Step 1: Understanding the Concept:
This problem asks for the sum of an infinite series containing inverse tangent functions. The standard method for these series is to manipulate the $n$-th term into the difference of two inverse tangents. This creates a telescoping series where almost all intermediate terms cancel out, leaving only boundary terms to evaluate the limit.
Step 2: Key Formula or Approach:
1. Identify the general $n$-th term: $T_n = \tan^{-1}\left(\frac{2^{n-1}}{1 + 2^{2n-1}}\right)$.
2. Use the algebraic identity: $\tan^{-1}\left(\frac{x - y}{1 + xy}\right) = \tan^{-1}x - \tan^{-1}y$.
3. Express $T_n$ as $f(n) - f(n-1)$ to utilize telescoping cancellation.
Step 3: Detailed Explanation:
Let's analyze the argument of the $n$-th term:
\[ \frac{2^{n-1}}{1 + 2^{2n-1}} \]
We want to express the numerator as a difference $x - y$ and the denominator as $1 + xy$.
Notice that $2^{2n-1} = 2^n \cdot 2^{n-1}$.
So, let $x = 2^n$ and $y = 2^{n-1}$. Let's verify the numerator:
$x - y = 2^n - 2^{n-1} = 2^{n-1}(2 - 1) = 2^{n-1}$. This matches perfectly!
Thus, the general term becomes:
\[ T_n = \tan^{-1}\left(\frac{2^n - 2^{n-1}}{1 + 2^n \cdot 2^{n-1}}\right) = \tan^{-1}(2^n) - \tan^{-1}(2^{n-1}) \]
Now, write out the sum of the first $n$ terms, $S_n = \sum_{k=1}^n T_k$:
\[ T_1 = \tan^{-1}(2^1) - \tan^{-1}(2^0) \]
\[ T_2 = \tan^{-1}(2^2) - \tan^{-1}(2^1) \]
\[ T_3 = \tan^{-1}(2^3) - \tan^{-1}(2^2) \]
\[ \vdots \]
\[ T_n = \tan^{-1}(2^n) - \tan^{-1}(2^{n-1}) \]
Adding these equations vertically, we observe massive cancellation (telescoping property):
\[ S_n = \tan^{-1}(2^n) - \tan^{-1}(2^0) \]
\[ S_n = \tan^{-1}(2^n) - \tan^{-1}(1) = \tan^{-1}(2^n) - \frac{\pi}{4} \]
To find the sum to infinity, we take the limit as $n \to \infty$:
\[ S_\infty = \lim_{n \to \infty} S_n = \lim_{n \to \infty} \left( \tan^{-1}(2^n) - \frac{\pi}{4} \right) \]
As $n \to \infty$, $2^n \to \infty$, and the limit of $\tan^{-1}(x)$ as $x \to \infty$ is $\frac{\pi}{2}$.
\[ S_\infty = \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4} \]
Step 4: Final Answer:
The sum to infinite terms is $\frac{\pi}{4}$.