Q117 / 160 · AP EAMCET 2022 (08 Jul Shift 1) · Mathematics
$\lim _{n \rightarrow \infty} \frac{1}{\sqrt{n}}\left[1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{4}}+\ldots+\frac{1}{\sqrt{n}}\right]=$
- A$\sqrt{2}$
- Correct answerB$2$
- C$\sqrt{2}-1$
- D$2 \sqrt{2}$
Solution
$\lim _{n \rightarrow \infty} \frac{1}{\sqrt{n}}\left[1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{4}}+\ldots \ldots \ldots \frac{1}{\sqrt{n}}\right]$
$\begin{aligned} & =\lim _{\mathrm{n} \rightarrow \infty} \frac{1}{\sqrt{\mathrm{n}}} \sum_{\mathrm{r}=1}^{\mathrm{n}} \frac{1}{\sqrt{\mathrm{r}}}=\lim _{\mathrm{n} \rightarrow \infty} \frac{1}{\sqrt{\mathrm{n}}} \sum_{\mathrm{r}=1}^{\mathrm{n}} \frac{1}{\left(\sqrt{\frac{\mathrm{r}}{\mathrm{n}}}\right) \times \sqrt{\mathrm{n}}} \\ & =\lim _{\mathrm{n} \rightarrow \infty}\left(\frac{1}{\mathrm{n}}\right) \sum_{\mathrm{r}=1}^{\mathrm{n}} \frac{1}{\sqrt{\frac{\mathrm{r}}{\mathrm{n}}}} \\ & =\int_0^1 \frac{1}{\sqrt{\mathrm{x}}} \mathrm{dx} ; \frac{\mathrm{r}}{\mathrm{n}}=\mathrm{x} \\ & =\left[\frac{\mathrm{x}^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}\right]_0^1=\left[2(\mathrm{x})^{\frac{1}{2}}\right]_0^1=2\end{aligned}$
Hence, the correct answer is option: 'B'.