let-u-n-1-i-lt-j-n-1-ij-1-find-a-equivalent-of-u-n-2-calculate-lim-n-u-n-

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) 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 .