lim _{n rightarrow infty}left{frac{1}{n+m}+fr | vedclass

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Question

(C) The given expression is $\lim _{n \rightarrow \infty} \sum_{k=1}^n \frac{1}{n+km}$.We can rewrite this as $\lim _{n \rightarrow \infty} \frac{1}{n

Vedclass · Accepted Answer

$\frac{\log _e(1+m)}{m}$

Vedclass · Answer

(C) The given expression is $\lim _{n \rightarrow \infty} \sum_{k=1}^n \frac{1}{n+km}$.We can rewrite this as $\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{k=1}^n \frac{1}{1+m(\frac{k}{n})}$.Using the definition of a definite integral as the limit of a sum,$\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{k=1}^n f(\frac{k}{n}) = \int_0^1 f(x) dx$.Here,$f(x) = \frac{1}{1+mx}$.Thus,the integral becomes $\int_0^1 \frac{1}{1+mx} dx$.Evaluating the integral: $\frac{1}{m} [\log _e(1+mx)]_0^1 = \frac{1}{m} (\log _e(1+m) - \log _e(1)) = \frac{\log _e(1+m)}{m}$.

$\lim _{n \rightarrow \infty}\left\{\frac{1}{n+m}+\frac{1}{n+2 m}+\frac{1}{n+3 m}+\ldots+\frac{1}{n+n m}\right\}=$

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