Question

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

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

Answer 1

Answer: (A), (B), (D)

$f_n\left(x\right)={\sum }_{j=1}^n{\tan }^{-1}\left(\frac{\left(x+j\right)-\left(x+j-1\right)}{1+\left(x+j\right)\left(x+j-1\right)}\right)$
$f_n\left(x\right)={\sum }_{j=1}^n\left[{\tan }^{-1}\left(x+j\right)-{\tan }^{-1}\left(x+j-1\right)\right]$
$f_n\left(x\right)={\tan }^{-1}\left(x+n\right)-{\tan }^{-1}x$
$\therefore \tan \left(f_n\left(x\right)\right)=\tan \left[{\tan }^{-1}\left(x+n\right)-{\tan }^{-1}x\right]$
$\tan \left(f_n\left(x\right)\right)=\frac{\left(x+n\right)-x}{1+x\left(x+n\right)}$
$\tan \left(f_n\left(x\right)\right)=\frac{n}{1+x^2+nx}$
$\therefore {\sec }^2\left(f_n\left(x\right)\right)=1+{\tan }^2\left(f_n\left(x\right)\right)$
${\sec }^2\left(f_n\left(x\right)\right)=1+{\left(\frac{n}{1+x^2+nx}\right)}^2$
${\lim }_{x\to \infty }{\sec }^2\left(f_n\left(x\right)\right)={\lim }_{x\to \infty }1+{\left(\frac{n}{1+x^2+nx}\right)}^2=1$