calculate Σ_(n=0) ^∞ ((n(−1)^n )/((2n+1)^2 (n+3)))


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Question Number 136404 by mathmax by abdo last updated on 21/Mar/21

$calculate \sum_{n = 0}^{\infty} \frac{n {(- 1)}^{n}}{{(2 n + 1)}^{2} (n + 3)}$
	Answered by mindispower last updated on 21/Mar/21
	$−(1/2){Σ_(n≥0) (((−1)^n )/((2n+1)(n+3)))−Σ_(n≥0) (((−1)^n )/((2n+1)^2 ))}=−(1/2)A A=Σ(2/5).(((−1)^n )/(2n+1))−(1/5)Σ_(n≥0) (((−1)^n )/(n+3)) =(2/5)Σ_(n≥0) (((−1)^n )/(2n+1))−(1/5)Σ(((−1)^(n+3) )/(n+3)) =(2/5)tan^(−1) (1)−(1/5)Σ_(n≥2) (((−1)^n )/(n+1))−G =(π/(10))−(1/5)(ln(2)−(1/2))−G we get −(π/(20))+((ln(2))/(10))−(1/(20))+(G/2),G catalan constante$
	$- \frac{1}{2} {\sum_{n ⩾ 0} \frac{{(- 1)}^{n}}{(2 n + 1) (n + 3)} - \sum_{n ⩾ 0} \frac{{(- 1)}^{n}}{{(2 n + 1)}^{2}}} = - \frac{1}{2} A$ $A = Σ \frac{2}{5} . \frac{{(- 1)}^{n}}{2 n + 1} - \frac{1}{5} \sum_{n ⩾ 0} \frac{{(- 1)}^{n}}{n + 3}$ $= \frac{2}{5} \sum_{n ⩾ 0} \frac{{(- 1)}^{n}}{2 n + 1} - \frac{1}{5} Σ \frac{{(- 1)}^{n + 3}}{n + 3}$ $= \frac{2}{5} \tan^{- 1} (1) - \frac{1}{5} \sum_{n ⩾ 2} \frac{{(- 1)}^{n}}{n + 1} - G$ $= \frac{π}{10} - \frac{1}{5} (l n (2) - \frac{1}{2}) - G$ $w e g e t - \frac{π}{20} + \frac{l n (2)}{10} - \frac{1}{20} + \frac{G}{2}, G c a t a l a n c o n s t a n t e$
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