Question

\[\lim_{n \to \infty} \left\{ \frac{1}{2n + 1} + \frac{1}{2n + 2} + . . . + \frac{1}{2n + n} \right\}\] is equal to

Options

• \[\ln\left( \frac{1}{3} \right)\]
• \[\ln\left( \frac{2}{3} \right)\]
• \[\ln\left( \frac{3}{2} \right)\]
• \[\ln\left( \frac{4}{3} \right)\]

Answer 1

`ln(3/2)`

\[\lim_{n \to \infty} \left\{ \frac{1}{2n + 1} + \frac{1}{2n + 2} + . . . . . . . . . . + \frac{1}{2n + n} \right\}\]
\[ = \lim_{n \to \infty} \sum\nolimits_{r = 1}^n \frac{1}{2n + r}\]
\[ = {lim}_{n \to \infty} \frac{1}{n} \sum\nolimits_{r = 1}^n \frac{1}{2 + \frac{r}{n}}\]
\[let \frac{r}{n} = x\]
\[ = \int_0^\infty \frac{1}{2 + x} d x\]
\[ = \left[ \log\left( 2 + x \right) \right]_0^\infty \]
\[ = \log3 - \log2\]
\[ = log\frac{3}{2}\]
\[ = \ln\left( \frac{3}{2} \right)\]