Question

$\underset{n\to \infty }{lim}[\frac{1}{n+1}+\frac{1}{n+2}+....+\frac{1}{6n}]$

• A. $\log 2$
• B. $\log (1+\sqrt{5})$
• C. $\log 6$
• D. 0

Answer 1

The correct option is C $\log 6$

$\underset{n\to \infty }{lim}[\frac{1}{n+1}+\frac{1}{n+2}+....+\frac{1}{6n}]$ $=\underset{n\to \infty }{lim}\frac{1}{n}[\frac{n}{n+1}+\frac{n}{n+2}+....+\frac{n}{n+5n}]$

$=\underset{n\to \infty }{lim}\frac{1}{n}[\frac{n}{n+1}+\frac{n}{n+2}+....+\frac{n}{n+5n}]=\underset{n\to \infty }{lim}\sum _{r=1}^{r=5n}\frac{1}{n}\left(\frac{1}{1+\frac{r}{n}}\right)$
taking $\frac{1}{n}=dx$ and $\frac{r}{n}=x$

$={\int }_0^5\frac{1}{1+x}dx=\left[\ln \right(1+x){]}_0^5=\ln 6-\ln 1=\ln 6$