Question:

Step 1: Simplify the general term of the series. Let $A = \tan^{-1}\sqrt{k}$ and $B = \tan^{-1}\sqrt{k-1}$. Then $\tan A = \sqrt{k}$ and $\tan B = \sqrt{k-1}$. From these, we can deduce $\sin A = \frac{\sqrt{k}}{\sqrt{k+1}}$, $\cos A = \frac{1}{\sqrt{k+1}}$, $\sin B = \frac{\sqrt{k-1}}{\sqrt{k}}$, and $\cos B = \frac{1}{\sqrt{k}}$. Using the sine subtraction formula: $$ \sin(A-B) = \sin A \cos B - \cos A \sin B $$ $$ = \frac{\sqrt{k}}{\sqrt{k+1}} \cdot \frac{1}{\sqrt{k}} - \frac{1}{\sqrt{k+1}} \cdot \frac{\sqrt{k-1}}{\sqrt{k}} $$ $$ = \frac{1}{\sqrt{k+1}} - \frac{\sqrt{k-1}}{\sqrt{k(k+1)}} = \frac{\sqrt{k} - \sqrt{k-1}}{\sqrt{k(k+1)}} $$ Therefore, the general term of the series is $\sin^{-1}\left(\dfrac{\sqrt{k}-\sqrt{k-1}}{\sqrt{k(k+1)}}\right) = \tan^{-1}\sqrt{k} - \tan^{-1}\sqrt{k-1}$. Step 2: Sum the telescoping series. The series is given by $\theta = \sum_{k=1}^{\infty}\left(\tan^{-1}\sqrt{k} - \tan^{-1}\sqrt{k-1}\right)$. This is a telescoping series. Let $S_N$ be the $N$-th partial sum: $$ S_N = \sum_{k=1}^{N}\left(\tan^{-1}\sqrt{k} - \tan^{-1}\sqrt{k-1}\right) $$ $$ S_N = \left(\tan^{-1}\sqrt{1} - \tan^{-1}\sqrt{0}\right) + \left(\tan^{-1}\sqrt{2} - \tan^{-1}\sqrt{1}\right) + \dots + \left(\tan^{-1}\sqrt{N} - \tan^{-1}\sqrt{N-1}\right) $$ All intermediate terms cancel, leaving: $$ S_N = \tan^{-1}\sqrt{N} - \tan^{-1}\sqrt{0} = \tan^{-1}\sqrt{N} - 0 = \tan^{-1}\sqrt{N} $$ To find $\theta$, we take the limit as $N \to \infty$: $$ \theta = \lim_{N\to\infty} S_N = \lim_{N\to\infty} \tan^{-1}\sqrt{N} = \frac{\pi}{2} $$ Step 3: Evaluate $\tan\dfrac{\theta}{2}$. Given $\theta = \dfrac{\pi}{2}$, we have: $$ \tan\frac{\theta}{2} = \tan\frac{\pi/2}{2} = \tan\frac{\pi}{4} = 1 $$ Step 4: Evaluate $\lim_{x \to 0}\left(1 + \dfrac{x}{\tan x}\right)^{\frac{2}{x-\theta}}$. Substitute $\theta = \dfrac{\pi}{2}$: $$ \lim_{x \to 0}\left(1 + \dfrac{x}{\tan x}\right)^{\frac{2}{x-\pi/2}} $$ As $x \to 0$, the term $\dfrac{x}{\tan x} \to 1$. Thus, the base of the expression approaches $1+1=2$. The exponent approaches $\lim_{x \to 0} \dfrac{2}{x-\pi/2} = \dfrac{2}{0-\pi/2} = -\dfrac{4}{\pi}$. Therefore, the limit is: $$ 2^{-\frac{4}{\pi}} $$ Step 5: Evaluate $\sin\theta$. Given $\theta = \dfrac{\pi}{2}$, we have: $$ \sin\theta = \sin\frac{\pi}{2} = 1 $$ Step 6: Evaluate $\lim_{x \to \theta} \dfrac{x - \cos x - \theta}{x - \theta}$. Substitute $\theta = \dfrac{\pi}{2}$: $$ \lim_{x \to \pi/2} \dfrac{x - \cos x - \pi/2}{x - \pi/2} $$ This limit is of the indeterminate form $\frac{0}{0}$. Applying L'Hôpital's Rule: $$ \lim_{x \to \pi/2} \dfrac{\frac{d}{dx}(x - \cos x - \pi/2)}{\frac{d}{dx}(x - \pi/2)} = \lim_{x \to \pi/2} \dfrac{1 + \sin x}{1} = 1 + \sin\frac{\pi}{2} = 1 + 1 = 2 $$