Step 1: Understanding the Concept:
This problem deals with the summation of an infinite series involving inverse trigonometric functions. The standard method to solve such series is to transform the general term \( T_r \) into a difference of two consecutive terms of a sequence, i.e., \( T_r = f(r) - f(r-1) \). This creates a "telescoping series" where all intermediate terms cancel out, leaving only the first and the last (limiting) terms.
Step 2: Key Formula or Approach:
1. Use the identity: \( \tan^{-1} x - \tan^{-1} y = \tan^{-1} \left( \frac{x - y}{1 + xy} \right) \).
2. Rewrite the argument \( \frac{1}{2r^2} \) to fit the form \( \frac{x - y}{1 + xy} \).
Step 3: Detailed Explanation:
Consider the general term \( T_r = \tan^{-1} \left( \frac{1}{2r^2} \right) \).
To make the denominator look like \( 1 + xy \), we first multiply the numerator and denominator by 2:
\[ T_r = \tan^{-1} \left( \frac{2}{4r^2} \right) \]
Now, let's write \( 4r^2 \) as \( 1 + (4r^2 - 1) \):
\[ T_r = \tan^{-1} \left( \frac{2}{1 + (2r - 1)(2r + 1)} \right) \]
Observe that the difference between the two factors in the denominator is:
\[ (2r + 1) - (2r - 1) = 2 \]
This matches the numerator perfectly. So we can write:
\[ T_r = \tan^{-1} \left( \frac{(2r + 1) - (2r - 1)}{1 + (2r + 1)(2r - 1)} \right) \]
Using the identity \( \tan^{-1} A - \tan^{-1} B = \tan^{-1} \frac{A - B}{1 + AB} \):
\[ T_r = \tan^{-1}(2r + 1) - \tan^{-1}(2r - 1) \]
Now, let's find the partial sum \( S_n = \sum_{r=1}^{n} T_r \):
\[ S_n = (\tan^{-1} 3 - \tan^{-1} 1) + (\tan^{-1} 5 - \tan^{-1} 3) + (\tan^{-1} 7 - \tan^{-1} 5) + \dots + (\tan^{-1}(2n + 1) - \tan^{-1}(2n - 1)) \]
The terms cancel out (telescope):
\[ S_n = \tan^{-1}(2n + 1) - \tan^{-1} 1 \]
As \( n \to \infty \), \( 2n+1 \to \infty \), and \( \tan^{-1}(\infty) = \pi/2 \).
\[ a = \lim_{n \to \infty} S_n = \frac{\pi}{2} - \tan^{-1}(1) = \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4} \]
We need to find \( \tan a \):
\[ \tan a = \tan \left( \frac{\pi}{4} \right) = 1 \]
Step 4: Final Answer:
The sum of the infinite series evaluates to \( \pi/4 \). Thus, the tangent of that sum is 1.