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

Question

Aicharya · Accepted Answer

Given, $\begin{aligned} & =\lim _{n \rightarrow \infty} \frac{1}{n}\left(\sum_{k=1}^n \frac{1}{1+m\left(\frac{k}{n}\right)}\right) \ & =\int_0^1 \frac{1}{1+m x} d x=\frac{1}{m} \int_0^1 \frac{m}{1+m x} d x \ & =\frac{1}{m}\left[\log _e(1+m x)\right]_0^1=\frac{1}{m}\left[\log _e(1+m)-\log (1)\right] \ & =\frac{\log _e(1+m)}{m} \end{aligned}$

\(\lim _{n \rightarrow \infty}\left\{\frac{1}{n+m}+\frac{1}{n+2 m}+\frac{1}{n+3 m}+\ldots+\frac{1}{n+n…

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