Let $a_n$ denote the angle opposite to the side of length $4n^2$ units in an integer right angled triangle with lengths of sides of the triangle being $4n^2, 4n^4+1$ and $4n^4-1$ where $n \in N$. Then find the value of $\lim_{p \to \infty} \sum_{n=1}^p a_n$
(A) $\pi/2$
(B)  $\pi/4$
(C) $\pi $
(D) $\pi/3$
1. **Identify the sides of the triangle:**
   Given the sides of the triangle are \(4n^2\), \(4n^4 + 1\), and \(4n^4 - 1\). We need to verify that these sides form a right-angled triangle. 

2. **Verify the Pythagorean theorem:**
   For a right-angled triangle, the sum of the squares of the two shorter sides should equal the square of the hypotenuse. Let's assume \(4n^4 + 1\) is the hypotenuse.
   \[
   (4n^2)^2 + (4n^4 - 1)^2 = (4n^4 + 1)^2
   \]
   Calculate each term:
   \[
   (4n^2)^2 = 16n^4
   \]
   \[
   (4n^4 - 1)^2 = (4n^4)^2 - 2 \cdot 4n^4 \cdot 1 + 1^2 = 16n^8 - 8n^4 + 1
   \]
   \[
   (4n^4 + 1)^2 = (4n^4)^2 + 2 \cdot 4n^4 \cdot 1 + 1^2 = 16n^8 + 8n^4 + 1
   \]
   Sum the squares of the two shorter sides:
   \[
   16n^4 + 16n^8 - 8n^4 + 1 = 16n^8 + 8n^4 + 1
   \]
   This matches the square of the hypotenuse, confirming that the sides form a right-angled triangle.

3. **Determine the angle \(a_n\):**
   The angle \(a_n\) is opposite the side of length \(4n^2\). Using the tangent function:
   \[
   a_n = \arctan \left( \frac{4n^2}{4n^4 - 1} \right)
   \]

4. **Sum of the angles \(a_n\):**
   We need to find the sum \(\sum_{n=1}^{p} a_n\):
   \[
   \sum_{n=1}^{p} a_n = \sum_{n=1}^{p} \arctan \left( \frac{4n^2}{4n^4 - 1} \right)
   \]

5. **Simplify the arctangent sum:**
   Using the identity for the sum of arctangents:
   \[
   \arctan x + \arctan y = \arctan \left( \frac{x + y}{1 - xy} \right)
   \]
   We can express the sum in a telescoping series:
   \[
   \sum_{n=1}^{p} \arctan \left( \frac{4n^2}{4n^4 - 1} \right) = \frac{\pi}{2} - \arctan \left( \frac{2p+1}{2p(p+1)} \right)
   \]

6. **Evaluate the limit as \(p \to \infty\):**
   \[
   \lim_{p \to \infty} \left( \frac{\pi}{2} - \arctan \left( \frac{2p+1}{2p(p+1)} \right) \right)
   \]
   As \(p \to \infty\), \(\frac{2p+1}{2p(p+1)} \to 0\), and \(\arctan(0) = 0\):
   \[
   \lim_{p \to \infty} \left( \frac{\pi}{2} - 0 \right) = \frac{\pi}{2}
   \]

The final answer is \(\boxed{\frac{\pi}{2}}\)