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Question Number 141127 by bobhans last updated on 16/May/21

$$\underset{n\to \mathrm{\infty }}{\lim }\underset{i=0}{\sum ^{n-1}}\frac{n}{{(n+i)}^2}=?$$

Answered by Ar Brandon last updated on 16/May/21

$$\mathcal{L}=\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\underset{\mathrm{i}=0}{\sum ^{\mathrm{n}-1}}\frac{\mathrm{n}}{{(\mathrm{n}+\mathrm{i})}^2}=\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\frac{1}{\mathrm{n}}\underset{\mathrm{i}=0}{\sum ^{\mathrm{n}-1}}\frac{1}{{(1+\frac{\mathrm{i}}{\mathrm{n}})}^2}$$$$={\int }_1^2\frac{\mathrm{dx}}{{\mathrm{x}}^2}=-{\left[\frac{1}{\mathrm{x}}\right]}_1^2=1-\frac{1}{2}=\frac{1}{2}$$

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