$\underset{\mathrm{n}\to \mathrm{\infty }}{lim} \left(\frac{1}{\mathrm{n}+1}+\frac{1}{\mathrm{n}+2}+\dots +\frac{1}{6\mathrm{n}}\right)$ is equal to

Solution:

$\begin{array}{l}\underset{\mathrm{n}\to \mathrm{\infty }}{lim} \left(\frac{1}{\mathrm{n}+1}\right.\left.+\frac{1}{\mathrm{n}+2}+\dots +\frac{1}{6\mathrm{n}}\right)\\ =\underset{\mathrm{n}\to \mathrm{\infty }}{lim} \left(\frac{1}{\mathrm{n}+1}+\frac{1}{\mathrm{n}+2}+\dots +\frac{1}{\mathrm{n}+5\mathrm{n}}\right)\\ =\underset{\mathrm{n}\to \mathrm{\infty }}{lim} \sum _{\mathrm{r}=1}^{5\mathrm{n}} \left(\frac{1}{\mathrm{n}+\mathrm{r}}\right)=\underset{\mathrm{n}\to \mathrm{\infty }}{lim} \frac{1}{\mathrm{n}}\sum _{\mathrm{r}=1}^{5\mathrm{n}} \left(\frac{1}{1+\frac{\mathrm{r}}{\mathrm{n}}}\right)\end{array}$

$\because \text{ Lower limit of }\mathrm{r}=1$

$$\begin{gathered} \therefore \text{ Lower limit of integration }=\underset{\mathrm{n}\to \mathrm{\infty }}{lim} \frac{1}{\mathrm{n}}=0 \\ \because \text{ Upper limit of }\mathrm{r}=5\mathrm{n} \\ \therefore \text{ Upper limit of integration }=\underset{\mathrm{n}\to \mathrm{\infty }}{lim} \frac{5\mathrm{n}}{\mathrm{n}}=5 \mathrm{from} \mathrm{eq}\left(\mathrm{i}\right) \\ {\int }_0^5 \frac{1}{1+\mathrm{x}}\mathrm{dx}=[\log (1+\mathrm{x}){]}_0^5=\log 6-\log 1=\log 6 \end{gathered}$$