Question

For any positive integer $n$, define $f_n:(0,\infty )\to R$ as
$f_n\left(x\right)=\sum _{j=1}^n{\tan }^{-1}\left(\frac{1}{1+(x+j)(x+j-1)}\right)$ for all $xϵ(0,\infty )$.
(Here, the inverse triagonometric function ${\tan }^{-1}x$ assume values in $(-\frac{\pi }{2},\frac{\pi }{2})$ )
Then, which of the following statement(s) is (are) TRUE?

• A. $\sum _{j=1}^5{\tan }^2\left(f_j\right(0\left)\right)=55$
• B. $\sum _{j=1}^{10}(1+f_j(0\left)\right){\sec }^2\left(f_j\right(0\left)\right)=10$
• C. For any fixed positive integer $n,\underset{x\to \infty }{lim}\tan \left(f_n\right(x\left)\right)=\frac{1}{n}$
• D. For any fixed positive integer $n,\underset{x\to \infty }{lim}{\sec }^2\left(f_n\right(x\left)\right)=1$

Answer 1

Answer: (A), (D)

$\sum _{j=1}^5{\tan }^2\left(f_j\right(0\left)\right)=55$
For any fixed positive integer $n,\underset{x\to \infty }{lim}{\sec }^2\left(f_n\right(x\left)\right)=1$

$f_n\left(x\right)=\sum _{j=1}^n{\tan }^{-1}\left(\frac{(x+j)-(x+j-1)}{1+(x+j)(x+j-1)}\right)$
$f_n\left(x\right)=\sum _{j=1}^n\left[{\tan }^{-1}\right(x+j)-{\tan }^{-1}(x+j-1\left)\right]$
$f_n\left(x\right)={\tan }^{-1}(x+n)-{\tan }^{-1}x$
$\therefore \tan \left(f_n\right(x\left)\right)=\tan \left[{\tan }^{-1}\right(x+n)-{\tan }^{-1}x]$

$\tan \left(f_n\right(x\left)\right)=\frac{(x+n)-x}{1+x(x+n)}$

$\tan \left(f_n\right(x\left)\right)=\frac{n}{1+x^2+nx}$

$\tan \left(f_n\right(0\left)\right)=n$

${\tan }^2\left(f_n\right(0\left)\right)=n^2$

$\sum _{n=0}^5\tan \left(f_n\right(0\left)\right)=\sum _{n=0}^5{n}^2=55$

$\therefore {\sec }^2\left(f_n\right(x\left)\right)=1+{\tan }^2\left(f_n\right(x\left)\right)$

${\sec }^2\left(f_n\right(x\left)\right)=1+{\left(\frac{n}{1+x^2+nx}\right)}^2$

$\therefore \underset{x\to \infty }{lim}{\sec }^2\left(f_n\right(x\left)\right)=1$