$\lim\limits_{n \rightarrow \infty}\left\{\frac{1}{n^2} \sec ^2 \frac{1}{n^2}+\frac{2}{n^2} \sec ^2 \frac{4}{n^2}+...+\frac{1}{n} \sec ^2 1\right\}$ equals
Answer & explanation
Correct answer: option 1
We have,
$\lim\limits_{n \rightarrow \infty}\left\{\frac{1}{n^2} \sec ^2 \frac{1}{n^2}+\frac{2}{n^2} \sec ^2 \frac{4}{n^2}+...+\frac{1}{n} \sec ^2 \frac{n^2}{n^2}\right\}$
$=\lim\limits_{n \rightarrow \infty} \sum\limits_{r=1}^n \frac{r}{n^2} \sec ^2 \frac{r^2}{n^2}$
$=\lim\limits_{n \rightarrow \infty} \sum\limits_{r=1}^n\left\{\frac{r}{n} \sec ^2\left(\frac{r}{n}\right)^2\right\} \frac{1}{n}$
$=\int\limits_0^1 x \sec ^2 x^2 d x=\frac{1}{2}\left[\tan x^2\right]_0^1=\frac{1}{2} \tan 1$