Problem Analysis #4

[THOUGHT] The limit is $\lim_{n\to\infty} \int_0^n \left(1 + \frac{205}{69n} x \right)^{-69n} e^{-\frac{6}{7}x} \, dx$.

Pointwise, as $n \to \infty$, $\left(1 + \frac{205}{69n} x \right)^{-69n} \to e^{-205 x}$.

Thus, the integrand converges pointwise to $e^{-205 x} e^{-\frac{6}{7} x} = e^{-\left(205 + \frac{6}{7}\right) x}$.

The limit of the integral is $\int_0^\infty e^{-\left(205 + \frac{6}{7}\right) x} \, dx = \frac{1}{205 + \frac{6}{7}} = \frac{1}{\frac{1435 + 6}{7}} = \frac{7}{1441}$.

Verify $\gcd(7, 1441) = 1$ and compute $p^2 q = 7^2 \times 1441 = 49 \times 1441$. [/THOUGHT] [PYTHON] import math print(math.gcd(7, 1441)) print(49 * 1441) [/PYTHON]

[THOUGHT] The calculation confirms $\gcd(7,1441)=1$ and $7^2 \times 1441 = 70609$. The limit is $\frac{7}{1441}$, so $p=7$, $q=1441$, and $p^2 q = 70609$. [/THOUGHT] $\boxed{70609}$

Explanation

The reasoning begins by analyzing the pointwise limit of the integrand as $n \to \infty$. For fixed $x$, the term $\left(1 + \frac{205}{69n} x \right)^{-69n}$ converges to $e^{-205 x}$, since $\left(1 + \frac{a}{n}\right)^{n b} \to e^{a b}$ with $a = \frac{205}{69} x$ and $b = -69$, yielding $a b = -205 x$.

Thus, the integrand $\left(1 + \frac{205}{69n} x \right)^{-69n} e^{-\frac{6}{7} x}$ converges pointwise to $e^{-205 x} e^{-\frac{6}{7} x} = e^{-\left(205 + \frac{6}{7}\right) x}$.

The upper limit of integration is $n$, which approaches infinity, so the limit of the integral equals $\int_0^\infty e^{-\left(205 + \frac{6}{7}\right) x} \, dx$.

Compute the exponent: $205 + \frac{6}{7} = \frac{205 \cdot 7 + 6}{7} = \frac{1435 + 6}{7} = \frac{1441}{7}$.

The integral evaluates to $\int_0^\infty e^{-\frac{1441}{7} x} \, dx = \frac{7}{1441}$.

Here, $\frac{7}{1441}$ is in lowest terms, as verified by $\gcd(7, 1441) = 1$.

Thus, $p = 7$ and $q = 1441$, so $p^2 q = 7^2 \cdot 1441 = 49 \cdot 1441 = 70609$.

Justification for interchanging limit and integral relies on the integrand being non-negative and the pointwise convergence to an integrable function, allowing application of the monotone convergence theorem.