prove-that-lim-n-k-1-n-n-2-n-6-k-1-

Question Number 192274 by York12 last updated on 13/May/23

p r o v e t h a t \lim_{n \to \infty} (\underset{k = 1}{\sum^{n}} \frac{n^{2}}{\sqrt{n^{6} + k}}) = 1

Answered by aleks041103 last updated on 14/May/23

\underset{k = 1}{\sum^{n}} \frac{n^{2}}{\sqrt{n^{6} + n}} < \underset{k = 1}{\sum^{n}} \frac{n^{2}}{\sqrt{n^{6} + k}} < \underset{k = 1}{\sum^{n}} \frac{n^{2}}{\sqrt{n^{6} + 0}}

\frac{n^{3}}{n^{3} \sqrt{1 + n^{- 5}}} < \underset{k = 1}{\sum^{n}} \frac{n^{2}}{\sqrt{n^{6} + k}} < \frac{n^{3}}{n^{3}}

\Rightarrow \frac{1}{\sqrt{1 + \frac{1}{n^{5}}}} < \underset{k = 1}{\sum^{n}} \frac{n^{2}}{\sqrt{n^{6} + k}} < 1

\Rightarrow \underset{n \to \infty}{l i m} \frac{1}{\sqrt{1 + \frac{1}{n^{5}}}} ⩽ \underset{n \to \infty}{l i m} (\underset{k = 1}{\sum^{n}} \frac{n^{2}}{\sqrt{n^{6} + k}}) ⩽ 1

\Rightarrow 1 ⩽ \underset{n \to \infty}{l i m} (\underset{k = 1}{\sum^{n}} \frac{n^{2}}{\sqrt{n^{6} + k}}) ⩽ 1

\Rightarrow \underset{n \to \infty}{l i m} (\underset{k = 1}{\sum^{n}} \frac{n^{2}}{\sqrt{n^{6} + k}}) = 1

Commented by York12 last updated on 14/May/23

\lim_{n \to \infty} [\underset{k = 1}{\sum^{n}} (\frac{n^{2}}{\sqrt{n^{6} + k}})] = \lim_{n \to \infty} [n^{2} \underset{k = 1}{\sum^{n}} (\frac{1}{\sqrt{n^{6} + k}})] \to [I]

L e t T_{k} = \frac{1}{\sqrt{n^{6} + k}} ⇛ k = \frac{1}{T_{k}} - n^{6}

I = \lim_{n \to \infty} [n^{2} \underset{k = 1}{\sum^{n}} (\frac{1}{\sqrt{n^{6} + k}})]

\underset{k = 1}{\sum^{n}} (\frac{1}{\sqrt{n^{6} + \sqrt{n^{6} + \sqrt{n^{6} + \sqrt{n^{6} + \dots}}}}})

= \lim_{n \to \infty} [n^{2} \underset{k = 1}{\sum^{n}} (\frac{2}{1 + \sqrt{1 + 4 n^{6}}})] = \lim_{n \to \infty} [\frac{2 n^{3}}{1 + \sqrt{1 + 4 n^{6}}}]

= \frac{2}{\sqrt{4}} = 1 \to ({T h a t}^{'} s i t)

(BY Y O R K)

Commented by York12 last updated on 14/May/23

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