Question

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

Answer 1

Now,

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

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

$=\underset{n\to \infty }{lim}\sum _{r=1}^n\frac{1}{n}\left(\frac{1}{1+\frac{r}{n}}\right)$

$=\underset{0}{\overset{1}{\int }}\frac{1}{1+x}dx$ [ Using definition of definite integral]

$={\left|\log |\right(1+x\left)|\right|}_{x=0}^{x=1}$

$=\log 2$.