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Question Number 122848 by ZiYangLee last updated on 20/Nov/20

$$\underset{n\to \mathrm{\infty }}{\lim }(\frac{1}{n^2}+\frac{2}{n^2}+\frac{3}{n^2}+......+\frac{n}{n^2})=?$$

Answered by nico last updated on 20/Nov/20

$$=\underset{x\to 0}{\lim }\frac{1}{n}\left(\underset{k=1}{\sum ^n}\frac{k}{n}\right)={\int }_0^{1}xdx=\frac{1}{2}$$

Answered by mindispower last updated on 20/Nov/20

$$=\sum _{0⩽k⩽n}\frac{k}{n^2}=-\frac{\partial }{\partial x}\frac{1}{n^2}\mathrm{\Sigma }{e}^{kx}{\mid }_{x=0}$$$$=\frac{\partial }{\partial x}.\frac{1}{n^2}.\frac{1-{\left(e^x\right)}^{n+1}}{1-e^x}$$$$=\frac{1}{n^2}.\frac{\partial }{\partial x}.e^{\frac{nx}{2}}.\frac{sh\left(\frac{n+1}{2}x\right)}{sh\left(\frac{x}{2}\right)}{\mid }_{x=0}$$$$$$$$=\frac{1}{n^2}\underset{x\to 0}{\lim }[\frac{n}{2}{e}^{\frac{nx}{2}}.\frac{sh\left(\frac{n+1}{2}x\right)}{sh\left(\frac{x}{2}\right)}+e^{n\frac{x}{2}}.\left[\frac{\frac{n+1}{2}ch\left(\frac{n+1}{2}x\right)sh\left(\frac{x}{2}\right)-\frac{1}{2}ch\left(\frac{x}{2}\right)sh\left(\frac{n+1}{2}x\right)}{{sh}^2\left(\frac{x}{2}\right)}\right]$$$$=\frac{1}{n^2},\frac{n}{2}.\frac{n+1}{2}.\frac{2}{}+1.0$$$$=\underset{n\to \mathrm{\infty }}{\lim }.\frac{1}{n^2}.\frac{n(n+1)}{2}=\frac{1}{2}$$

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