Practicing Success
$\lim\limits_{n \rightarrow \infty}\left\{\frac{1}{n}+\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{3 n}\right\}=$ |
log 2 log 3 log 5 0 |
log 3 |
Let $S =\lim\limits_{n \rightarrow \infty}\left\{\frac{1}{n}+\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{n+2 n}\right\}$ $\Rightarrow =\lim\limits_{n \rightarrow \infty}\left\{\sum\limits_{r=0}^{2 n} \frac{1}{n+r}\right\}$ $\Rightarrow S =\lim\limits_{n \rightarrow \infty}\left\{\frac{1}{n} \sum\limits_{r=0}^{2 n} \frac{n}{n+r}\right\}=\lim\limits_{n \rightarrow \infty}\left\{\frac{1}{n} \sum\limits_{r=0}^{2 n} \frac{1}{1+(r / n)}\right\}$ Now, Lower limit = $\lim\limits_{n \rightarrow \infty} \frac{r}{n}$ $=\lim\limits_{n \rightarrow \infty} \frac{0}{n}=0$ [∵ r = 0 for the first term] Upper Limit = $\lim\limits_{n \rightarrow 0} \frac{r}{n}$ $=\lim\limits_{n \rightarrow \infty} \frac{2 n}{n}=2$ [∵ r = 2n for the last term] ∴ $S=\lim\limits_{n \rightarrow \infty}\left\{\frac{1}{n} \sum\limits_{r=0}^{2 n} \frac{1}{1+(r / n)}\right\}$ $\Rightarrow S=\int\limits_0^2 \frac{1}{1+x} d x=[\log (1+x)]_0^2=\log 3-\log 1=\log 3$ |