Question

$$\begin{gathered} f_n\left(x\right)=\underset{j=1}{\overset{n}{∑}}{\tan }^{-1}\left(\frac{\left(x+j\right)-\left(x+j-1\right)}{1+\left(x+j\right)\left(x+j-1\right)}\right) \\ {f}_n\left(x\right)=\underset{j=1}{\overset{n}{∑}}\left[{\tan }^{-1}\right(x+j)-{\tan }^{-1}(x+j-1\left)\right] \end{gathered}$$
f_n(x) = tan^–1(x + n) – tan^–1x
$∴$ tan(f_n(x)) = tan[tan^–1(x + n) – tan^–1x]
$$\begin{gathered} \tan \left(f_n\right(x\left)\right)Â =\frac{(x+n)-x}{1+x(x+n)} \\ \tan \left(f_n\right(x\left)\right)Â =\frac{n}{1+x^{2}nx} \end{gathered}$$
$$\begin{gathered} ∴se{c}^2\left(f_n\right(x\left)\right) = 1 + {\tan }^2\left(f_n\right(x\left)\right) \\  se{c}^2\left(f_n\right(x\left)\right) = 1+{\left(\frac{n}{1+x^2+nx}\right)}^2 \\ \underset{x→∞}{lim }se{c}^2\left(f_n\right(x\left)\right)=\underset{x→∞}{lim1 }+\left(\frac{n}{1+x^2+nx}\right)=1 \end{gathered}$$