Subject: Mathematics — Limits
Question: For any positive integer $n$, define
$f_{\eta}:(0,\infty) \rightarrow \mathbb{R}$ as
$f_{n}(x) = \sum_{j = 1}^{n}\mspace{2mu}\tan^{- 1}\left( \frac{1}{1 + (x + j)(x + j - 1)} \right)$
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, for any fixed positive integer $n$, find the value of
$\lim_{x \rightarrow \infty}\mspace{2mu}\sec^{2}\left( f_{n}(x) \right)$
Answer: 1
Solution: $f_{n}(x) = \sum_{j = 1}^{n}\mspace{2mu}\mspace{2mu}\tan^{- 1}\left( \frac{1}{1 + (x + j)(x + j - 1)} \right)$\
$${= \sum_{j = 1}^{n}\mspace{2mu}\mspace{2mu}\tan^{- 1}\left\lbrack \frac{(x + j) - (x + j - 1)}{1 + (x + j)(x + j - 1)} \right\rbrack
}{= \sum_{j = 1}^{n}\mspace{2mu}\mspace{2mu}\left\lbrack \tan^{- 1}(x + j) - \tan^{- 1}(x + j - 1) \right\rbrack
}{\Rightarrow f_{n}(x) = \tan^{- 1}(x + n) - \tan^{- 1}(x)
}{= \tan^{- 1}\left( \frac{n}{1 + x(n + x)} \right) \Rightarrow f_{n}^{'}(x) = \frac{1}{1 + (x + n)^{2}} - \frac{1}{1 + x^{2}}}$$
and
$f_{n}(0) = \tan^{- 1}(n),\therefore\tan^{2}\left( \tan^{- 1}n \right) = n^{2}$\
Here $x = 0$ is not in the given domain, i.e., $x \in (0,\infty)$.\
$\lim_{x \rightarrow \infty}\mspace{2mu} tan\left( f_{n}(x) \right) = \lim_{x \rightarrow \infty}\mspace{2mu}\left( \frac{n}{1 + x(n + x)} \right) = 0$\
$\lim_{x \rightarrow \infty}\mspace{2mu}\sec^{2}\left( f_{n}(x) \right) = \lim_{n \rightarrow \infty}\mspace{2mu} 1 + \tan^{2}\left( f_{n}(x) \right)$\
$$= 1 + \lim_{x \rightarrow \infty}\mspace{2mu}\tan^{2}\left( f_{n}(x) \right) = 1$$