Question

For any positive integer $n$, define $f_n : (0 ,\infty) \to \mathbb R$ as $f_n (x) = \sum^{n}_{j = 1} \tan^{-1} \left(\frac{1}{1 + (x +j (x+ j -1)} \right) $ for all $x \in (0, \infty ).$
(Here, the inverse trigonometric function $tan^{-1} x$ 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. $\sum^{10}_{j=1} (1 + f'_j (0)) \sec^2 (f_j (0)) = 1 0$
• 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)

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