Question

$L\underset{n\to \infty }{lim}t\frac{1}{n}[\frac{1}{n+1}+\frac{2}{n+2}+....+\frac{3n}{4n}]$

• A. $3-\ln 4$
• B. $3+\ln 4$
• C. $3\ln 4$
• D. None of these
  

Answer 1

The correct option is A $3-\ln 4$

Let $S=\frac{lim}{n\to \infty }\frac{1}{n}(\frac{1}{n+1}+\frac{2}{n+2}+...+\frac{3n}{4n})$
$=\frac{lim}{n\to \infty }\frac{1}{n}\sum _{s=1}^{3n}\frac{\left(\frac{s}{3n}\right)3}{1+\left(\frac{s}{3n}\right)\left(3\right)}$
$=3{\int }_0^1\frac{3x}{1+3x}dx=3{\int }_0^1(1-\frac{1}{1+3x})dx$
$=3[1-{\left(\frac{ln(1+3x)}{3}\right)}_0^1]\Rightarrow S=(3-ln4)$