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Question Number 39443 by rahul 19 last updated on 06/Jul/18

$$\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }[\frac{1}{{\mathrm{n}}^2+1}+\frac{2}{{\mathrm{n}}^2+2}+\frac{3}{{\mathrm{n}}^2+3}+....+\frac{1}{\mathrm{n}+1}]=?$$

Commented by tanmay.chaudhury50@gmail.com last updated on 06/Jul/18

$$verygood...thisisthewaytosolveit...$$

Commented by math khazana by abdo last updated on 06/Jul/18

$$S_n=\sum _{k=1}^n\frac{k}{n^2+k}wehave1⩽k⩽n\Rightarrow$$$$1+n^2⩽n^2+k⩽n^2+n\Rightarrow \frac{1}{n^2+n}⩽\frac{1}{n^2+k}⩽\frac{1}{1+n^2}$$$$\Rightarrow \sum _{k=1}^n\frac{k}{n^2+n}⩽\sum _{k=1}^n\frac{k}{n^2+k}⩽\sum _{k=1}^n\frac{k}{n^2+1}\Rightarrow$$$$\frac{1}{n^2+n}\frac{n^2+n}{2}⩽S_n⩽\frac{1}{n^2+1}\frac{n^2+n}{2}\Rightarrow$$$$\frac{1}{2}⩽S_n⩽\frac{n^2+n}{2(n^2+1)}\Rightarrow {lim}_{n\to +\mathrm{\infty }}{S}_n=\frac{1}{2}.$$

Commented by abdo mathsup 649 cc last updated on 07/Jul/18

$$thanks.$$

Answered by ajfour last updated on 06/Jul/18

$$Let{n}^2=N$$$$L=\underset{N\to \mathrm{\infty }}{\lim }N\underset{r=1}{\sum ^{\sqrt{N}}}\frac{\left(r/N\right)\left(\frac{1}{N}\right)}{1+\frac{r}{N}}$$$$=\underset{N\to \mathrm{\infty }}{\lim }N{\int }_0^{1/\sqrt{N}}\frac{xdx}{1+x}$$$$=\underset{N\to \mathrm{\infty }}{\lim }N[\frac{1}{\sqrt{N}}-\ln (1+\frac{1}{\sqrt{N}})]$$$$=\underset{N\to \mathrm{\infty }}{\lim }N[\frac{1}{\sqrt{N}}-\frac{1}{\sqrt{N}}+\frac{1}{2N}-....]$$$$\Rightarrow L=\frac{1}{2}.$$

Commented by rahul 19 last updated on 06/Jul/18

Ans. given is 1/2

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