calculate Σ_(n=2) ^∞ ((2n+1)/(n^4 −n^2 ))


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Question Number 45240 by maxmathsup by imad last updated on 10/Oct/18

$c a l c u l a t e \sum_{n = 2}^{\infty} \frac{2 n + 1}{n^{4} - n^{2}}$
Commented by maxmathsup by imad last updated on 11/Oct/18
$let decompose F(x)=((2x+1)/(x^4 −x^2 )) =((2x+1)/(x^2 (x−1)(x+1))) ⇒ F(x)=(a/x) +(b/x^2 ) +(c/(x−1)) +(d/(x+1)) b =lim_(x→0) x^2 F(x) =−1 c =lim_(x→1) (x−1)F(x)=(3/2) d =lim_(x→−1) (x+1)F(x) =((−1)/(−2)) =(1/2) ⇒F(x)=(a/x) −(1/x^2 ) +(3/(2(x−1))) +(1/(2(x+1))) F(2) = (5/(4.3)) =(5/(12)) =(a/2) −(1/4) +(3/2) +(1/6) ⇒5 =6a−3 +18 +2 ⇒5=6a +17 ⇒ 6a=−12 ⇒a =−2 ⇒F(x)=−(2/x) −(1/x^2 ) +(3/(2(x−1))) +(1/(2(x+1))) let S_n =Σ_(k=2) ^∞ ((2k+1)/(k^4 −k^2 )) ⇒S_n =Σ_(k=2) ^n F(k) =−2 Σ_(k=2) ^n (1/k) −Σ_(k=2) ^n (1/k^2 ) +(3/2) Σ_(k=2) ^n (1/(k−1)) +(1/2)Σ_(k=2) ^n (1/(k+1)) but Σ_(k=2) ^n (1/k)=H_n −(3/2) Σ_(k=2) ^n =ξ_n (2)−1 Σ_(k=2) ^n (1/(k−1)) =Σ_(k=1) ^(n−1) (1/k) =H_(n−1) Σ_(k=2) ^n (1/(k+1)) =Σ_(k=3) ^(n+1) (1/k)=H_(n+1) −1−(1/2)−(1/3) =H_(n+1) −(3/2) −(1/3) =H_(n+1) −((11)/6) ⇒ S_n =−2H_n +3 +ξ_n (2)−1 +(3/2) H_(n−1) +(1/2) H_(n+1) −((11)/(12)) =−2H_n +(3/2) H_(n−1) +(1/2) H_(n+1) +ξ_n (2)+2−((11)/(12)) ⇒ S_n =−2(ln(n) +γ +o((1/n)))+(3/2)( ln(n−1)+γ +o((1/n)))+(1/2)(ln(n)+γ +o((1/n)))+ξ_n (2)+((13)/(12)) =−2ln(n) +(3/2)ln(n−1) +(1/2)ln(n) +o((1/n)) +ξ_n (2)+((13)/(12)) =ln((√((n−1)^3 ))(√n))−ln(n^2 ) +o((1/n))+ξ_n (2)+((13)/(12)) =ln{(((n−1)(√(n(n−1))))/n^2 )}+o((1/n))+ξ_n (2)+((13)/(12)) ⇒ lim_(n→+∞) S_n =ξ(2)+((13)/(12)) =(π^2 /6) +((13)/(12)) .$
$l e t d e c o m p o s e F (x) = \frac{2 x + 1}{x^{4} - x^{2}} = \frac{2 x + 1}{x^{2} (x - 1) (x + 1)} \Rightarrow$ $F (x) = \frac{a}{x} + \frac{b}{x^{2}} + \frac{c}{x - 1} + \frac{d}{x + 1}$ $b = {l i m}_{x \to 0} x^{2} F (x) = - 1$ $c = {l i m}_{x \to 1} (x - 1) F (x) = \frac{3}{2}$ $d = {l i m}_{x \to - 1} (x + 1) F (x) = \frac{- 