$\lim\limits_{n \rightarrow \infty}\left\{\frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3}+...+\frac{1}{n+n}\right\}$, is equal to
Answer & explanation
Correct answer: option 3
We have,
$\lim\limits_{n \rightarrow \infty}\left\{\frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3}+...+\frac{1}{n+n}\right\}$
$=\lim\limits_{n \rightarrow \infty} \sum\limits_{r=1}^n \frac{1}{n+r}=\frac{1}{n} \lim\limits_{n \rightarrow \infty} \sum\limits_{r=1}^n \frac{1}{1+(r / n)}$
$=\int\limits_0^1 \frac{1}{1+x} d x=[\log (1+x)]_0^1=\log 2-\log 1=\log 2$