lim_(x−oo) (1/(n(√n))) Σ_(k=1) ^n E((√(k)))


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Question Number 155710 by SANOGO last updated on 03/Oct/21

$li \underset{x - o o}{m} \frac{1}{n \sqrt{n}} \underset{k = 1}{\sum^{n}} E (\sqrt{k)}$
Commented by yeti123 last updated on 03/Oct/21

$\lim_{x \to \infty} \frac{1}{n \sqrt{n}} \underset{k = 1}{\sum^{n}} E (\sqrt{k}) = \frac{1}{n \sqrt{n}} \underset{k = 1}{\sum^{n}} E (\sqrt{k})$
	Answered by puissant last updated on 03/Oct/21

	$L = \lim_{n \to + \infty} \frac{1}{n \sqrt{n}} \underset{k = 1}{\sum^{n}} E (\sqrt{k})$ $L = \lim_{n \to + \infty} \frac{1}{n} \underset{k = 1}{\sum^{n}} \frac{1}{\sqrt{n}} E (\sqrt{k})$ $= \lim_{n \to + \infty} \frac{1}{n} \underset{k = 1}{\sum^{n}} E (\sqrt{\frac{k}{n}})$ $Q u i e s t s o u s l a f o r m e \frac{b - a}{n} \underset{k = 1}{\sum^{n}} f (a + k \frac{b - a}{n})$ $i l s^{'} a g i t d e l a s o m m e d e r i e m a n n o n a d o n c :$ $\lim_{n \to + \infty} \frac{b - a}{n} \underset{k = 1}{\sum^{n}} f (a + k \frac{b - a}{n}) = \int_{a}^{b} f (x) d x$ $\Rightarrow L = \int_{0}^{1} E (x) d x = 0 (1 - 0) = 0 . .$ $∴∵ L = \lim_{n \to + \infty} \frac{1}{n \sqrt{n}} \underset{k = 1}{\sum^{n}} E (\sqrt{k}) = 0 . .$ $. . . . . . . . . . . L e p u i s s a n t . . . . . . . . . . .$
	Commented by SANOGO last updated on 03/Oct/21

	$m e r c i b i e n m o n f r e r e$
	Commented by Kamel last updated on 04/Oct/21

	$M r . P u i s s a n t y E (x) \neq E (x y)$
	Commented by puissant last updated on 04/Oct/21

	$T h a n k s M r K a m e l . .$ $M r S a n o g o D e s o l \overset{´}{e e} . .$
	Answered by Kamel last updated on 04/Oct/21
	$S_n =Σ_(k=1) ^n [(√k)]=Σ_(k=1) ^n (√k)−Σ_(k=1) ^n {(√k)} ∀1≤k≤n 0≤ Σ_(k=1) ^n {(√k)}<n ⇒0≤(1/( n(√n)))Σ_(k=1) ^n {(√k)}<(1/( (√n))) ∴ lim_(n→+∞) (1/(n(√n)))Σ_(k=1) ^n [(√k)]=lim_(n→+∞) (1/n)Σ_(k=1) ^n (√(k/n))=∫_0 ^1 (√x)dx =(2/3)$
	$S_{n} = \underset{k = 1}{\sum^{n}} [\sqrt{k}] = \underset{k = 1}{\sum^{n}} \sqrt{k} - \underset{k = 1}{\sum^{n}} {\sqrt{k}}$ $\forall 1 ⩽ k ⩽ n 0 ⩽ \underset{k = 1}{\sum^{n}} {\sqrt{k}} < n \Rightarrow 0 ⩽ \frac{1}{n \sqrt{n}} \underset{k = 1}{\sum^{n}} {\sqrt{k}} < \frac{1}{\sqrt{n}}$ $∴ \underset{n \to + \infty}{l i m} \frac{1}{n \sqrt{n}} \underset{k = 1}{\sum^{n}} [\sqrt{k}] = \underset{n \to + \infty}{l i m} \frac{1}{n} \underset{k = 1}{\sum^{n}} \sqrt{\frac{k}{n}} = \int_{0}^{1} \sqrt{x} d x$ $= \frac{2}{3}$
	Commented by SANOGO last updated on 04/Oct/21

	$merci bien$
	Answered by mindispower last updated on 04/Oct/21

	$x - 1 < E (x) ⩽ x$ $\Rightarrow Σ \frac{1}{n \sqrt{n}} (\sqrt{k} - 1) ⩽ \underset{k = 1}{\sum^{n}} \frac{E (\sqrt{k})}{n \sqrt{n}} < \underset{k = 1}{\sum^{n}} \frac{1}{n \sqrt{n}} \sqrt{k})$ $Σ \frac{1}{n} \sqrt{\frac{k}{n}} - \frac{1}{\sqrt{n}} ⩽ S ⩽ \int_{0}^{1} \sqrt{x} d x$ $\int_{0}^{1} \sqrt{x} d x - \lim_{n \to \infty} \frac{1}{\sqrt{n}} ⩽ S ⩽ \int_{0}^{1} \sqrt{x} d x$ $\frac{3}{2} ⩽ S ⩽ \frac{3}{2}$
	Commented by SANOGO last updated on 04/Oct/21

	$m e r c i b i e n$
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