1}{- 2} = \frac{1}{2} \Rightarrow F (x) = \frac{a}{x} - \frac{1}{x^{2}} + \frac{3}{2 (x - 1)} + \frac{1}{2 (x + 1)}$ $F (2) = \frac{5}{4.3} = \frac{5}{12} = \frac{a}{2} - \frac{1}{4} + \frac{3}{2} + \frac{1}{6} \Rightarrow 5 = 6 a - 3 + 18 + 2 \Rightarrow 5 = 6 a + 17 \Rightarrow$ $6 a = - 12 \Rightarrow a = - 2 \Rightarrow F (x) = - \frac{2}{x} - \frac{1}{x^{2}} + \frac{3}{2 (x - 1)} + \frac{1}{2 (x + 1)}$ $l e t S_{n} = \sum_{k = 2}^{\infty} \frac{2 k + 1}{k^{4} - k^{2}} \Rightarrow S_{n} = \sum_{k = 2}^{n} F (k)$ $= - 2 \sum_{k = 2}^{n} \frac{1}{k} - \sum_{k = 2}^{n} \frac{1}{k^{2}} + \frac{3}{2} \sum_{k = 2}^{n} \frac{1}{k - 1} + \frac{1}{2} \sum_{k = 2}^{n} \frac{1}{k + 1} b u t$ $\sum_{k = 2}^{n} \frac{1}{k} = H_{n} - \frac{3}{2}$ $\sum_{k = 2}^{n} = ξ_{n} (2) - 1$ $\sum_{k = 2}^{n} \frac{1}{k - 1} = \sum_{k = 1}^{n - 1} \frac{1}{k} = H_{n - 1}$ $\sum_{k = 2}^{n} \frac{1}{k + 1} = \sum_{k = 3}^{n + 1} \frac{1}{k} = H_{n + 1} - 1 - \frac{1}{2} - \frac{1}{3} = H_{n + 1} - \frac{3}{2} - \frac{1}{3} = H_{n + 1} - \frac{11}{6} \Rightarrow$ $S_{n} = - 2 H_{n} + 3 + ξ_{n} (2) - 1 + \frac{3}{2} H_{n - 1} + \frac{1}{2} H_{n + 1} - \frac{11}{12}$ $= - 2 H_{n} + \frac{3}{2} H_{n - 1} + \frac{1}{2} H_{n + 1} + ξ_{n} (2) + 2 - \frac{11}{12} \Rightarrow$ $S_{n} = - 2 (l n (n) + γ + o (\frac{1}{n})) + \frac{3}{2} (l n (n - 1) + γ + o (\frac{1}{n})) + \frac{1}{2} (l n (n) + γ + o (\frac{1}{n})) + ξ_{n} (2) + \frac{13}{12}$ $= - 2 l n (n) + \frac{3}{2} l n (n - 1) + \frac{1}{2} l n (n) + o (\frac{1}{n}) + ξ_{n} (2) + \frac{13}{12}$ $= l n (\sqrt{{(n - 1)}^{3}} \sqrt{n}) - l n (n^{2}) + o (\frac{1}{n}) + ξ_{n} (2) + \frac{13}{12}$ $= l n {\frac{(n - 1) \sqrt{n (n - 1)}}{n^{2}}} + o (\frac{1}{n}) + ξ_{n} (2) + \frac{13}{12} \Rightarrow$ ${l i m}_{n \to + \infty} S_{n} = ξ (2) + \frac{13}{12} = \frac{π^{2}}{6} + \frac{13}{12} .$
	Answered by ajfour last updated on 11/Oct/18

