Question

$\displaystyle \lim _{n \rightarrow \infty}\left[\frac{1}{1+n}+\frac{1}{2+n}+\frac{1}{3+n}+\ldots+\frac{1}{2 n}\right]$ is equal to

• A. $\log _e 2$
• B. $\log _e\left(\frac{2}{3}\right)$
• C. 0
• D. $\log _e\left(\frac{3}{2}\right)$

Answer 1

Answer: (A)

$\displaystyle \lim _{n \rightarrow \infty}\left(\frac{1}{1+n}+\ldots+\frac{1}{n+n}\right)=\displaystyle\lim _{n \rightarrow \infty} \displaystyle\sum_{r=1}^n \frac{1}{n+r} $
$=\displaystyle\lim _{n \rightarrow \infty} \sum_{r=1}^n \frac{1}{n}\left(\frac{1}{1+\frac{r}{n}}\right) $
$=\int\limits_0^1 \frac{1}{1+x} d x=\left[\ell \ln (1+x]_0^1=\ell n 2\right.$