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Question Number 122152 by liberty last updated on 14/Nov/20

(1) \lim_{n \to \infty} (\frac{1}{\sqrt{n^{2} + 1}} + \frac{1}{\sqrt{n^{2} + 2}} + \frac{1}{\sqrt{n^{2} + 3}} + \dots + \frac{1}{\sqrt{n^{2} + n + 1}}) = ?

(2) \lim_{n \to \infty} \frac{{(2 \sqrt[n]{n} - 1)}^{n}}{n^{2}} = ?

Answered by benjo_mathlover last updated on 14/Nov/20

(1) \lim_{n \to \infty} (\frac{1}{\sqrt{n^{2} + 1}} + \frac{1}{\sqrt{n^{2} + 2}} + \frac{1}{\sqrt{n^{2} + 3}} + \dots + \frac{1}{\sqrt{n^{2} + n + 1}})

n o t e t h a t

\frac{n + 1}{\sqrt{n^{2} + n + 1}} ⩽ \frac{1}{\sqrt{n^{2} + 1}} + \frac{1}{\sqrt{n^{2} + 2}} + \dots + \frac{1}{\sqrt{n^{2} + n + 1}} ⩽ \frac{n + 1}{\sqrt{n^{2} + 1}}

T h i s a n d t h e s q u e e z e l a w f o r s e q u e n c e s

i m p l y t h a t t h e l i m i t i s 1 .

\lim_{n \to \infty} \frac{n (1 + \frac{1}{n})}{n \sqrt{1 + \frac{1}{n} + \frac{1}{n^{2}}}} = \lim_{n \to \infty} \frac{n (1 + \frac{1}{n})}{n \sqrt{1 + \frac{1}{n^{2}}}} = 1 .

Commented by liberty last updated on 14/Nov/20

thanks

Answered by mathmax by abdo last updated on 14/Nov/20

U_{n} = \frac{{(2 n^{\frac{1}{n}} - 1)}^{n}}{n^{2}} \Rightarrow U_{n} = \frac{{({2 n}^{\frac{1}{n}} - 1)}^{n}}{{(n^{\frac{2}{n}})}^{n}} = {(\frac{{2 n}^{\frac{1}{n}} - 1}{n^{\frac{2}{n}}})}^{n}

= {(2 n^{- \frac{1}{n}} - n^{- \frac{2}{n}})}^{n} \Rightarrow U_{n} = e^{nlog (2 n^{- \frac{1}{n}} - n^{\frac{- 2}{n}})}

v_{n} = nlog (2 . n^{- \frac{1}{n}} - n^{- \frac{2}{n}}) \Rightarrow v_{n} = nlog ({2 n}^{- \frac{1}{n}} {1 - \frac{1}{2} n^{- \frac{1}{n}}})

= nlog (2 n^{- \frac{1}{n}}) + n \log (1 - \frac{1}{{2 n}^{\frac{1}{n}}}) = nlog 2 - logn + nlog (1 - \frac{1}{{2 n}^{\frac{1}{n}}})

{2 n}^{\frac{1}{n}} = 2 e^{\frac{logn}{n}} \to 2 \Rightarrow \log (1 - \frac{1}{{2 n}^{\frac{1}{n}}}) \to - nlog 2 \Rightarrow \lim_{n \to + \infty} v_{n} = - \infty \Rightarrow

\lim_{n \to + \infty} U_{n} = e^{- \infty} = 0