let u_n =Σ_(k=1) ^(n−1) ((n−k)/(n−k+1)) find a equivalent of u


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Question Number 40898 by abdo.msup.com last updated on 28/Jul/18

$l e t u_{n} = \sum_{k = 1}^{n - 1} \frac{n - k}{n - k + 1}$ $f i n d a e q u i v a l e n t o f u_{n} (n \to + \infty)$
Commented by maxmathsup by imad last updated on 29/Jul/18

$t h a n k s i u n d e r s t a n d . . . .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 29/Jul/18
	$T_k =1−(1/(n+1−k)) T_1 =1−(1/n) T_2 =1−(1/(n−1)) ... ... T_(n−1) =1−(1/2) so u_n =(n−1)−{(1/2)+(1/3)+...+(1/n)} to find u_n =(n−1)−s value of u_n n→∞ 1>(1/2)>(1/n) 1>(1/3)>(1/n) 1>(1/4)>(1/n) ..... ...... so adding n−1>s>((n−1)/n) −(n−1)<−s<−(((n−1)/n)) (n−1)−(n−1)<(n−1)−s<(n−1)−(((n−1)/n)) 0<u_n <n−1−1+(1/n) 0<u_n <n+(1/n)−2 0<u_n <((√n) −(1/(√n_ )))^2 when n→∞ ∞>u_n >0 pls check....$
	$T_{k} = 1 - \frac{1}{n + 1 - k}$ $T_{1} = 1 - \frac{1}{n}$ $T_{2} = 1 - \frac{1}{n - 1}$ $. . .$ $. . .$ $T_{n - 1} = 1 - \frac{1}{2}$ $s o u_{n} = (n - 1) - {\frac{1}{2} + \frac{1}{3} + . . . + \frac{1}{n}} t o f i n d$ $u_{n} = (n - 1) - s$ $v a l u e o f u_{n} n \to \infty$ $1 > \frac{1}{2} > \frac{1}{n}$ $1 > \frac{1}{3} > \frac{1}{n}$ $1 > \frac{1}{4} > \frac{1}{n}$ $. . . . .$ $. . . . . .$ $s o a d d i n g n - 1 > s > \frac{n - 1}{n}$ $- (n - 1) < - s < - (\frac{n - 1}{n})$ $(n - 1) - (n - 1) < (n - 1) - s < (n - 1) - (\frac{n - 1}{n})$ $0 < u_{n} < n - 1 - 1 + \frac{1}{n}$ $0 < u_{n} < n + \frac{1}{n} - 2$ $0 < u_{n} < {(\sqrt{n} - \frac{1}{\sqrt{n_{}}})}^{2}$ $w h e n n \to \infty \infty > u_{n} > 0$ $p l s c h e c k . . . .$
	Answered by math khazana by abdo last updated on 29/Jul/18

	$w e h a v e u_{n} = \sum_{k = 1}^{n - 1} \frac{n - k + 1 - 1}{n - k + 1}$ $= \sum_{k = 1}^{n - 1} (1) - \sum_{k = 1}^{n - 1} \frac{1}{n - k + 1} b u t$ $\sum_{k = 1}^{n - 1} (1) = n - 1 a n d$ $\sum_{k = 1}^{n - 1} \frac{1}{n - k + 1} =_{n - k = i} \sum_{i = n - 1}^{1} \frac{1}{i + 1} = \sum_{i = 1}^{n - 1} \frac{1}{i + 1}$ $= \sum_{i = 2}^{n} \frac{1}{i} = H_{n} - 1 \Rightarrow$ $u_{n} = n - 1 - H_{n} + 1 = n - H_{n} b u t$ $H_{n} = l n (n) + γ + o (\frac{1}{n}) \Rightarrow u_{n} = n - l n (n) - γ + o (\frac{1}{n}) \Rightarrow$ $u_{n} \sim n - l n (n) (n \to + \infty)$
	Commented by math khazana by abdo last updated on 29/Jul/18

	$H_{n} = \sum_{k = 1}^{n} \frac{1}{k}$
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