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Question Number 174951 by infinityaction last updated on 14/Aug/22

$$\underset{n\to \mathrm{\infty }}{\lim }\frac{1}{n}[1+\sqrt{2}{+}^3\sqrt{3}+..{.}^n\sqrt{n}]$$

Answered by TheHoneyCat last updated on 16/Aug/22

$${\mathrm{There}}^{\prime }\mathrm{s}\mathrm{a}\mathrm{theorem}\mathrm{that}\mathrm{sates}\mathrm{that}$$$$U_n\underset{n\to \mathrm{\infty }}{\to }l\in \stackrel{\mathrm{\_}}{\mathbb{R}}\Rightarrow {\lim }_{n\to \mathrm{\infty }}\underset{k=0}{\sum ^n}{U}_k=l$$$$$$$$\mathrm{Now}$$$${}^n\sqrt{n}=n^{1/n}=\exp \left(\frac{\ln n}{n}\right)$$$$\mathrm{since}\ln n/n\to 0$$$${}^n\sqrt{n}\to \exp \left(0\right)=1$$$$$$$$$$$$\mathrm{Thus}:$$$$\mathrm{your}\mathrm{limit}\mathrm{is}1.$$

Commented by TheHoneyCat last updated on 16/Aug/22

$$\mathrm{Woops},{\mathrm{didn}}^{\prime }\mathrm{t}\mathrm{pay}\mathrm{attention}$$$$\mathrm{the}\mathrm{theorem}\mathrm{states}\mathrm{that}$$$${\lim }_{n\to \mathrm{\infty }}\frac{1}{n}\underset{k=0}{\sum ^n}{U}_k=l$$$$\mathrm{of}\mathrm{course}...\mathrm{Otherwise}\mathrm{it}{\mathrm{wouldn}}^{\prime }\mathrm{t}\mathrm{work}...$$$$\mathrm{Sorry}$$

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