Practicing Success
$\lim\limits_{n \rightarrow \infty}\left\{\frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3}+...+\frac{1}{n+n}\right\}$, is equal to |
$\log_e 3$ 0 $\log_e 2$ 1 |
$\log_e 2$ |
We have, $\lim\limits_{n \rightarrow \infty}\left\{\frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3}+...+\frac{1}{n+n}\right\}$ $=\lim\limits_{n \rightarrow \infty} \sum\limits_{r=1}^n \frac{1}{n+r}=\frac{1}{n} \lim\limits_{n \rightarrow \infty} \sum\limits_{r=1}^n \frac{1}{1+(r / n)}$ $=\int\limits_0^1 \frac{1}{1+x} d x=[\log (1+x)]_0^1=\log 2-\log 1=\log 2$ |