If $a$ and $b$ are positive integers such that $ba$, then $\lim _{n \rightarrow \infty}\left[\frac{1}{n a}+\frac{1}{n a+1}+\frac{1}{n a+2}+\ldots+\frac{1}{n b}\right]=$

Question

If $a$ and $b$ are positive integers such that $b>a$, then $\lim _{n \rightarrow \infty}\left[\frac{1}{n a}+\frac{1}{n a+1}+\frac{1}{n a+2}+\ldots+\frac{1}{n b}\right]=$

Aicharya · Accepted Answer

The given limit $ \begin{aligned} & \lim _{x ightarrow \infty}\left[\frac{1}{n a}+\frac{1}{n a+1}+\frac{1}{n a+2}+\ldots+\frac{1}{n b}ight] \ & =\lim _{x ightarrow \infty}\left[\frac{1}{n a}+\frac{1}{n a+1}+\frac{1}{n a+2}+\ldots+\frac{1}{n a+n(b-a)}ight] \ & =\lim _{x ightarrow \infty} \sum_{r=0}^{(b-a) n} \frac{1}{n a+r}=\lim _{x ightarrow \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} $

If $a$ and $b$ are positive integers such that $b>a$, then $\lim _{n \rightarrow \infty}\left[\frac{1}{n…

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