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. $\log _e2$
• B. $\log _e\left(\frac{2}{3}\right)$
• C. 0
• D. $\log _e\left(\frac{3}{2}\right)$

Answer 1

Answer: (A)

$\log _e2$

Answer 2

$\lim_{n\to\infty} (\frac{1}{1+n}+…+\frac{1}{n+n})= \lim_{n\to\infty} ∑^n_{r=1}\frac{1}{n+r}$
$=\lim_{n\to\infty} ∑^n_{r=1}\frac{1}{n}(\frac{1}{1+\frac{r}{n}})$
$= ∫_0^1 \frac{1}{1+x}dx=[ℓln(1+x]_0^1 ​=ℓn2$