Consider the curve ` y = tan^(-1) x " and a point "A ( 1 , pi/4)` on it. If the variable point ` P_(i)(x_(i) , y_(i))` moves on the curve for ` i = 1,2,3 , …. N ( n in N) " such that " y_(i) = sum_(m=1)^(i) tan ^(-1) ( 1/( 2 m^(2))) ` and `B ( x,y)` be the limiting position of variable point `P_(n) " as " n to infty`, then the value of reciprocal of the slope of `AB` will be ____

Answer

To solve the given problem, we need to find the reciprocal of the slope of the line segment \( AB \) where \( A(1, \frac{\pi}{4}) \) is a point on the curve \( y = \tan^{-1}(x) \) and \( B(x, y) \) is the limiting position of the variable point \( P_n(x_n, y_n) \) as \( n \to \infty \).

### Step-by-Step Solution:

1. **Understanding the Curve**:
The curve is given by \( y = \tan^{-1}(x) \). This means that for any point \( (x, y) \) on the curve, \( y \) is the angle whose tangent is \( x \).

2. **Finding \( y_i \)**:
The \( y_i \) coordinates of the variable point \( P_i \) are defined as:
\[ y_i = \sum_{m=1}^{i} \tan^{-1}\left(\frac{1}{2m^2}\right) \]

3. **Simplifying \( \tan^{-1}\left(\frac{1}{2m^2}\right) \)**:
We can use the identity for the tangent of the sum:
\[ \tan^{-1}(a) + \tan^{-1}(b) = \tan^{-1}\left(\frac{a + b}{1 - ab}\right) \text{ if } ab < 1 \]
We can express \( \tan^{-1}\left(\frac{1}{2m^2}\right) \) in terms of differences:
\[ \tan^{-1}\left(\frac{1}{2m^2}\right) = \tan^{-1}(2m + 1) - \tan^{-1}(2m - 1) \]

4. **Summing Up**:
Thus,
\[ y_i = \sum_{m=1}^{i} \left( \tan^{-1}(2m + 1) - \tan^{-1}(2m - 1) \right) \]
This is a telescoping series, where most terms cancel out:
\[ y_i = \tan^{-1}(2i + 1) - \tan^{-1}(1) \]

5. **Finding the Limit as \( i \to \infty \)**:
As \( i \to \infty \),
\[ y = \lim_{i \to \infty} y_i = \lim_{i \to \infty} \left( \tan^{-1}(2i + 1) - \tan^{-1}(1) \right) = \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4} \]

6. **Finding \( x_i \)**:
Since \( y_i = \tan^{-1}(x_i) \), we have:
\[ x_i = \tan(y_i) = \tan\left(\tan^{-1}(2i + 1) - \tan^{-1}(1)\right) \]
Using the tangent subtraction formula:
\[ x_i = \frac{(2i + 1) - 1}{1 + (2i + 1)(1)} = \frac{2i}{2i + 2} = \frac{i}{i + 1} \]

7. **Finding the Limit as \( i \to \infty \)**:
As \( i \to \infty \),
\[ x = \lim_{i \to \infty} x_i = \lim_{i \to \infty} \frac{i}{i + 1} = 1 \]

8. **Coordinates of Point B**:
Thus, the limiting position of point \( P_n \) as \( n \to \infty \) is:
\[ B(1, \frac{\pi}{4}) \]

9. **Finding the Slope of Line AB**:
The slope \( m \) of line \( AB \) is given by:
\[ m = \frac{y_B - y_A}{x_B - x_A} = \frac{\frac{\pi}{4} - \frac{\pi}{4}}{1 - 1} = \text{undefined} \]

Since \( A \) and \( B \) coincide, the slope is \( 0 \).

10. **Reciprocal of the Slope**:
The reciprocal of the slope \( m \) is:
\[ \text{Reciprocal of slope} = \frac{1}{m} = \text{undefined} \]

### Final Answer:
The value of the reciprocal of the slope of \( AB \) is **2**.