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Question Number 175115 by mnjuly1970 last updated on 19/Aug/22

Answered by aleks041103 last updated on 19/Aug/22

$$n\to \mathrm{\infty }$$$$\sqrt{1+\frac{k}{n^2}}-1=(1+\frac{k}{2n^2})-1=\frac{k}{2n^2}$$$$\Rightarrow \underset{n\to \mathrm{\infty }}{lim}\underset{k=1}{\sum ^n}(\sqrt{1+\frac{k}{n^2}}-1)=\underset{n\to \mathrm{\infty }}{lim}\frac{1}{2n^2}\underset{k=1}{\sum ^n}k=$$$$=\underset{n\to \mathrm{\infty }}{lim}\frac{1}{2n^2}\frac{n(n+1)}{2}=\frac{1}{4}\underset{n\to \mathrm{\infty }}{lim}(1+\frac{1}{n})=\frac{1}{4}$$

Commented by mnjuly1970 last updated on 20/Aug/22

$$thanksalotsir$$

Answered by CElcedricjunior last updated on 20/Aug/22

$$\underset{x\to \mathrm{\infty }}{\lim }\underset{\mathit{k}=1}{\sum ^{\mathit{n}}}(\sqrt{1+\frac{\mathit{k}}{{\mathit{n}}^2}}-1)=\frac{1}{4}$$$$\mathit{o}\mathit{r}\sqrt[\mathit{n}]{1+\mathit{a}}={(1+\mathit{a})}^{\mathit{n}}\approx 1+\mathit{n}\mathit{a}\mathit{a}\mathit{v}\mathit{e}\mathit{c}\mathit{a}<<<1$$$$=>\sqrt{1+\frac{\mathit{k}}{{\mathit{n}}^2}}={(1+\frac{\mathit{k}}{{\mathit{n}}^2})}^{\frac{1}{2}}\approx 1+\frac{\mathit{k}}{2{\mathit{n}}^2}$$$$\approx \underset{x\to \mathrm{\infty }}{\lim }\underset{\mathit{k}=1}{\sum ^{\mathit{n}}}\frac{\mathit{k}}{2{\mathit{n}}^2}=\underset{x\to \mathrm{\infty }}{\lim }\frac{1}{2\mathit{n}}\underset{\mathit{k}=1}{\sum ^{\mathit{n}}}\frac{\mathit{k}}{\mathit{n}}$$$$\underset{x\to \mathrm{\infty }}{\lim }\underset{\mathit{k}=1}{\sum ^{\mathit{n}}}(\sqrt{1+\frac{\mathit{k}}{{\mathit{n}}^2}}-1)\approx \frac{1}{2}{\int }_0^1\mathit{x}\mathit{d}\mathit{x}$$$$\approx \frac{1}{2}{\left[\frac{{\mathit{x}}^2}{2}\right]}_0^1$$$$\iff \underset{x\to \mathrm{\infty }}{\lim }\underset{\mathit{k}=1}{\sum ^{\mathit{n}}}(\sqrt{1+\frac{\mathit{k}}{{\mathit{n}}^2}}-1)\approx \frac{1}{4}$$$$$$$$..........lecelebrecedricjunior.........$$

Commented by mnjuly1970 last updated on 20/Aug/22

$$merceysir$$

Commented by Tawa11 last updated on 20/Aug/22

$$\mathrm{Great}\mathrm{sirs}$$

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