find-sum-of-the-series-n-0-1-n-2n-1-2n-3-

Question Number 80708 by M±th+et£s last updated on 05/Feb/20

f i n d s u m o f t h e s e r i e s

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

Commented by abdomathmax last updated on 05/Feb/20

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

\Rightarrow 2 S = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{2 n + 1} - \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{2 n + 3} w e [h a v e

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

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

= \frac{1}{3} + \sum_{p = 2}^{\infty} \frac{{(- 1)}^{p - 1}}{2 p + 1} = \frac{1}{3} - \sum_{p = 2}^{\infty} \frac{{(- 1)}^{p}}{2 p + 1}

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

= \frac{1}{3} - \frac{π}{4} + \frac{2}{3} = 1 - \frac{π}{4} \Rightarrow 2 S = \frac{π}{2} - 1

S = \frac{π}{4} - \frac{1}{2}

Commented by M±th+et£s last updated on 05/Feb/20

n i c e s o l u t i o n s i r

Commented by mathmax by abdo last updated on 05/Feb/20

t h a n k s .

Answered by mind is power last updated on 05/Feb/20

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

\frac{1}{1 + x^{2}} = \sum_{k ⩾ 0} {(- 1)}^{k} x^{2 k}

\Rightarrow a r c t a n (x) = \sum_{k ⩾ 0} \frac{{(- 1)}^{k}}{2 k + 1} x^{2 k + 1}

\Rightarrow \sum_{k ⩾ 0} \frac{{(- 1)}^{k}}{2 k + 1} = a r c t a n (1) = \frac{π}{4}

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

= - \frac{π}{4} + 1

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

= \frac{1}{2} {\frac{π}{4} - (- \frac{π}{4} + 1)} = \frac{π}{4} - \frac{1}{2}

Commented by mr W last updated on 05/Feb/20

w o n d e r f u l s o l u t i o n s i r!

Commented by M±th+et£s last updated on 05/Feb/20

g o d b l e s s y o u s i r . t h a n k y o u