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

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

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

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(A) The given limit is $L = \lim_{n 	o \infty} \sum_{r=0}^{n(b-a)} \frac{1}{na + r}$.We can rewrite this as $L = \lim_{n 	o \infty} \sum_{r=0}^{n(b-a)} \frac{1}{n(a + \frac{r}{n})}$.This is a Riemann sum of the form $\lim_{n 	o \infty} \frac{1}{n} \sum_{r=0}^{n(b-a)} f(\frac{r}{n})$,where $f(x) = \frac{1}{a+x}$.Converting this to a definite integral,we have $L = \int_{0}^{b-a} \frac{1}{a+x} dx$.Evaluating the integral,we get $L = [\log(a+x)]_{0}^{b-a}$.Substituting the limits,$L = \log(a + (b-a)) - \log(a + 0) = \log(b) - \log(a) = \log(\frac{b}{a})$.

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

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