Suppose the following limit is given:
where and are positive integers which are relatively prime. Find .
oindent
limₙ→∞ ∫₀ⁿ (1 + 205/(69n) x)^(-69n) e^(-6/7 x) dx = p/q
And then find p²q where p and q are coprime.
First, I should analyze the integral. The integrand is a product of two functions: (1 + a/n x)^(-bn) and e^{-cx}, where a, b, c are constants. As n becomes large, (1 + a/(n) x)^{n} tends to e^{a x}. So here, with exponent -69n, it should approach e^{-69*(205/69)x} = e^{-205x}. Wait, let me check that.
Wait, more accurately: (1 + k/n x)^{-bn} ≈ e^{-b k x} as n→∞. Here, k is 205/(69n) multiplied by x? Hmm, maybe I need to re-express it correctly.
Let me write the term inside the integral:
(1 + (205)/(69n) x)^(-69n)
This can be rewritten as [ (1 + (205x)/(69n))^{69n} ]^{-1}
As n→∞, (1 + a/n)^{bn} ≈ e^{ab}. So here, a = 205x/69 and b = 69. Therefore:
(1 + (205x)/(69n))^{69n} → e^{205x}
So the entire term becomes [e^{205x}]^{-1} = e^{-205x}.
Therefore, as n→∞, the integrand approximates to e^{-205x} * e^{-6/7 x} = e^{-(205 + 6/7)x}.
But wait, that seems too straightforward. Let me double-check:
The term is (1 + a/(n) x)^{-bn}, where a = 205/69 and b=69. So as n→∞, this tends to e^{-a b x} = e^{-(205/69)*69 x} = e^{-205x}. Yes, that's correct.
So the integral becomes approximately ∫₀^∞ e^{-(205 + 6/7)x} dx. But wait, let me compute this:
The integrand is e^{-(205 + 6/7)x}, so integrating from 0 to ∞ gives 1/(205 + 6/7).
Compute 205 + 6/7: that's (205*7 +6)/7 = (1435 +6)/7=1441/7.
So the integral is 1/(1441/7) = 7/1441. Therefore, p/q =7/1441.
But wait, are 7 and 1441 coprime? Let's check: 1441 ÷7=205.857... So no, 1441 is not divisible by 7. Wait, let me compute 7*205=1435, so 1441-1435=6, so remainder 6. Therefore, 7 and 1441 are coprime.
Thus p=7, q=1441, so p²q =7² 1441=491441.
Compute 49*1441:
First compute 50*1441=72050
Subtract 1441: 72050 -1441=70609.
Wait, let me verify that calculation again. Alternatively, compute 49 *1441:
Compute 1441*49:
Break it down as (1400 +41)49 =140049 +41*49
140049=1400(50-1)=70,000 -1,400=68,600.
4149: 4049=1,960; 1*49=49. So total 1,960+49=2,009.
Add them together: 68,600 +2,009=70,609.
So p²q=70,609.
But wait, let me make sure that the integral approximation is correct. Because as n approaches infinity, the upper limit of integration goes to infinity, but does the convergence hold?
Yes, because for large x, the integrand decays exponentially, so the integral converges.
Therefore, the final answer should be 70609. </think>
[THOUGHT] To solve the given limit involving an integral, we recognize that as ( n \to \infty ), the term ( \left(1 + \frac{205}{69n}x\right)^{-69n} ) approximates to ( e^{-205x} ). Thus, the integrand simplifies to ( e^{-(205 + 6/7)x} ), and the integral becomes ( \int_{0}^{\infty} e^{-(205 + 6/7)x} dx = \frac{1}{205 + 6/7} = \frac{7}{1441} ). Since 7 and 1441 are coprime, ( p = 7 ) and ( q = 1441 ), leading to ( p^2q = 7^2 \times 1441 = 49 \times 1441 = 70609 ). [/THOUGHT]
\boxed{70609}