lim-n-1-n-1-1-2-1-3-1-n-

Question Number 149151 by mathdanisur last updated on 03/Aug/21

\underset{n \to \infty}{l i m} \frac{1}{\sqrt{n}} (1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \dots + \frac{1}{\sqrt{n}}) = ?

Answered by Kamel last updated on 03/Aug/21

L = \underset{n \to \infty}{l i m} \frac{1}{\sqrt{n}} (1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \dots + \frac{1}{\sqrt{n}})

= \underset{n \to + \infty}{l i m} \frac{1}{\sqrt{n}} \underset{k = 1}{\sum^{n}} \frac{1}{\sqrt{k}} = \underset{n \to + \infty}{l i m} \frac{1}{n} \underset{k = 1}{\sum^{n}} \sqrt{\frac{n}{k}} = \int_{0}^{1} \frac{d x}{\sqrt{x}} = 2

Commented by mathdanisur last updated on 03/Aug/21

T h a n k Y o u S e r

Answered by puissant last updated on 03/Aug/21

= \lim_{n \to \infty} \frac{1}{\sqrt{n}} \underset{k = 1}{\sum^{n}} \sqrt{\frac{1}{k}}

= \lim_{n \to \infty} \frac{1}{n} \underset{k = 1}{\sum^{n}} \sqrt{\frac{n}{k}} = f (\frac{k}{n})

d^{'} apr \overset{`}{e s} riemann, on a :

= \int_{0}^{1} \frac{1}{\sqrt{x}} dx = \int_{0}^{1} x^{- \frac{1}{2}} dx = 2 {[\sqrt{x}]}_{0}^{1} = 2 . .

Commented by mathdanisur last updated on 03/Aug/21

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Answered by Rachid last updated on 03/Aug/21

= \lim_{n \to + \infty} \frac{1}{\sqrt{n}} \underset{i = 1}{\sum^{n}} \frac{1}{\sqrt{i}} = \lim_{n \to + \infty} \frac{1}{n} \underset{i = 1}{\sum^{n}} \frac{\sqrt{n}}{\sqrt{i}}

= \lim_{n \to + \infty} \frac{1}{n} \underset{i = 1}{\sum^{n}} \frac{1}{\sqrt{\frac{i}{n}}} = \underset{0}{\int^{1}} \frac{1}{\sqrt{x}} d x = 2

Commented by mathdanisur last updated on 03/Aug/21

T h a n k Y o u S e r

Answered by dumitrel last updated on 03/Aug/21

u s e \frac{1}{\sqrt{n + 1}} < \sqrt{n + 1} - \sqrt{n} < \frac{1}{\sqrt{n}}

Commented by mathdanisur last updated on 03/Aug/21

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