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


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Question Number 45598 by maxmathsup by imad last updated on 14/Oct/18

$c a l c u l a t e \sum_{n = 1}^{\infty} \frac{2 n + 1}{n^{2} (4 n^{2} - 1)}$
Commented by maxmathsup by imad last updated on 15/Oct/18
$let S_n =Σ_(k=1) ^n ((2k+1)/(k^2 (4k^2 −1))) we have S_n =Σ_(k=1) ^n ((2k+1)/(k^2 (2k+1)(2k−1))) =Σ_(k=1) ^n (1/(k^2 (2k−1))) let decompose F(x)=(1/(x^2 (2x−1))) F(x)=(a/x) +(b/x^2 ) +(c/(2x−1)) b =lim_(x→0) x^2 F(x) =−1 c =lim_(x→(1/2)) (2x−1)F(x) =4 ⇒F(x)=(a/x) −(1/x^2 ) +(4/(2x−1)) F(1)=1 =a−1 +4 =a+3 ⇒a=−2 ⇒F(x)=−(2/x) −(1/x^2 ) +(4/(2x−1)) ⇒ Σ_(k=1) ^n F(k) =−2 Σ_(k=1) ^n (1/k) −Σ_(k=1) ^n (1/k^2 ) +4 Σ_(k=1) ^n (1/(2k−1)) but Σ_(k=1) ^n (1/k) =H_n Σ_(k=1) ^n (1/k^2 ) =ξ_n (2) Σ_(k=1) ^n (1/(2k−1)) = 1 +(1/3) +(1/5) +...+(1/(2n−1)) =1+(1/2) +(1/3)+(1/4) +(1/5) +....+(1/(2n−1)) +(1/(2n)) −(1/2) (1+(1/2) +....+(1/n)) =H_(2n) −(1/2)H_n ⇒ S_n = −2 H_n −ξ_n (2)+4 H_(2n) −2H_n =4{ H_(2n) −H_n }−ξ_n (2) but H_(2n) =ln(2n)+γ +o((1/n)) and H_n =ln(n)+γ +o((1/n)) ⇒ H_(2n) −H_n =ln(((2n)/n))+o((1/n)) →ln(2)(n→+∞) also ξ_n (2)→ξ(2) =(π^2 /6) (n→+∞) ⇒ S=lim_(n→+∞) S_n = 4ln(2)−(π^2 /6) .$
$l e t S_{n} = \sum_{k = 1}^{n} \frac{2 k + 1}{k^{2} (4 k^{2} - 1)} w e h a v e S_{n} = \sum_{k = 1}^{n} \frac{2 k + 1}{k^{2} (2 k + 1) (2 k - 1)}$ $= \sum_{k = 1}^{n} \frac{1}{k^{2} (2 k - 1)} l e t d e c o m p o s e F (x) = \frac{1}{x^{2} (2 x - 1)}$ $F (x) = \frac{a}{x} + \frac{b}{x^{2}} + \frac{c}{2 x - 1}$ $b = {l i m}_{x \to 0} x^{2} F (x) = - 1$ $c = {l i m}_{x \to \frac{1}{2}} (2 x - 1) F (x) = 4 \Rightarrow F (x) = \frac{a}{x} - \frac{1}{x^{2}} + \frac{4}{2 x - 1}$ $F (1) = 1 = a - 1 + 4 = a + 3 \Rightarrow a = - 2 \Rightarrow F (x) = - \frac{2}{x} - \frac{1}{x^{2}} + \frac{4}{2 x - 1} \Rightarrow$ $\sum_{k = 1}^{n} F (k) = - 2 \sum_{k = 1}^{n} \frac{1}{k} - \sum_{k = 1}^{n} \frac{1}{k^{2}} + 4 \sum_{k = 1}^{n} \frac{1}{2 k - 1} b u t$ $\sum_{k = 1}^{n} \frac{1}{k} = H_{n}$ $\sum_{k = 1}^{n} \frac{1}{k^{2}} = ξ_{n} (2)$ $\sum_{k = 1}^{n} \frac{1}{2 k - 1} = 1 + \frac{1}{3} + \frac{1}{5} + . . . + \frac{1}{2 n - 1}$ $= 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + . . . . + \frac{1}{2 n - 1} + \frac{1}{2 n} - \frac{1}{2} (1 + \frac{1}{2} + . . . . + \frac{1}{n})$ $= H_{2 n} - \frac{1}{2} H_{n} \Rightarrow S_{n} = - 2 H_{n} - ξ_{n} (2) + 4 H_{2 n} - 2 H_{n}$ $= 4 {H_{2 n} - H_{n}} - ξ_{n} (2) b u t$ $H_{2 n} = l n (2 n) + γ + o (\frac{1}{n}) a n d H_{n} = l n (n) + γ + o (\frac{1}{n}) \Rightarrow$ $H_{2 n} - H_{n} = l n (\frac{2 n}{n}) + o (\frac{1}{n}) \to l n (2) (n \to + \infty) a l s o$ $ξ_{n} (2) \to ξ (2) = \frac{π^{2}}{6} (n \to + \infty) \Rightarrow S = {l i m}_{n \to + \infty} S_{n} = 4 l n (2) - \frac{π^{2}}{6} .$
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