	$\underset{n = 2}{\sum^{\infty}} \frac{2 n + 1}{n^{2} (n - 1) (n + 1)}$ $= Σ \frac{(n + 1) + (n - 1)}{(n - 1) n n (n + 1)} + \frac{1}{2} Σ \frac{(n + 1) - (n - 1)}{(n - 1) n n (n + 1)}$ $= Σ \frac{1}{(n - 1) n^{2}} + Σ \frac{1}{n^{2} (n + 1)}$ $+ \frac{1}{2} Σ \frac{1}{(n - 1) n^{2}} - \frac{1}{2} Σ \frac{1}{n^{2} (n + 1)}$ $= \frac{3}{2} Σ \frac{1}{(n - 1) n^{2}} + \frac{1}{2} Σ \frac{1}{n^{2} (n + 1)}$ $= \frac{3}{4} Σ (\frac{1}{(n - 1) n} - \frac{1}{n^{2}}) + \frac{1}{4} Σ (\frac{1}{n^{2}} - \frac{1}{n (n + 1)})$ $= \frac{3}{4} Σ (\frac{1}{n - 1} - \frac{1}{n}) - \frac{1}{4} Σ (\frac{1}{n} - \frac{1}{n + 1})$ $- \frac{1}{2} Σ \frac{1}{n^{2}}$ $= \frac{3}{4} (1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + . .) - \frac{1}{4} (\frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} + . .)$ $- \frac{1}{2} (\frac{1}{4} + \frac{1}{9} + . . .)$ $= \frac{3}{4} - \frac{1}{8} - \frac{1}{2} (\frac{1}{4} + \frac{1}{9} + \frac{1}{16} + . .)$ $\ln (1 - x) = - (x + \frac{x^{2}}{2} + \frac{x^{3}}{3} + . . .)$ $\int_{0}^{1} \frac{\ln (1 - x)}{x} d x = - (1 + \frac{1}{4} + \frac{1}{9} + \frac{1}{16} + . .)$ $s o$ $\underset{n = 2}{\sum^{\infty}} \frac{2 n + 1}{n^{4} - n^{2}} = \frac{3}{4} - \frac{1}{8} - \frac{1}{2} [1 + \int_{0}^{1} \frac{\ln (1 - x)}{x} d x]$ $. . . . . (c a n t s o l v e t h e i n t e g r a l, s i r) .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 11/Oct/18
	$((2n+1)/(n^2 (n+1)(n−1)))=(a/n)+(b/n^2 )+(c/(n+1))+(d/(n−1)) 2n+1=an(n+1)(n−1)+b(n+1)(n−1)+cn^2 (n−1)+dn^2 (n+1) put n=0 1=b(−1) b=−1 put n−1=0 3=d×1×2 d=(3/2) put n+1=0 −2+1=c×(−1)^2 ×(−2) −1=−2c c=(1/2) 2n+1=an(n+1)(n−1)+b(n+1)(n−1)+cn^2 (n−1)+dn^2 (n+1) 2n+1=a(n^3 −n)+(−1)(n^2 −1)+(1/2)n^2 (n−1)+(3/2)n^2 (n+1) 2n+1=n^3 (a+(1/2)+(3/2))+n^2 (−1−(1/2)+(3/2))+n(−a)+1 a+2=0 a=−2 =Σ_(n=2) ^∞ (((−2)/n)+((−1)/n^2 )+((1/2)/(n+1))+((3/2)/(n−1))) =S_1 +S_2 +S_3 +S_4 S_1 =−2Σ_(n=2) ^∞ (1/n) S_1 =−2{((1/1)+(1/2)+(1/3)+...∞)−(1/1)} =−2(γ−1)=−2γ+1 γ=Eulers constant S_2 =−1Σ_(n=2) ^∞ (1/n^2 ) =−1{((1/1^2 )+(1/2^2 )+(1/3^2 )+..∞)−(1/1)} =−((π^2 /6)−1)=1−(π^2 /6) S_3 =(1/2)Σ_(n=2) ^∞ (1/(n+1)) =(1/2){((1/1)+(1/2)+(1/3)+...∞)−((1/1)+(1/2))} =(1/2)(γ−(3/2))=(γ/2)−(3/4) S_4 =(3/2)Σ_(n=2) ^∞ (1/(n−1)) (3/2)((1/1)+(1/2)+(1/3)+...) =((3γ)/2) so reauired ans is S_1 +S_2 +S_3 +S_4 =(−2γ+1)+(1−(π^2 /6))+((γ/2)−(3/4))+(((3γ)/2)) =−2γ+2γ+1+1−(3/4)−(π^2 /6) =(5/4)−(π^2 /6) pls check...$
