[THOUGHT] To solve the given limit involving the integral, we recognize that as $n \to \infty$, the term $\left(1 + \frac{205}{69n}x\right)^{-69n}$ approaches $e^{-205x}$. The integrand then simplifies to $e^{-205x} \cdot e^{-\frac{6}{7}x} = e^{-\frac{1441}{7}x}$. The integral from 0 to $\infty$ of $e^{-\frac{1441}{7}x} \, dx$ evaluates to $\frac{7}{1441}$. Since 7 and 1441 are coprime (1441 factors into 11 and 131, both primes), we find $p = 7$ and $q = 1441$. Thus, $p^2 q = 7^2 \times 1441$.

To verify, we compute $7^2 \times 1441$ using Python. [/THOUGHT] [PYTHON]

Calculate p^2 * q where p = 7 and q = 1441

p = 7 q = 1441 result = p**2 * q print(result) [/PYTHON]

[THOUGHT] The limit simplifies to the integral of $e^{-\frac{1441}{7}x}$ from 0 to $\infty$, which evaluates to $\frac{7}{1441}$. Since 7 and 1441 are coprime, $p = 7$ and $q = 1441$. The Python computation confirms that $p^2 q = 7^2 \times 1441 = 70609$. Thus, the final answer is:

[/THOUGHT]