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

$$\mathrm{L}=\underset{\mathrm{n}\to \propto }{\lim }[\frac{1}{{\mathrm{n}}^2+\mathrm{n}+1}+\frac{1}{{\mathrm{n}}^2+\mathrm{n}+2}+...+\frac{1}{{\mathrm{n}}^2+\mathrm{n}+\mathrm{n}}]$$

Answered by mindispower last updated on 21/Mar/21

$$\sum _{k⩽n}\frac{1}{n^2+n+n}=\frac{1}{n+2}⩽\underset{k=1}{\sum ^n}\frac{1}{n^2+n+k}⩽\sum _{k⩽n}\frac{1}{n^2+n+1}$$$$=\frac{n}{n^2+n+1}$$

Answered by mnjuly1970 last updated on 21/Mar/21

$${lim}_{n\to \mathrm{\infty }}\frac{n}{n^2+2n}⩽L⩽{lim}_{\to \mathrm{\infty }}\frac{n}{n^2+n+1}$$$$L=0$$

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