$\lim _{n \rightarrow \infty}\left(\frac{1}{n}+\frac{1}{n+1}+\ldots+\frac{1}{3 n}\right)$ is equal to
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$\lim _{n \rightarrow \infty}\left(\frac{1}{n}+\frac{1}{n+1}+\ldots+\frac{1}{3 n}\right)$ is equal to

(A) $\log 2$

(B) $\log 3$

(C) $\log 5$

(D) 0

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SOLUTION —

$\begin{array}{l}\lim _{n \rightarrow \infty}\left(\frac{1}{n}+\frac{1}{n+1}+\ldots+\frac{1}{n+2 n}\right) \\\quad=\sum_{r=0}^{2 n} \frac{1}{n+r}=\frac{1}{n} \sum_{r=0}^{2 n} \frac{1}{1+\frac{r}{n}} \\\quad=\int_0^2 \frac{1}{1+x} d x=[\log (1+x)]_0^2=\log 3\end{array}$

So, The correct option will be (B).

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