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Question Number 99697 by Ar Brandon last updated on 22/Jun/20

$$\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\underset{\mathrm{k}=0}{\sum ^{2\mathrm{n}}}\frac{\mathrm{k}}{\mathrm{k}+{\mathrm{n}}^2}$$

Commented by MWSuSon last updated on 22/Jun/20

just dropping a comment so that I'll get notified when someone solves it. if only the k in the denominator was k^2��

Answered by Ar Brandon last updated on 23/Jun/20

$$l=\lim \underset{\mathrm{k}=0}{\sum ^{2\mathrm{n}}}\frac{\mathrm{k}}{\mathrm{k}+{\mathrm{n}}^2}$$$$0⩽\mathrm{k}⩽2\mathrm{n}\Rightarrow {\mathrm{n}}^2⩽\mathrm{k}+{\mathrm{n}}^2⩽2\mathrm{n}+{\mathrm{n}}^2\Rightarrow \frac{1}{2\mathrm{n}+{\mathrm{n}}^2}⩽\frac{1}{\mathrm{k}+{\mathrm{n}}^2}⩽\frac{1}{{\mathrm{n}}^2}$$$$\Rightarrow \frac{\mathrm{k}}{2\mathrm{n}+{\mathrm{n}}^2}⩽\frac{\mathrm{k}}{\mathrm{k}+{\mathrm{n}}^2}⩽\frac{\mathrm{k}}{{\mathrm{n}}^2}\Rightarrow \underset{\mathrm{k}=0}{\sum ^{2\mathrm{n}}}\frac{\mathrm{k}}{2\mathrm{n}+{\mathrm{n}}^2}⩽\underset{\mathrm{k}=0}{\sum ^{2\mathrm{n}}}\frac{\mathrm{k}}{\mathrm{k}+{\mathrm{n}}^2}⩽\underset{\mathrm{k}=0}{\sum ^{2\mathrm{n}}}\frac{\mathrm{k}}{{\mathrm{n}}^2}$$$$\Rightarrow \frac{\mathrm{n}(2\mathrm{n}+1)}{2\mathrm{n}+{\mathrm{n}}^2}⩽\underset{\mathrm{k}=0}{\sum ^{2\mathrm{n}}}\frac{\mathrm{k}}{\mathrm{k}+{\mathrm{n}}^2}⩽\frac{\mathrm{n}(2\mathrm{n}+1)}{{\mathrm{n}}^2}\Rightarrow \frac{2+\frac{1}{\mathrm{n}}}{1+\frac{2}{\mathrm{n}}}⩽\underset{\mathrm{k}=0}{\sum ^{2\mathrm{n}}}\frac{\mathrm{k}}{\mathrm{k}+{\mathrm{n}}^2}⩽\frac{2+\frac{1}{\mathrm{n}}}{1}$$$$\Rightarrow 2⩽\lim \underset{\mathrm{k}=0}{\sum ^{2\mathrm{n}}}\frac{\mathrm{k}}{\mathrm{k}+{\mathrm{n}}^2}⩽2\Rightarrow \lim \underset{\mathrm{k}=0}{\sum ^{2\mathrm{n}}}\frac{\mathrm{k}}{\mathrm{k}+{\mathrm{n}}^2}=2$$

Answered by Ar Brandon last updated on 22/Jun/20

Commented by Ar Brandon last updated on 22/Jun/20

I got this. Am I right ?

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