If k in N then lim _{n rightarrow infty}left[ | vedclass

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Question

(B) The given limit is $L = \lim _{n \rightarrow \infty} \sum_{r=1}^{(k-1)n} \frac{1}{n+r}$.This can be rewritten as $L = \lim _{n \rightarrow \infty}

Vedclass · Accepted Answer

$\log k$

Vedclass · Answer

(B) The given limit is $L = \lim _{n \rightarrow \infty} \sum_{r=1}^{(k-1)n} \frac{1}{n+r}$.This can be rewritten as $L = \lim _{n \rightarrow \infty} \sum_{r=1}^{(k-1)n} \frac{1}{n(1 + \frac{r}{n})}$.Using the definition of a definite integral as the limit of a sum,$\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^{m n} f(\frac{r}{n}) = \int_{0}^{m} f(x) dx$.Here,$m = k-1$ and $f(x) = \frac{1}{1+x}$.Thus,$L = \int_{0}^{k-1} \frac{1}{1+x} dx$.Evaluating the integral,we get $L = [\log(1+x)]_{0}^{k-1}$.Substituting the limits,$L = \log(1 + k - 1) - \log(1 + 0) = \log(k) - \log(1) = \log(k)$.Therefore,the correct option is $B$.

If $k \in N$ then $\lim _{n \rightarrow \infty}\left[\frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3}+\ldots+\frac{1}{k n}\right]=$

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