MCQ
$\lim\limits_{\text{n}\rightarrow\infty}\Big\{\frac{1}{2\text{n}+1}+\frac{1}{2\text{n}+2}+\ .....+\frac{1}{2\text{n}+\text{n}}\Big\}$ is equal to:
  • A
    $\ln\Big(\frac{1}{3}\Big)$
  • B
    $\ln\Big(\frac{2}{3}\Big)$
  • $\ln\Big(\frac{3}{2}\Big)$
  • D
    $\ln\Big(\frac{4}{3}\Big)$

Answer

Correct option: C.
$\ln\Big(\frac{3}{2}\Big)$
$\lim\limits_{\text{n}\rightarrow\infty}\Big\{\frac{1}{2\text{n}+1}+\frac{1}{2\text{n}+2}+\ .....\ +\frac{1}{2\text{n}+\text{n}}\Big\}$

$=\lim\limits_{\text{n}\rightarrow\infty}\sum\limits^\text{n}_{\text{r}=1}\frac{1}{2\text{n}+\text{r}}$

$=\lim\limits_{\text{n}\rightarrow\infty}\frac{1}{\text{n}}\sum\limits^\text{n}_{\text{r}=1}\frac{1}{2+\frac{\text{r}}{\text{n}}}$

let $\frac{\text{r}}{\text{n}}=\text{x}$

$=\int\limits^\infty_0\frac{1}{2+\text{x}}\text{dx}$

$=\Big[\log\big(2+\text{x}\big)\Big]^\infty_0$

$=\log3-\log2$

$=\log\frac{3}{2}$

$=\ln\Big(\frac{3}{2}\Big)$

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