mathop {lim }limits_{n to infty } left[ {frac | vedclass

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Question

(B) The given expression is $\mathop {\lim }\limits_{n 	o \infty } \sum\limits_{r = 1}^n {\frac{r}{{{n^2}}}{{\sec }^2}\frac{{{r^2}}}{{{n^2}}}} $. Thi

Vedclass · Accepted Answer

$\frac{1}{2}	an 1$

Vedclass · Answer

(B) The given expression is $\mathop {\lim }\limits_{n 	o \infty } \sum\limits_{r = 1}^n {\frac{r}{{{n^2}}}{{\sec }^2}\frac{{{r^2}}}{{{n^2}}}} $. This can be rewritten as $\mathop {\lim }\limits_{n 	o \infty } \frac{1}{n}\sum\limits_{r = 1}^n {\frac{r}{n}{{\sec }^2}\left( {\frac{r}{n}} ight)^2} $. Using the definition of a definite integral as the limit of a sum,$\mathop {\lim }\limits_{n 	o \infty } \frac{1}{n}\sum\limits_{r = 1}^n {f\left( {\frac{r}{n}} ight)} = \int_0^1 {f(x)dx} $. Here,$f(x) = x \sec^2(x^2)$. So,the limit is equal to $\int_0^1 {x{{\sec }^2}{x^2}dx} $. Let $t = x^2$,then $dt = 2x dx$,or $x dx = \frac{1}{2} dt$. When $x = 0$,$t = 0$. When $x = 1$,$t = 1$. The integral becomes $\frac{1}{2} \int_0^1 {\sec^2 t dt} = \frac{1}{2} [	an t]_0^1 = \frac{1}{2} 	an 1$.

$\mathop {\lim }\limits_{n \to \infty } \left[ {\frac{1}{{{n^2}}}{{\sec }^2}\frac{1}{{{n^2}}} + \frac{2}{{{n^2}}}{{\sec }^2}\frac{4}{{{n^2}}} + ..... + \frac{1}{n}{{\sec }^2}1} \right]$ equals

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