calculate Σ_(n=2) ^∞ (1/(n^3 −n))


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Question Number 45237 by maxmathsup by imad last updated on 10/Oct/18

$c a l c u l a t e \sum_{n = 2}^{\infty} \frac{1}{n^{3} - n}$
Commented by maxmathsup by imad last updated on 11/Oct/18
$let decompose F(x)=(1/(x^3 −x)) =(1/(x(x−1)(x+1))) F(x)=(a/x) +(b/(x−1)) +(c/(x+1)) a =lim_(x→0) xF(x) =−1 b =lim_(x→1) (x−1)F(x)=(1/2) c =lim_(x→−1) (x+1)F(x) =(1/2) ⇒F(x)=−(1/x) +(1/(2(x−1))) +(1/(2(x+1))) let S_n =Σ_(k=2) ^n (1/(k^3 −k)) ⇒S_n =Σ_(k=2) ^n F(k) =−Σ_(k=2) ^n (1/k) +(1/2)Σ_(k=2) ^n (1/(k−1)) +(1/2)Σ_(k=2) ^n (1/(k+1)) but Σ_(k=2) ^n (1/k) =H_n −1 Σ_(k=2) ^n (1/(k−1)) =Σ_(k=1) ^(n−1) (1/k) =H_(n−1) Σ_(k=2) ^n (1/(k+1)) =Σ_(k=3) ^(n+1) (1/k)=H_(n+1) −(3/2) ⇒S_n =−H_(n+1) +1+(1/2) H_(n−1) +(1/2)H_(n+1) −(3/4) ⇒ S_n =(1/2)(ln(n−1)+γ +ln(n+1)+γ +o((1/n)))−ln(n)−γ +o((1/n)) +(1/4) =ln((√(n−1))(√(n+1)))−ln(n)+o((1/n))+(1/4) =ln{((√(n^2 −1))/n)}+o((1/n))+(1/4) →(1/4) (n→+∞) so lim_(n→+∞) S_n =(1/4) .$
$l e t d e c o m p o s e F (x) = \frac{1}{x^{3} - x} = \frac{1}{x (x - 1) (x + 1)}$ $F (x) = \frac{a}{x} + \frac{b}{x - 1} + \frac{c}{x + 1}$ $a = {l i m}_{x \to 0} x F (x) = - 1$ $b = {l i m}_{x \to 1} (x - 1) F (x) = \frac{1}{2}$ $c = {l i m}_{x \to - 1} (x + 1) F (x) = \frac{1}{2} \Rightarrow F (x) = - \frac{1}{x} + \frac{1}{2 (x - 1)} + \frac{1}{2 (x + 1)}$ $l e t S_{n} = \sum_{k = 2}^{n} \frac{1}{k^{3} - k} \Rightarrow S_{n} = \sum_{k = 2}^{n} F (k) = - \sum_{k = 2}^{n} \frac{1}{k} + \frac{1}{2} \sum_{k = 2}^{n} \frac{1}{k - 1} + \frac{1}{2} \sum_{k = 2}^{n} \frac{1}{k + 1}$ $b u t \sum_{k = 2}^{n} \frac{1}{k} = H_{n} - 1$ $\sum_{k = 2}^{n} \frac{1}{k - 1} = \sum_{k = 1}^{n - 1} \frac{1}{k} = H_{n - 1}$ $\sum_{k = 2}^{n} \frac{1}{k + 1} = \sum_{k = 3}^{n + 1} \frac{1}{k} = H_{n + 1} - \frac{3}{2} \Rightarrow S_{n} = - H_{n + 1} + 1 + \frac{1}{2} H_{n - 1} + \frac{1}{2} H_{n + 1} - \frac{3}{4}$ $\Rightarrow S_{n} = \frac{1}{2} (l n (n - 1) + γ + l n (n + 1) + γ + o (\frac{1}{n})) - l n (n) - γ + o (\frac{1}{n}) + \frac{1}{4}$ $= l n (\sqrt{n - 1} \sqrt{n + 1}) - l n (n) + o (\frac{1}{n}) + \frac{1}{4}$ $= l n {\frac{\sqrt{n^{2} - 1}}{n}} + o (\frac{1}{n}) + \frac{1}{4} \to \frac{1}{4} (n \to + \infty) s o {l i m}_{n \to + \infty} S_{n} = \frac{1}{4} .$
	Answered by Smail last updated on 11/Oct/18

	$\frac{1}{n (n^{2} - 1)} = \frac{1}{n (n - 1) (n + 1)} = \frac{a}{n} + \frac{b}{n - 1} + \frac{c}{n + 1}$ $a = - 1; b = \frac{1}{2}; c = \frac{1}{2}$ $\underset{n = 2}{\sum^{\infty}} \frac{1}{n^{3} - n} = \underset{n = 2}{\sum^{\infty}} (- \frac{1}{n} + \frac{1}{2 (n - 1)} + \frac{1}{2 (n + 1)})$ $= \frac{1}{2} \underset{n = 2}{\sum^{\infty}} \frac{1}{n - 1} + \frac{1}{2} \underset{n = 2}{\sum^{\infty}} \frac{1}{n + 1} - \underset{n = 2}{\sum^{\infty}} \frac{1}{n}$ $= \frac{1}{2} (1 + \frac{1}{2} + \frac{1}{3} + . . .) + \frac{1}{2} (\frac{1}{3} + \frac{1}{4} + . . .) - (\frac{1}{2} + \frac{1}{3} + . . .)$ $= \frac{1}{2} (1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{3} + \frac{1}{4} + \frac{1}{4} + . . .) - (\frac{1}{2} + \frac{1}{3} + . . .)$ $= \frac{1}{2} (1 + \frac{1}{2}) + (\frac{1}{3} + \frac{1}{4} + \frac{1}{5} + . . .) - \frac{1}{2} - (\frac{1}{3} + \frac{1}{4} + \frac{1}{5} + . . .)$ $= \frac{1}{2} (1 + \frac{1}{2}) - \frac{1}{2} = \frac{1}{4}$
	Commented by Tawa1 last updated on 11/Oct/18

	$Sir please what is the sum of nth term of \frac{1}{1.2.3} + \frac{1}{4.5.6} + \frac{1}{7.8.9} + . . . = ?$
	Commented by maxmathsup by imad last updated on 11/Oct/18

	$y o u r a n s w e r i s c o r r e c t s i r s m a 3 l t b a n k s . .$
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