Question

$\underset{n\to \infty }{lim}\frac{1}{n}\left(\frac{1}{n+1}+\frac{2}{n+2}+\frac{3}{n+3}+...+\frac{3n}{4n}\right)$

Answer 1

Given, $\underset{n\to \infty }{lim}\frac{1}{n}\left(\frac{1}{n+1}+\frac{2}{n+2}+\frac{3}{n+3}+...+\frac{3n}{4n}\right)$

$=\underset{n\to \infty }{lim}\frac{1}{n}(\frac{1/n}{1+1/n}+\frac{2/n}{1+2/n}........+\frac{4-1}{4})$

$=\underset{n\to \infty }{lim}\frac{1}{n}\sum _{r=1}^{3n}\frac{r}{n+r}$

$=\underset{n\to \infty }{lim}\sum _{r=1}^{3n}\frac{1}{n}\frac{r/n}{1+r/n}$

$={\int }_{x=0}^{x=1}\frac{x}{1+x}dx$

$=x-ln(1+x){|}_0^1=1-ln2$