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

$$\left(1\right)\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }(\frac{1}{\sqrt{{\mathrm{n}}^2+1}}+\frac{1}{\sqrt{{\mathrm{n}}^2+2}}+\frac{1}{\sqrt{{\mathrm{n}}^2+3}}+...+\frac{1}{\sqrt{{\mathrm{n}}^2+\mathrm{n}+1}})=?$$$$\left(2\right)\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\frac{{(2\sqrt[\mathrm{n}]{\mathrm{n}}-1)}^{\mathrm{n}}}{{\mathrm{n}}^2}=?$$

Answered by benjo_mathlover last updated on 14/Nov/20

$$\left(1\right)\underset{n\to \mathrm{\infty }}{\lim }(\frac{1}{\sqrt{n^2+1}}+\frac{1}{\sqrt{n^2+2}}+\frac{1}{\sqrt{n^2+3}}+...+\frac{1}{\sqrt{n^2+n+1}})$$$$notethat$$$$\frac{n+1}{\sqrt{n^2+n+1}}⩽\frac{1}{\sqrt{n^2+1}}+\frac{1}{\sqrt{n^2+2}}+...+\frac{1}{\sqrt{n^2+n+1}}⩽\frac{n+1}{\sqrt{n^2+1}}$$$$Thisandthesqueezelawforsequences$$$$implythatthelimitis1.$$$$\underset{n\to \mathrm{\infty }}{\lim }\frac{n(1+\frac{1}{n})}{n\sqrt{1+\frac{1}{n}+\frac{1}{n^2}}}=\underset{n\to \mathrm{\infty }}{\lim }\frac{n(1+\frac{1}{n})}{n\sqrt{1+\frac{1}{n^2}}}=1.$$

Commented by liberty last updated on 14/Nov/20

$$\mathrm{thanks}$$

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

$${\mathrm{U}}_{\mathrm{n}}=\frac{{(2{\mathrm{n}}^{\frac{1}{\mathrm{n}}}-1)}^{\mathrm{n}}}{{\mathrm{n}}^2}\Rightarrow {\mathrm{U}}_{\mathrm{n}}=\frac{{({2\mathrm{n}}^{\frac{1}{\mathrm{n}}}-1)}^{\mathrm{n}}}{{\left({\mathrm{n}}^{\frac{2}{\mathrm{n}}}\right)}^{\mathrm{n}}}={\left(\frac{{2\mathrm{n}}^{\frac{1}{\mathrm{n}}}-1}{{\mathrm{n}}^{\frac{2}{\mathrm{n}}}}\right)}^{\mathrm{n}}$$$$={(2{\mathrm{n}}^{-\frac{1}{\mathrm{n}}}-{\mathrm{n}}^{-\frac{2}{\mathrm{n}}})}^{\mathrm{n}}\Rightarrow {\mathrm{U}}_{\mathrm{n}}={\mathrm{e}}^{\mathrm{nlog}(2{\mathrm{n}}^{-\frac{1}{\mathrm{n}}}-{\mathrm{n}}^{\frac{-2}{\mathrm{n}}})}$$$${\mathrm{v}}_{\mathrm{n}}=\mathrm{nlog}(2.{\mathrm{n}}^{-\frac{1}{\mathrm{n}}}-{\mathrm{n}}^{-\frac{2}{\mathrm{n}}})\Rightarrow {\mathrm{v}}_{\mathrm{n}}=\mathrm{nlog}\left({2\mathrm{n}}^{-\frac{1}{\mathrm{n}}}\{1-\frac{1}{2}{\mathrm{n}}^{-\frac{1}{\mathrm{n}}}\}\right)$$$$=\mathrm{nlog}\left(2{\mathrm{n}}^{-\frac{1}{\mathrm{n}}}\right)+\mathrm{n}\log (1-\frac{1}{{2\mathrm{n}}^{\frac{1}{\mathrm{n}}}})=\mathrm{nlog}2-\mathrm{logn}+\mathrm{nlog}(1-\frac{1}{{2\mathrm{n}}^{\frac{1}{\mathrm{n}}}})$$$${2\mathrm{n}}^{\frac{1}{\mathrm{n}}}=2{\mathrm{e}}^{\frac{\mathrm{logn}}{\mathrm{n}}}\to 2\Rightarrow \log (1-\frac{1}{{2\mathrm{n}}^{\frac{1}{\mathrm{n}}}})\to -\mathrm{nlog}2\Rightarrow {\lim }_{\mathrm{n}\to +\mathrm{\infty }}{\mathrm{v}}_{\mathrm{n}}=-\mathrm{\infty }\Rightarrow$$$${\lim }_{\mathrm{n}\to +\mathrm{\infty }}{\mathrm{U}}_{\mathrm{n}}={\mathrm{e}}^{-\mathrm{\infty }}=0$$

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