[THOUGHT] I will compute the expected limit by recognizing the integrand limit and then confirm the final arithmetic with Python. [/THOUGHT] [PYTHON]

Compute p^2 * q where the expected limit is 7/1441

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

[THOUGHT] For each fixed (x), as (n\to\infty), [ \left(1+\frac{205}{69n}x\right)^{-69n} \to e^{-205x}. ] Thus the integrand converges pointwise to [ e^{-205x} e^{-\frac{6}{7}x} = e^{-\left(205+\frac{6}{7}\right)x} = e^{-\frac{1441}{7}x}. ] One can dominate the integrand by an exponentially decaying function independent of (n), so by dominated convergence the limit of the integrals equals the integral of the limit: [ \lim_{n\to\infty} \int_0^n \left(1+\frac{205}{69n}x\right)^{-69n} e^{-\frac{6}{7}x},dx = \int_0^\infty e^{-\frac{1441}{7}x},dx = \frac{1}{\frac{1441}{7}} = \frac{7}{1441}. ] Thus (p=7), (q=1441), so [ p^2 q = 7^2\cdot 1441 = 70609. ] [/THOUGHT] \boxed{70609}