limn→∞ 1n+1+1n+2+…+16n is equal to

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

1. A

$\mathrm{log}2$

2. B

$\mathrm{log}3$

3. C

$\mathrm{log}5$

4. D

$\mathrm{log}6$

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### Solution:

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

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