Question

$\underset{n\to \infty }{lim}\frac{1}{n}(\frac{1}{n+1}+\frac{2}{n+2}+\cdots +\frac{3n}{4n})$ is equal to

• A. $\log 4$
• B. $-\log 4$
• C. $1-\log 4$
• D. None of the above

Answer 1

Answer: (D)

None of the above

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

$=\underset{n\to \infty }{lim}\frac{1}{n}(\frac{\frac{1}{n}}{1+\frac{1}{n}}+\frac{\frac{2}{n}}{1+\frac{2}{n}}+\cdots +\frac{\frac{3n}{n}}{1+\frac{3n}{n}})$

$=\underset{n\to \infty }{lim}\frac{1}{n}\sum _{r=1}^{3n}\left(\frac{\frac{r}{n}}{1-\frac{r}{n}}\right)$

$={\int }_0^3\frac{x}{1+x}dx$

$={\int }_0^3\frac{x+1-1}{1+x}dx$

$={\int }_0^{3}dx-{\int }_0^3\frac{1}{1+x}dx$

$={[x-\log (x+1)]}_0^3$

$=3-\log 4$