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If $a$ and $b$ are positive integers such that $b>a$, then $\lim _{n \rightarrow \infty}\left[\frac{1}{n a}+\frac{1}{n a+1}+\frac{1}{n a+2}+\ldots+\frac{1}{n b}\right]=$
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Verified Answer
The correct answer is:
$\log \left(\frac{b}{a}\right)$
The given limit
$$
\begin{aligned}
& \lim _{x \rightarrow \infty}\left[\frac{1}{n a}+\frac{1}{n a+1}+\frac{1}{n a+2}+\ldots+\frac{1}{n b}\right] \\
& =\lim _{x \rightarrow \infty}\left[\frac{1}{n a}+\frac{1}{n a+1}+\frac{1}{n a+2}+\ldots+\frac{1}{n a+n(b-a)}\right] \\
& =\lim _{x \rightarrow \infty} \sum_{r=0}^{(b-a) n} \frac{1}{n a+r}=\lim _{x \rightarrow \infty} \frac{1}{n} \sum_{r=0}^{(b-a) n} \frac{1}{a+\frac{r}{n}} \\
& =\int_0^{(b-a)} \frac{d x}{a+x} \\
& =[\log (a+x)]_0^{b-a}=\log (a+b-a)-\log (a+0) \\
& =\log b-\log a=\log \frac{b}{a}
\end{aligned}
$$
$$
\begin{aligned}
& \lim _{x \rightarrow \infty}\left[\frac{1}{n a}+\frac{1}{n a+1}+\frac{1}{n a+2}+\ldots+\frac{1}{n b}\right] \\
& =\lim _{x \rightarrow \infty}\left[\frac{1}{n a}+\frac{1}{n a+1}+\frac{1}{n a+2}+\ldots+\frac{1}{n a+n(b-a)}\right] \\
& =\lim _{x \rightarrow \infty} \sum_{r=0}^{(b-a) n} \frac{1}{n a+r}=\lim _{x \rightarrow \infty} \frac{1}{n} \sum_{r=0}^{(b-a) n} \frac{1}{a+\frac{r}{n}} \\
& =\int_0^{(b-a)} \frac{d x}{a+x} \\
& =[\log (a+x)]_0^{b-a}=\log (a+b-a)-\log (a+0) \\
& =\log b-\log a=\log \frac{b}{a}
\end{aligned}
$$
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