Σ_(n≥1) ^∞ (1/(n(2n+1)^2 ))=?


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Question Number 160289 by amin96 last updated on 27/Nov/21

$\underset{n ⩾ 1}{\sum^{\infty}} \frac{1}{n {(2 n + 1)}^{2}} = ?$
	Answered by qaz last updated on 27/Nov/21

	$\underset{n = 1}{\sum^{\infty}} \frac{1}{n {(2 n + 1)}^{2}}$ $= \underset{n = 1}{\sum^{\infty}} (\frac{1}{n} - \frac{1}{n + \frac{1}{2}} - \frac{2}{{(2 n + 1)}^{2}})$ $= H_{1 / 2} - 2 ((1 - 2^{- 2}) ζ (2) - 1)$ $= 4 - 2 \ln 2 - \frac{3}{2} ζ (2)$
	Answered by mathmax by abdo last updated on 28/Nov/21
	$we decompose f(x)=(1/(x(2x+1)^2 )) ⇒f(x)=(a/x)+(b/(2x+1))+(c/((2x+1)^2 )) a=xf(x)∣_(x=0) =1 c=(2x+1)^2 f(x)∣_(x=−(1/2)) =−2 lim_(x→+∞) xf(x)=0=a+(b/2) ⇒(b/2)=−a ⇒b=−2 ⇒ f(x)=(1/x)−(2/(2x+1))−(2/((2x+1)^2 )) S_n =Σ_(k=1) ^n (1/(k(2k+1)^2 )) ⇒S_n =Σ_(k=1) ^n f(k)=Σ_(k=1) ^n ((1/k)−(2/(2k+1))−(2/((2k+1)^2 ))) =Σ_(k=1) ^n (1/k)−2Σ_(k=1) ^n (1/(2k+1))−2Σ_(k=1) ^n (1/((2k+1)^2 )) Σ_(k=1) ^n (1/k)=H_n Σ_(k=1) ^n (1/(2k+1))=(1/3)+(1/5)+...+(1/(2n+1))=1+(1/2)+(1/3)+(1/4)+....+(1/(2n))+(1/(2n+1)) −1−(1/2)−(1/4)−...−(1/(2n))=H_(2n+1) −1−(1/2)H_n Σ_(k=1) ^n (1/((2k+1)^2 ))→Σ_(k=1) ^∞ (1/((2k+1)^2 ))=(π^2 /8)−1 due to Σ (1/k^2 )=(1/4)Σ(1/k^2 )+Σ_(k=0) ^∞ (1/((2k+1)^2 )) ⇒Σ_(k=0) ^∞ (1/((2k+1)^2 ))=(3/4)(π^2 /6)=(π^2 /8) Σ_(k=1) ^n (1/k)−2Σ_(k=1) ^n (1/(2k+1))=H_n −2{H_(2n+1) −1−(1/2)H_n } =2H_n −2H_(2n+1) +2∼2(ln(n)+γ)−2(ln(2n+1)+γ)+2 =2ln((n/(2n+1)))+2→2ln((1/2))+2=2−2ln2 ⇒ lim_(n→+∞) S_n =2−2ln2−2{(π^2 /8)−1} =2−2ln2−(π^2 /4) +2=4−2ln2−(π^2 /4)$
	$we decompose f (x) = \frac{1}{x {(2 x + 1)}^{2}} \Rightarrow f (x) = \frac{a}{x} + \frac{b}{2 x + 1} + \frac{c}{{(2 x + 1)}^{2}}$ $a = xf (x) ∣_{x = 0} = 1$ $c = {(2 x + 1)}^{2} f (x) ∣_{x = - \frac{1}{2}} = - 2$ $\lim_{x \to + \infty} xf (x) = 0 = a + \frac{b}{2} \Rightarrow \frac{b}{2} = - a \Rightarrow b = - 2 \Rightarrow$ $f (x) = \frac{1}{x} - \frac{2}{2 x + 1} - \frac{2}{{(2 x + 1)}^{2}}$ $S_{n} = \sum_{k = 1}^{n} \frac{1}{k {(2 k + 1)}^{2}} \Rightarrow S_{n} = \sum_{k = 1}^{n} f (k) = \sum_{k = 1}^{n} (\frac{1}{k} - \frac{2}{2 k + 1} - \frac{2}{{(2 k + 1)}^{2}})$ $= \sum_{k = 1}^{n} \frac{1}{k} - 2 \sum_{k = 1}^{n} \frac{1}{2 k + 1} - 2 \sum_{k = 1}^{n} \frac{1}{{(2 k + 1)}^{2}}$ $\sum_{k = 1}^{n} \frac{1}{k} = H_{n}$ $\sum_{k = 1}^{n} \frac{1}{2 k + 1} = \frac{1}{3} + \frac{1}{5} + . . . + \frac{1}{2 n + 1} = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + . . . . + \frac{1}{2 n} + \frac{1}{2 n + 1}$ $- 1 - \frac{1}{2} - \frac{1}{4} - . . . - \frac{1}{2 n} = H_{2 n + 1} - 1 - \frac{1}{2} H_{n}$ $\sum_{k = 1}^{n} \frac{1}{{(2 k + 1)}^{2}} \to \sum_{k = 1}^{\infty} \frac{1}{{(2 k + 1)}^{2}} = \frac{π^{2}}{8} - 1 due to$ $Σ \frac{1}{k^{2}} = \frac{1}{4} Σ \frac{1}{k^{2}} + \sum_{k = 0}^{\infty} \frac{1}{{(2 k + 1)}^{2}} \Rightarrow \sum_{k = 0}^{\infty} \frac{1}{{(2 k + 1)}^{2}} = \frac{3}{4} \frac{π^{2}}{6} = \frac{π^{2}}{8}$ $\sum_{k = 1}^{n} \frac{1}{k} - 2 \sum_{k = 1}^{n} \frac{1}{2 k + 1} = H_{n} - 2 {H_{2 n + 1} - 1 - \frac{1}{2} H_{n}}$ $= {2 H}_{n} - {2 H}_{2 n + 1} + 2 \sim 2 (\ln (n) + γ) - 2 (\ln (2 n + 1) + γ) + 2$ $= 2 \ln (\frac{n}{2 n + 1}) + 2 \to 2 \ln (\frac{1}{2}) + 2 = 2 - 2 \ln 2 \Rightarrow$ $\lim_{n \to + \infty} S_{n} = 2 - 2 \ln 2 - 2 {\frac{π^{2}}{8} - 1}$ $= 2 - 2 \ln 2 - \frac{π^{2}}{4} + 2 = 4 - 2 \ln 2 - \frac{π^{2}}{4}$
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