let u_n = Σ_(1≤i<j≤n) (1/(√(ij))) 1) find a equivalent of u_n 2)calculate lim_(n→+∞) u


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Question Number 43003 by abdo.msup.com last updated on 06/Sep/18

$l e t u_{n} = \sum_{1 ⩽ i < j ⩽ n} \frac{1}{\sqrt{i j}}$ $1) f i n d a e q u i v a l e n t o f u_{n}$ $2) c a l c u l a t e {l i m}_{n \to + \infty} u_{n}$
	Answered by maxmathsup by imad last updated on 07/Sep/18
	$1) we have (Σ_(i=1) ^n (1/(√i)))^2 = Σ_(i=1) ^n (1/i) +2 Σ_(1≤i<j≤n) (1/(√i)) (1/(√j)) = H_n + 2u_n ⇒u_n =(1/2){ (Σ_(i=1) ^n (1/(√i)))^2 −H_n ) by we have provedthat Σ_(i=1) ^n (1/(√i)) ∼ 2(√n)(n→+∞) and H_n = ln(n) +γ +o((1/n)) ⇒ u_n ∼ (1/2){ 4n −ln(n)−γ +o((1/n))} ⇒ u_n ∼ 2n −ln((√n)) −(γ/2) +o((1/n)) 2) we have u_n ∼ 2n −ln((√n)) −(γ/2) +o((1/n)) but lim_(n→+∞) 2n−ln((√n)) =lim_(n→+∞) n(2−((ln(n))/(2n))) =lim_(n→+∞) (2n) =+∞ ⇒ lim_(n→+∞) u_n =+∞ .$
	$1) w e h a v e {(\sum_{i = 1}^{n} \frac{1}{\sqrt{i}})}^{2} = \sum_{i = 1}^{n} \frac{1}{i} + 2 \sum_{1 ⩽ i < j ⩽ n} \frac{1}{\sqrt{i}} \frac{1}{\sqrt{j}}$ $= H_{n} + 2 u_{n} \Rightarrow u_{n} = \frac{1}{2} {{(\sum_{i = 1}^{n} \frac{1}{\sqrt{i}})}^{2} - H_{n}) b y w e h a v e p r o v e d t h a t$ $\sum_{i = 1}^{n} \frac{1}{\sqrt{i}} \sim 2 \sqrt{n} (n \to + \infty) a n d H_{n} = l n (n) + γ + o (\frac{1}{n}) \Rightarrow$ $u_{n} \sim \frac{1}{2} {4 n - l n (n) - γ + o (\frac{1}{n})} \Rightarrow u_{n} \sim 2 n - l n (\sqrt{n}) - \frac{γ}{2} + o (\frac{1}{n})$ $2) w e h a v e u_{n} \sim 2 n - l n (\sqrt{n}) - \frac{γ}{2} + o (\frac{1}{n}) b u t$ ${l i m}_{n \to + \infty} 2 n - l n (\sqrt{n}) = {l i m}_{n \to + \infty} n (2 - \frac{l n (n)}{2 n}) = {l i m}_{n \to + \infty} (2 n) = + \infty \Rightarrow$ ${l i m}_{n \to + \infty} u_{n} = + \infty .$
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