_n[1n^2^21n^2+2n^2^24n^2++1n^2 1] equals:

Question

$\displaystyle\lim_{n	o\infty}\left[\dfrac{1}{n^2}\sec^2\dfrac{1}{n^2}+\dfrac{2}{n^2}\sec^2\dfrac{4}{n^2}+\cdots+\dfrac{1}{n}\sec^2 1ight]$ equals:

Accepted Answer

$$\mathop {\lim }\limits_{n \to \infty } \left[ {{1 \over {{n^2}}}{{\sec }^2}{1 \over {{n^2}}} + {2 \over {{n^2}}}{{\sec }^2}{4 \over {{n^2}}} + {3 \over {{n^2}}}se{c^2}{9 \over {{n^2}}} + ... + {1 \over n}{{\sec }^2}1} \right]$$

$$ = \mathop {\lim }\limits_{n \to \infty } {r \over {{n^2}}}{\sec ^2}{{{r^2}} \over {{n^2}}} = \mathop {\lim }\limits_{n \to \infty } {1 \over n}.{\pi  \over n}{\sec ^2}{{{r^2}} \over {{n^2}}}$$

$$ \Rightarrow $$ Given limit is equal to value of integral

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\int\limits_0^1 {x{{\sec }^2}} \,{x^2}\,dx$$

or $${1 \over 2}\int\limits_0^1 {2x\,\sec } \,\,{x^2}dx = {1 \over 2}\int\limits_0^1 {{{\sec }^2}} tdl$$

$$\left[ {} \right.$$ put $$\left. {{x^2} = t} \right]$$

$$ = {1 \over 2}\left( {\tan t} \right)_0^1 = {1 \over 2}\tan \,1.$$

Solutions · 1