Question

The value of $\underset{n\to \infty }{lim}\frac{1}{n}\sum _{j=1}^n\frac{(2j-1)+8n}{(2j-1)+4n}$ is equal to

• A. $1+2{\log }_e\left(\frac{3}{2}\right)$
• B. $2-{\log }_e\left(\frac{2}{3}\right)$
• C. $3+2{\log }_e\left(\frac{2}{3}\right)$
• D. $5+{\log }_e\left(\frac{3}{2}\right)$

Answer 1

The correct option is A $1+2{\log }_e\left(\frac{3}{2}\right)$

$\underset{n\to \infty }{lim}\frac{1}{n}\sum _{j=1}^n\frac{(2j-1)+8n}{(2j-1)+4n}$
$=\underset{n\to \infty }{lim}\frac{1}{n}\sum _{j=1}^n\frac{(2\frac{j}{n}-\frac{1}{n})+8}{(2\frac{j}{n}-\frac{1}{n})+4}$
$=\underset{0}{\overset{1}{\int }}\left(\frac{2x+8}{2x+4}\right)dx$
$=\underset{0}{\overset{1}{\int }}(1+\frac{2}{x+2})dx$
$=1+2{\log }_e\left(\frac{3}{2}\right)$