find sum of the series Σ_(n=0) ^∞ (((−1)^n )/((2n+1)(2n+3)))


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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
$S=Σ_(n=0) ^∞ (((−1)^n )/((2n+1)(2n+3))) =(1/2)Σ_(n=0) ^∞ (−1)^n {(1/(2n+1))−(1/(2n+3))} ⇒2S=Σ_(n=0) ^∞ (((−1)^n )/(2n+1)) −Σ_(n=0) ^∞ (((−1)^n )/(2n+3)) we[have Σ_(n=0) ^∞ (((−1)^n )/(2n+1)) =(π/4) Σ_(n=0) ^∞ (((−1)^n )/(2n+3)) =(1/3) +Σ_(n=1) ^∞ (((−1)^n )/(2n+3)) (n=p−1) =(1/3)+Σ_(p=2) ^∞ (((−1)^(p−1) )/(2p+1)) =(1/3) −Σ_(p=2) ^∞ (((−1)^p )/(2p+1)) =(1/3)−{Σ_(p=0) ^∞ (((−1)^p )/(2p+1))−(1−(1/3))} =(1/3)−(π/4) +(2/3)=1−(π/4) ⇒2S=(π/2)−1 S=(π/4)−(1/2)$
$S = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{(2 n + 1) (2 n + 3)} = \frac{1}{2} \sum_{n = 0}^{\infty} {(- 1)}^{n} {\frac{1}{2 n + 1} - \frac{1}{2 n + 3}}$ $\Rightarrow 2 S = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{2 n + 1} - \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{2 n + 3} w e [h a v e$ $\sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{2 n + 1} = \frac{π}{4}$ $\sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{2 n + 3} = \frac{1}{3} + \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{2 n + 3} (n = p - 1)$ $= \frac{1}{3} + \sum_{p = 2}^{\infty} \frac{{(- 1)}^{p - 1}}{2 p + 1} = \frac{1}{3} - \sum_{p = 2}^{\infty} \frac{{(- 1)}^{p}}{2 p + 1}$ $= \frac{1}{3} - {\sum_{p = 0}^{\infty} \frac{{(- 1)}^{p}}{2 p + 1} - (1 - \frac{1}{3})}$ $= \frac{1}{3} - \frac{π}{4} + \frac{2}{3} = 1 - \frac{π}{4} \Rightarrow 2 S = \frac{π}{2} - 1$ $S = \frac{π}{4} - \frac{1}{2}$
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
	$Σ_(n=0) ^(+∞) (((−1)^n )/(2n+1)) (1/(1+x^2 ))=Σ_(k≥0) (−1)^k x^(2k) ⇒arctan(x)=Σ_(k≥0) (((−1)^k )/(2k+1))x^(2k+1) ⇒Σ_(k≥0) (((−1)^k )/(2k+1))=arctan(1)=(π/4) Σ_(k≥0) (((−1)^k )/(2k+3))=Σ_(k≥1) (((−1)^(k−1) )/(2k+1))⇒Σ_(k≥0) (((−1)^k )/(2k+3))=−(Σ_(k≥0) (((−1)^k )/(2k+1))−1) =−(π/4)+1 Σ_(n≥0) (((−1)^n )/((2n+1)(2n+3)))=(1/2){Σ_(n≥0) (((−1)^n )/(2n+1))−Σ(((−1)^n )/(2n+3))} =(1/2){(π/4)−(−(π/4)+1)}=(π/4)−(1/2)$
	$\underset{n = 0}{\sum^{+ \infty}} \frac{{(- 1)}^{n}}{2 n + 1}$ $\frac{1}{1 + x^{2}} = \sum_{k ⩾ 0} {(- 1)}^{k} x^{2 k}$ $\Rightarrow a r c t a n (x) = \sum_{k ⩾ 0} \frac{{(- 1)}^{k}}{2 k + 1} x^{2 k + 1}$ $\Rightarrow \sum_{k ⩾ 0} \frac{{(- 1)}^{k}}{2 k + 1} = a r c t a n (1) = \frac{π}{4}$ $\sum_{k ⩾ 0} \frac{{(- 1)}^{k}}{2 k + 3} = \sum_{k ⩾ 1} \frac{{(- 1)}^{k - 1}}{2 k + 1} \Rightarrow \sum_{k ⩾ 0} \frac{{(- 1)}^{k}}{2 k + 3} = - (\sum_{k ⩾ 0} \frac{{(- 1)}^{k}}{2 k + 1} - 1)$ $= - \frac{π}{4} + 1$ $\sum_{n ⩾ 0} \frac{{(- 1)}^{n}}{(2 n + 1) (2 n + 3)} = \frac{1}{2} {\sum_{n ⩾ 0} \frac{{(- 1)}^{n}}{2 n + 1} - Σ \frac{{(- 1)}^{n}}{2 n + 3}}$ $= \frac{1}{2} {\frac{π}{4} - (- \frac{π}{4} + 1)} = \frac{π}{4} - \frac{1}{2}$
	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

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