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Question Number 200413 by Mingma last updated on 18/Nov/23

Commented by Mingma last updated on 18/Nov/23
Evaluate!
	Answered by witcher3 last updated on 18/Nov/23

	$x - \frac{x^{3}}{6} ⩽ \sin (x) ⩽ x$ ${(\frac{\ln (k)}{k} - \frac{1}{6} {(\frac{\ln (k)}{k})}_{= v_{n}}^{3})}^{\frac{1}{n}} ⩽ {(\underset{k = 2}{\sum^{n}} \sin (\frac{\ln (k)}{k}))}^{\frac{1}{n}} ⩽ {(\underset{k = 2}{\sum^{n}} \frac{\ln (k)}{k})}^{\frac{1}{n}} = u_{n}$ $\ln (u_{n}) = \frac{1}{n} \ln (\underset{k = 2}{\sum^{n}} \frac{\ln (k)}{k}) ⩽ \frac{1}{n} \ln (nln (n)) \to 0$ $\ln (v_{n}) = \frac{1}{n} \ln (\underset{k ⩾ 2}{\sum^{n}} \frac{\ln (k)}{k} - \frac{1}{6} {(\frac{\ln (k)}{k})}^{3})$ $\underset{k ⩾ 2}{\sum^{n}} \frac{\ln (k)}{k} - \frac{1}{6} {(\frac{\ln (k)}{k})}^{3} ⩾ \underset{k = 2}{\sum^{n}} \frac{5}{6} \frac{\ln (k)}{k} . . . (E)$ $\exists n \in N \forall n ⩾ N \underset{k = 2}{\sum^{n}} \frac{\ln (k)}{k} . \frac{5}{6} ⩾ 1$ $since \sum_{k ⩾ 2} \frac{\ln (k)}{k} diverge in + \infty$ $E \Rightarrow \underset{k ⩾ 2}{\sum^{n}} (\frac{\ln (k)}{k} - \frac{1}{6} {(\frac{\ln (k)}{k})}^{3}) ⩾ 1, \forall n ⩾ N$ $\Rightarrow \frac{1}{n} \ln (\underset{k ⩾ 2}{\sum^{n}} \frac{lnk}{k} - \frac{1}{6} {(\frac{lnk}{k})}^{3}) ⩾ 0, \forall n ⩾ N$ $\Rightarrow \forall n ⩾ N$ $0 ⩽ \frac{1}{n} \ln {(Σ \sin (\frac{lnk}{k}))}^{} ⩽ \ln (u_{n}) \to 0$ $\Rightarrow \frac{1}{n} \ln (Σ \sin (\frac{\ln (k)}{k})) \to 0$ $\Leftrightarrow {(\sum_{k ⩾ 2} \sin (\frac{\ln (k)}{k}))}^{\frac{1}{n}} \to 1$
	Commented by Mingma last updated on 19/Nov/23
	Perfect ��
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