Mathematics

$$n\overset{Lt}{\rightarrow}\infty \displaystyle \{\frac{1}{n+1}+\frac{1}{n+2}+\ldots+\frac{1}{6n}\}=$$


ANSWER

log 6


SOLUTION

$$n\overset{Lt}{\rightarrow}\infty\sum_{r=1}^{5n}\dfrac{1}{n+r}=n\overset{Lt}{\rightarrow}\infty\sum_{r=1}^{5n}\dfrac{1}{n}\left [ \dfrac{1}{1+\left ( \dfrac{r}{n} \right )}\right ]$$
It is in the form
$$n\overset{Lt}{\rightarrow}\infty\sum_{r=1}^{n}\dfrac{1}{n}f\left ( \dfrac{r}{n} \right )=\int_{0}^{1}f(x)dx$$
So, $$n\overset{Lt}{\rightarrow}\infty\sum_{r=1}^{5n}\dfrac{1}{n}f\left ( \dfrac{r}{n} \right )=\int_{0}^{5}\dfrac{1}{1+x}dx$$
$$=log (1+x)\int_{0}^{5}$$
$$=log (6)-log (1)$$
$$=log 6$$

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Single Correct Medium Published on 17th 09, 2020
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