L-lim-n-1-n-2-n-1-1-n-2-n-2-1-n-2-n-n-

Question Number 136356 by SOMEDAVONG last updated on 21/Mar/21

$$\mathrm{L}=\underset{\mathrm{n}\rightarrow\propto} {\mathrm{lim}}\left[\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{2}} +\mathrm{n}+\mathrm{1}}\:+\:\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{2}} +\mathrm{n}+\mathrm{2}}\:+\:…+\:\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{2}} +\mathrm{n}+\mathrm{n}}\right] \\ $$

Answered by mindispower last updated on 21/Mar/21

$$\underset{{k}\leqslant{n}} {\sum}\frac{\mathrm{1}}{{n}^{\mathrm{2}} +{n}+{n}}=\frac{\mathrm{1}}{{n}+\mathrm{2}}\leqslant\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{2}} +{n}+{k}}\leqslant\underset{{k}\leqslant{n}} {\sum}\frac{\mathrm{1}}{{n}^{\mathrm{2}} +{n}+\mathrm{1}} \\ $$$$=\frac{{n}}{{n}^{\mathrm{2}} +{n}+\mathrm{1}} \\ $$

Answered by mnjuly1970 last updated on 21/Mar/21

$$\:{lim}_{{n}\rightarrow\infty} \:\frac{{n}}{{n}^{\mathrm{2}} +\mathrm{2}{n}}\leqslant{L}\leqslant{lim}_{\rightarrow\infty} \frac{{n}}{{n}^{\mathrm{2}} +{n}+\mathrm{1}} \\ $$$$\:\:\:\:\:\:\:\:{L}=\mathrm{0} \\ $$

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