lim_(n→+∞) (1/(√n)) Σ


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Question Number 96602 by Ar Brandon last updated on 03/Jun/20

$\lim_{n \to + \infty} \frac{1}{\sqrt{n}} \underset{k = 1}{\sum^{n}} \sqrt{k}$
	Answered by Sourav mridha last updated on 03/Jun/20

	$b y i n s p e c t i o n \underset{k = 1}{\sum^{n}} \sqrt{k} t h i s i s a$ $d i v e r g i n g s e r i e s a s n \Rightarrow \infty .$ $p r o o f :$ $u s i n g E ular - M a c l a u r i n$ $i n t e g r a t i o n f o r m u l a . . . .$ $\underset{k = 1}{\sum^{n}} \sqrt{k} = [\frac{16 n^{2} + 12 n - 5 \sqrt{n} + 1}{24 \sqrt{n}}]$ $a p p r o x i m a t e r e s u l t . . b u t o k k k . .$ $n o w, \lim_{n \to \infty} \frac{1}{\sqrt{n}} \underset{k = 1}{\sum^{n}} \sqrt{k}$ $= \lim_{n \to \infty} [\frac{16 n + 12 - \frac{5}{\sqrt{n}} + \frac{1}{n}}{24}]$ $= \infty . .$
	Answered by mathmax by abdo last updated on 03/Jun/20

	$S_{n} = \frac{1}{\sqrt{n}} \sum_{k = 1}^{n} \sqrt{k} \Rightarrow S_{n} = \sum_{k = 1}^{n} \sqrt{\frac{k}{n}} = n \times \frac{1}{n} \sum_{k = 1}^{n} \sqrt{\frac{k}{n}}$ $but \lim_{n \to + \infty} \frac{1}{n} \sqrt{\frac{k}{n}} = \int_{0}^{1} \sqrt{x} dx = \int_{0}^{1} x^{\frac{1}{2}} dx = {[\frac{2}{3} x^{\frac{3}{2}}]}_{0}^{1} = \frac{2}{3} \Rightarrow$ $\lim_{n \to + \infty} S_{n} = \lim_{n \to + \infty} \frac{2 n}{3} = + \infty$
	Answered by maths mind last updated on 03/Jun/20

	$\sqrt{k} ⩾ 1$ $\Rightarrow \frac{1}{\sqrt{n}} \underset{k = 1}{\sum^{n}} \sqrt{k} ⩾ \frac{n}{\sqrt{n}} = \sqrt{n}$
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