	$\frac{2 n + 1}{n^{2} (n + 1) (n - 1)} = \frac{a}{n} + \frac{b}{n^{2}} + \frac{c}{n + 1} + \frac{d}{n - 1}$ $2 n + 1 = a n (n + 1) (n - 1) + b (n + 1) (n - 1) + {c n}^{2} (n - 1) + {d n}^{2} (n + 1)$ $p u t n = 0$ $1 = b (- 1) b = - 1$ $p u t n - 1 = 0$ $3 = d \times 1 \times 2 d = \frac{3}{2}$ $p u t n + 1 = 0$ $- 2 + 1 = c \times {(- 1)}^{2} \times (- 2)$ $- 1 = - 2 c c = \frac{1}{2}$ $2 n + 1 = a n (n + 1) (n - 1) + b (n + 1) (n - 1) + {c n}^{2} (n - 1) + {d n}^{2} (n + 1)$ $2 n + 1 = a (n^{3} - n) + (- 1) (n^{2} - 1) + \frac{1}{2} n^{2} (n - 1) + \frac{3}{2} n^{2} (n + 1)$ $2 n + 1 = n^{3} (a + \frac{1}{2} + \frac{3}{2}) + n^{2} (- 1 - \frac{1}{2} + \frac{3}{2}) + n (- a) + 1$ $a + 2 = 0 a = - 2$ $= \underset{n = 2}{\sum^{\infty}} (\frac{- 2}{n} + \frac{- 1}{n^{2}} + \frac{\frac{1}{2}}{n + 1} + \frac{\frac{3}{2}}{n - 1})$ $= S_{1} + S_{2} + S_{3} + S_{4}$ $S_{1} = - 2 \underset{n = 2}{\sum^{\infty}} \frac{1}{n}$ $S_{1} = - 2 {(\frac{1}{1} + \frac{1}{2} + \frac{1}{3} + . . . \infty) - \frac{1}{1}}$ $= - 2 (γ - 1) = - 2 γ + 1 γ = E u l e r s c o n s t a n t$ $S_{2} = - 1 \underset{n = 2}{\sum^{\infty}} \frac{1}{n^{2}}$ $= - 1 {(\frac{1}{1^{2}} + \frac{1}{2^{2}} + \frac{1}{3^{2}} + . . \infty) - \frac{1}{1}}$ $= - (\frac{π^{2}}{6} - 1) = 1 - \frac{π^{2}}{6}$ $S_{3} = \frac{1}{2} \underset{n = 2}{\sum^{\infty}} \frac{1}{n + 1}$ $= \frac{1}{2} {(\frac{1}{1} + \frac{1}{2} + \frac{1}{3} + . . . \infty) - (\frac{1}{1} + \frac{1}{2})}$ $= \frac{1}{2} (γ - \frac{3}{2}) = \frac{γ}{2} - \frac{3}{4}$ $S_{4} = \frac{3}{2} \underset{n = 2}{\sum^{\infty}} \frac{1}{n - 1}$ $\frac{3}{2} (\frac{1}{1} + \frac{1}{2} + \frac{1}{3} + . . .)$ $= \frac{3 γ}{2}$ $s o r e a u i r e d a n s i s S_{1} + S_{2} + S_{3} + S_{4}$ $= (- 2 γ + 1) + (1 - \frac{π^{2}}{6}) + (\frac{γ}{2} - \frac{3}{4}) + (\frac{3 γ}{2})$ $= - 2 γ + 2 γ + 1 + 1 - \frac{3}{4} - \frac{π^{2}}{6}$ $= \frac{5}{4} - \frac{π^{2}}{6} p l s c h e c k . . .$
	Commented by tanmay.chaudhury50@gmail.com last updated on 11/Oct/18

	Commented by tanmay.chaudhury50@gmail.com last updated on 11/Oct/18

	Commented by ajfour last updated on 11/Oct/18

	$s i r p l e a s e e x p l a i n m e h o w$ $\underset{n = 1}{\sum^{\infty}} \frac{1}{n^{2}} = \frac{π^{2}}{6} .$
	Commented by tanmay.chaudhury50@gmail.com last updated on 11/Oct/18

	Commented by tanmay.chaudhury50@gmail.com last updated on 11/Oct/18

	Commented by tanmay.chaudhury50@gmail.com last updated on 11/Oct/18

	Commented by tanmay.chaudhury50@gmail.com last updated on 11/Oct/18

	Commented by tanmay.chaudhury50@gmail.com last updated on 11/Oct/18

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