The user wants to evaluate the limit of an integral: [ L = \lim_{n\to\infty} \int_{0}^{n} \left(1+\frac{205}{69n}x\right)^{-69n} e^{-\frac{6}{7}x},dx ] The integrand converges pointwise to $e^{-205x} e^{-\frac{6}{7}x} = e^{-(205 + 6/7)x}$. Assuming the limit of the integral is the integral of the limit (Dominated Convergence Theorem applies here as the integrand is dominated by integrable functions for large $n$), the result should be: [ \int_{0}^{\infty} e^{-(205 + \frac{6}{7})x} , dx = \frac{1}{205 + \frac{6}{7}} = \frac{1}{\frac{1435+6}{7}} = \frac{7}{1441} ] Thus $p=7$ and $q=1441$. I need to verify if $p$ and $q$ are relatively prime and then calculate $p^2 q$. $1441 = 11 \times 131$. Since 7 is prime and does not divide 1441 (1441 = 7*205 + 6), they are coprime. The final calculation is $7^2 \times 1441 = 49 \times 1441$.

I will use Python to:

1. Numerically estimate the integral for a large value of $n$ to confirm the convergence to $7/1441$.
2. Check the primality/factors of 1441 to ensure $p, q$ are coprime.
3. Calculate $p^2 q$.

[PYTHON] import numpy as np from scipy.integrate import quad

def integrand(x, n): # (1 + 205/(69n) * x)**(-69n) * exp(-6/7 * x) # To avoid overflow/underflow with large powers, use log-exp # log_term = -69n * log(1 + 205x/(69*n)) # term = exp(log_term - 6/7 * x) base = 1 + 205 * x / (69 * n) if base <= 0: return 0 # Should not happen in [0, n] val = base**(-69 * n) * np.exp(-6/7 * x) return val

def limit_integral_approx(n): res, err = quad(integrand, 0, n, args=(n,)) return res

Calculate for a few increasing n

n_values = [100, 1000, 10000] results = [] for n in n_values: results.append(limit_integral_approx(n))

Theoretical limit

p = 7 q = 1441 theoretical = p / q

print(f"Approximations: {results}") print(f"Theoretical: {theoretical}")

Check GCD

import math gcd_val = math.gcd(p, q) print(f"GCD({p}, {q}) = {gcd_val}")

Final Answer

ans = p**2 * q print(f"p^2 * q = {ans}") [/PYTHON]

The limit is given by [ L = \lim_{n\to\infty} \int_{0}^{n} \left(1+\frac{205}{69n}x\right)^{-69n} e^{-\frac{6}{7}x},dx. ] Let $f_n(x) = \left(1+\frac{205}{69n}x\right)^{-69n} e^{-\frac{6}{7}x} \mathbf{1}_{[0,n]}(x)$. First, we find the pointwise limit of the integrand as $n \to \infty$. Recall that $\lim_{n\to\infty} \left(1+\frac{u}{n}\right)^n = e^u$. We can rewrite the first factor as: [ \left(1+\frac{205}{69n}x\right)^{-69n} = \left[ \left(1+\frac{\frac{205}{69}x}{n}\right)^n \right]^{-69}. ] As $n \to \infty$, the term inside the square brackets approaches $e^{\frac{205}{69}x}$. Thus, the first factor approaches $\left(e^{\frac{205}{69}x}\right)^{-69} = e^{-205x}$. The second factor is $e^{-\frac{6}{7}x}$. So the integrand converges pointwise to $e^{-205x} \cdot e^{-\frac{6}{7}x} = e^{-\left(205 + \frac{6}{7}\right)x}$.

Since the sequence of functions is dominated by the integrable function $e^{-\left(205 + \frac{6}{7}\right)x}$ (because $(1+y/n)^n$ increases to $e^y$, so its reciprocal decreases to $e^{-y}$), we can apply the Dominated Convergence Theorem: [ L = \int_{0}^{\infty} e^{-\left(205 + \frac{6}{7}\right)x},dx. ] Evaluating the integral: [ L = \frac{1}{205 + \frac{6}{7}} = \frac{1}{\frac{1435 + 6}{7}} = \frac{7}{1441}. ] Thus, $L = \frac{p}{q}$ with $p=7$ and $q=1441$. We check if $p$ and $q$ are relatively prime. $1441 = 11 \times 131$. Since 7 is prime and not equal to 11 or 131, $\gcd(7, 1441) = 1$. We need to find $p^2 q$: [ p^2 q = 7^2 \times 1441 = 49 \times 1441. ] Calculation: $49 \times 1441 = 49(1400 + 41) = 68600 + 2009 = 70609$.

\boxed{70609}