^{n + 2} \frac{{(- 1)}^{k}}{k} = \sum_{k = 1}^{n + 2} \frac{{(- 1)}^{k}}{k} + 1 \to 1 - \ln (2)

\sum_{k = 0}^{n} \frac{{(- 1)}^{k}}{2 k + 1} \to \frac{π}{4}

\sum_{k = 0}^{n} \frac{{(- 1)}^{k}}{2 k + 3} =_{k = p - 1} \sum_{p = 1}^{n + 1} \frac{{(- 1)}^{p - 1}}{2 p + 1} = - \sum_{p = 1}^{n + 1} \frac{{(- 1)}^{p}}{2 p + 1}

= - \sum_{p = 0}^{n + 1} \frac{{(- 1)}^{p}}{2 p + 1} + 1 = 1 - \frac{π}{4} \Rightarrow

S = \lim_{n \to + \infty} S_{n} = - \ln (2) + 1 - \ln (2) + \frac{2}{3} . \frac{π}{4} + 2 (1 - \frac{π}{4})

= 1 - 2 \ln (2) + \frac{π}{6} + 2 - \frac{π}{2} = 3 - 2 \ln (2) - \frac{π}{3}

Commented by mathdave last updated on 25/Sep/20

b e a u t i f u l a p p r o a c h e s b u t l i t t l e m i s t a k e

Commented by mathmax by abdo last updated on 25/Sep/20

i think there is no mistakes \dots!

Commented by Tawa11 last updated on 06/Sep/21

Great sir

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