Question
$$\mathop {\lim }\limits_{n \to \infty } \sum\limits_{r = 1}^{4n} {\frac{1}{{n + r}}} $$ is :
A.
$${\log _e}5$$
B.
$$0$$
C.
$${\log _e}4$$
D.
none of these
Answer :
$${\log _e}5$$
Solution :
$$\eqalign{
& {\text{Limit}} = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{r = 1}^{4n} {\frac{1}{n}.\frac{1}{{1 + \frac{r}{n}}}} = \int_\alpha ^\beta {\frac{1}{{1 + x}}dx} \cr
& {\text{where }}\alpha = \mathop {\lim }\limits_{n \to \infty } \frac{r}{n},\,{\text{when}}\,\,\,r = 1\,\,\,\,\,\,\,\, \Rightarrow \alpha = 0 \cr
& \beta = \mathop {\lim }\limits_{n \to \infty } \frac{r}{n},\,{\text{when}}\,\,\,r = 4n\,\,\,\, \Rightarrow \beta = 4 \cr
& \therefore {\text{ Limit}} = \int_0^4 {\frac{{dx}}{{1 + x}}} = \left[ {\log \left( {1 + x} \right)} \right]_0^4 = \log \,5 \cr} $$