if n>1 prove that 2ln(n)−ln(n+1)−ln(n−1)=(1/n^2 )+(1/(2n^4 ))+(1/(3n^6 ))+...=


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Question Number 79625 by M±th+et£s last updated on 26/Jan/20

$i f n > 1 p r o v e t h a t$ $2 l n (n) - l n (n + 1) - l n (n - 1) = \frac{1}{n^{2}} + \frac{1}{2 n^{4}} + \frac{1}{3 n^{6}} + . . . =$
	Answered by mind is power last updated on 27/Jan/20
	$⇔ln((n^2 /(n^2 −1)))=ln((1/(1−(1/n^2 ))))=−ln(1−(1/n^2 )) ∀n∈N−{0,1} (1/n)<1⇒(1/n^2 )<1 −ln(1−x)=Σ_(k≥1) (x^k /k),∀x∈]−1,1[,x=(1/n^2 )⇒ −ln(1−(1/n^2 ))=Σ_(k≥1) (1/(n^(2k) .k))=(1/n^2 )+(1/(2n^4 ))+..........$
	$\Leftrightarrow l n (\frac{n^{2}}{n^{2} - 1}) = l n (\frac{1}{1 - \frac{1}{n^{2}}}) = - l n (1 - \frac{1}{n^{2}})$ $\forall n \in N - {0, 1} \frac{1}{n} < 1 \Rightarrow \frac{1}{n^{2}} < 1$ $- l n (1 - x) = \sum_{k ⩾ 1} \frac{x^{k}}{k}, \forall x \in] - 1, 1 [, x = \frac{1}{n^{2}} \Rightarrow$ $- l n (1 - \frac{1}{n^{2}}) = \sum_{k ⩾ 1} \frac{1}{n^{2 k} . k} = \frac{1}{n^{2}} + \frac{1}{2 n^{4}} + . . . . . . . . . .$
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