Question:

Step 1: Simplify the general term using the identity $\tan^{-1}(a) - \tan^{-1}(b) = \tan^{-1}\left(\frac{a-b}{1+ab}\right)$. The general term is $\tan^{-1}\left(\dfrac{1}{k(k+1)+1}\right)$. Observe that the argument can be rewritten as: $$ \frac{1}{k(k+1)+1} = \frac{(k+1)-k}{1+k(k+1)} $$ Applying the identity, we have: $$ \tan^{-1}\left(\frac{1}{k(k+1)+1}\right) = \tan^{-1}(k+1) - \tan^{-1}(k) $$ Step 2: Evaluate the sum $S_n$. The sum $S_n$ is a telescoping series: $$ S_n = \sum_{k=1}^{n} \left[\tan^{-1}(k+1) - \tan^{-1}(k)\right] $$ $$ S_n = (\tan^{-1}(2) - \tan^{-1}(1)) + (\tan^{-1}(3) - \tan^{-1}(2)) + \cdots + (\tan^{-1}(n+1) - \tan^{-1}(n)) $$ $$ S_n = \tan^{-1}(n+1) - \tan^{-1}(1) $$ Since $\tan^{-1}(1) = \frac{\pi}{4}$, the sum simplifies to: $$ S_n = \tan^{-1}(n+1) - \frac{\pi}{4} $$ Step 3: Evaluate $S_{10}$. Substitute $n=10$ into the expression for $S_n$: $$ S_{10} = \tan^{-1}(10+1) - \frac{\pi}{4} = \tan^{-1}(11) - \frac{\pi}{4} $$ Using the identity $\tan^{-1}(x) = \frac{\pi}{2} - \tan^{-1}\left(\frac{1}{x}\right)$ for $x>0$: $$ \tan^{-1}(11) = \frac{\pi}{2} - \tan^{-1}\left(\frac{1}{11}\right) $$ Substitute this back into the expression for $S_{10}$: $$ S_{10} = \left(\frac{\pi}{2} - \tan^{-1}\left(\frac{1}{11}\right)\right) - \frac{\pi}{4} = \frac{\pi}{4} - \tan^{-1}\left(\frac{1}{11}\right) $$ Step 4: Evaluate $\displaystyle\lim_{n \to \infty} S_n$. $$ \lim_{n \to \infty} S_n = \lim_{n \to \infty} \left[\tan^{-1}(n+1) - \frac{\pi}{4}\right] $$ As $n \to \infty$, $n+1 \to \infty$, so $\tan^{-1}(n+1) \to \frac{\pi}{2}$. $$ \lim_{n \to \infty} S_n = \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4} $$ Step 5: Evaluate $5 + \displaystyle\sum_{n=1}^{62} \dfrac{1 + \tan S_n}{1 - \tan S_n}$. First, find $\tan S_n$: $$ \tan S_n = \tan\left(\tan^{-1}(n+1) - \frac{\pi}{4}\right) $$ Using the tangent subtraction formula $\tan(A-B) = \frac{\tan A - \tan B}{1+\tan A \tan B}$ with $A = \tan^{-1}(n+1)$ and $B = \frac{\pi}{4}$: $$ \tan S_n = \frac{\tan(\tan^{-1}(n+1)) - \tan(\frac{\pi}{4})}{1+\tan(\tan^{-1}(n+1))\tan(\frac{\pi}{4})} = \frac{(n+1) - 1}{1+(n+1)(1)} = \frac{n}{n+2} $$ Next, simplify the term $\dfrac{1 + \tan S_n}{1 - \tan S_n}$. This expression is equivalent to $\tan\left(S_n + \frac{\pi}{4}\right)$. $$ \frac{1 + \tan S_n}{1 - \tan S_n} = \tan\left(S_n + \frac{\pi}{4}\right) $$ Substitute the expression for $S_n$: $$ \tan\left(S_n + \frac{\pi}{4}\right) = \tan\left(\left(\tan^{-1}(n+1) - \frac{\pi}{4}\right) + \frac{\pi}{4}\right) = \tan(\tan^{-1}(n+1)) = n+1 $$ Now, evaluate the sum: $$ \sum_{n=1}^{62} (n+1) = (1+1) + (2+1) + \cdots + (62+1) = 2 + 3 + \cdots + 63 $$ This is the sum of an arithmetic series or can be calculated as: $$ \sum_{n=1}^{62} (n+1) = \sum_{j=2}^{63} j = \left(\sum_{j=1}^{63} j\right) - 1 = \frac{63 \cdot (63+1)}{2} - 1 = \frac{63 \cdot 64}{2} - 1 = 63 \cdot 32 - 1 = 2016 - 1 = 2015 $$ Finally, calculate the total value: $$ 5 + \sum_{n=1}^{62} \frac{1 + \tan S_n}{1 - \tan S_n} = 5 + 2015 = 2020 $$ Step 6: Evaluate $S_5$. Substitute $n=5$ into the expression for $S_n$: $$ S_5 = \tan^{-1}(5+1) - \frac{\pi}{4} = \tan^{-1}(6) - \frac{\pi}{4} $$