limlimits_n → ∞≤ft1/n+1/(n+1)+1/(n+2)+...+1/(3 n)=

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]

∴  $S=\lim\limits_{n \rightarrow \infty}\left\{\frac{1}{n} \sum\limits_{r=0}^{2 n} \frac{1}{1+(r / n)}\right\}$

$\Rightarrow S=\int\limits_0^2 \frac{1}{1+x} d x=[\log (1+x)]_0^2=\log 3-\log 1=\log 3$