Question

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

• A. $\frac{\log _{0}(m)}{m}$
• B. $\frac{\log _{0}(1+m)}{1+m}$
• C. $\frac{\log _{0}(1+m)}{m}$
• D. $\frac{\log _{0}(1+m)}{1-m}$

Answer 1

Answer: (C)

$\displaystyle \lim _{n \rightarrow \infty}\left\{\frac{1}{n+m}+\frac{1}{n+2 m}+\frac{1}{n+3 m}+\ldots+\frac{1}{n+n m}\right\}$
$=\displaystyle\lim _{n \rightarrow \infty} \frac{1}{n}\left\{\begin{array}{c}\frac{1}{1+\frac{m}{n}}+\frac{1}{1+2\left(\frac{m}{n}\right)}+\frac{1}{1+3\left(\frac{m}{n}\right)}+\ldots . \\ +\frac{1}{1+n\left(\frac{m}{n}\right)}\end{array}\right\}$
$=\displaystyle\lim _{n \rightarrow \infty} \frac{1}{n}\left(\sum_{k=1}^{n} \frac{1}{1+m\left(\frac{k}{n}\right)}\right)$
$=\int\limits_{0}^{1} \frac{1}{1+m x} d x=\frac{1}{m} \int\limits_{0}^{1} \frac{m}{1+m x} d x$
$=\frac{1}{m}\left[\log _{e}(1+m x)\right]_{0}^{1}=\frac{1}{m}\left[\log _{e}(1+m)-\log (1)\right]$
$=\frac{\log _{e}(1+m)}{m}$