If a and b are positive integers such that b | vedclass

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(A) The given limit is $L = \lim _{n \rightarrow \infty} \sum_{r=0}^{(b-a)n} \frac{1}{na+r}$.We can rewrite this as $L = \lim _{n \rightarrow \infty}

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$\log \left( \frac{b}{a} \right)$

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(A) The given limit is $L = \lim _{n \rightarrow \infty} \sum_{r=0}^{(b-a)n} \frac{1}{na+r}$.We can rewrite this as $L = \lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=0}^{(b-a)n} \frac{1}{a + \frac{r}{n}}$.Using the definition of a definite integral as the limit of a sum,$\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=0}^{kn} f\left( \frac{r}{n} \right) = \int_{0}^{k} f(x) dx$.Here,$f(x) = \frac{1}{a+x}$ and $k = b-a$.Thus,$L = \int_{0}^{b-a} \frac{1}{a+x} dx$.Evaluating the integral,we get $L = [\log(a+x)]_{0}^{b-a}$.$L = \log(a + b - a) - \log(a + 0) = \log(b) - \log(a) = \log \left( \frac{b}{a} \right)$.

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

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