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Question Number 130845 by mnjuly1970 last updated on 29/Jan/21

$$...nicecalculus...$$$$calculate::$$$${lim}_{n\to \mathrm{\infty }}\underset{k=1}{\sum ^n}\frac{k}{2n^2+k}=?$$$$$$

Answered by Ar Brandon last updated on 29/Jan/21

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

Commented by mnjuly1970 last updated on 29/Jan/21

$$himasterbrandonansweris\frac{1}{4}$$

Commented by mnjuly1970 last updated on 29/Jan/21

$$inthirdline\frac{n(n+1)}{2(2n^2+n)}⩽A_n⩽\frac{n(n+1)}{2(2n^2+1)}$$$$youransweriscorrect$$$${lim}_{n\to \mathrm{\infty }}{A}_n=\frac{1}{4}$$

Commented by Ar Brandon last updated on 29/Jan/21

$$\mathrm{Oh}!\mathrm{thanks}\mathrm{for}\mathrm{remark},\mathrm{Sir}$$😃

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