[High School Mathematics Daily Problem 5-1 and Initial Version Answer]
I asked DS to solve this problem a few days ago. DS was continuously testing the first few terms of the sequence and was unable to find a solution. Today, it actually managed to solve it! DS’s learning ability is indeed impressive![DS’s Answer]
DS’s answer also considered the variations I mentioned. However, such a solution is something only DS could come up with; humans do not possess such powerful computational abilities, relying instead on techniques!The total time taken by DS to think was 375 seconds. I believe that under the right conditions, it can already surpass 99% of high school students.[DS’s Thought Process] (Time taken: 375 seconds)Now, I need to derive the general term formula for this sequence. The problem states that a₁ = 3/2, and for n≥2, aₙ = (3n aₙ₋₁)/(2aₙ₋₁ + n -1). First, I need to observe the structure of this recursive relationship to see if it can be transformed into a known form, such as linear recursion, fractional recursion, or possibly using techniques like reciprocal transformation.First, I notice that this recursive formula is in fractional form, where the numerator is 3n multiplied by aₙ₋₁, and the denominator is 2aₙ₋₁ plus n minus 1. This fractional structure might be troublesome, so I want to see if I can simplify this expression by taking the reciprocal. That is, let bₙ = 1/aₙ, which might allow me to transform the recursive formula into a simpler form.Let me try. First, the original formula is aₙ = (3n aₙ₋₁)/(2aₙ₋₁ + n -1). If I take the reciprocal on both sides, I get:1/aₙ = (2aₙ₋₁ + n -1)/(3n aₙ₋₁) = [2aₙ₋₁/(3n aₙ₋₁)] + (n -1)/(3n aₙ₋₁)However, separating like this might not be ideal; perhaps I should directly handle the entire fraction. Let me rewrite it:Since aₙ = (3n aₙ₋₁)/(2aₙ₋₁ + n -1), taking the reciprocal on both sides gives:1/aₙ = (2aₙ₋₁ + n -1)/(3n aₙ₋₁) = [2aₙ₋₁/(3n aₙ₋₁)] + [(n -1)/(3n aₙ₋₁)]Simplifying, the first term is 2/(3n), and the second term is (n-1)/(3n aₙ₋₁). So:1/aₙ = 2/(3n) + (n -1)/(3n aₙ₋₁)However, this still doesn’t seem concise enough; perhaps I can further organize it. Or, I might need to substitute the entire expression using bₙ = 1/aₙ to obtain a recursive formula in terms of bₙ.Let bₙ = 1/aₙ, then the original formula becomes:aₙ = (3n aₙ₋₁)/(2aₙ₋₁ + n -1)Taking the reciprocal on both sides:1/aₙ = (2aₙ₋₁ + n -1)/(3n aₙ₋₁) = [2aₙ₋₁/(3n aₙ₋₁)] + (n -1)/(3n aₙ₋₁)That is:bₙ = 2/(3n) + (n -1)/(3n) * (1/aₙ₋₁) = 2/(3n) + (n -1)/(3n) * bₙ₋₁Thus, we obtain a recursive formula in terms of bₙ:bₙ = [2/(3n)] + [(n -1)/(3n)] * bₙ₋₁This expression might be further organized, or multiplied by some factor to make it a linear recursive formula. For example, it might be organized into the form of a linear non-homogeneous recursive formula.Let me multiply both sides by 3n to see if it helps:3n bₙ = 2 + (n -1) bₙ₋₁That is:3n bₙ – (n -1) bₙ₋₁ = 2This expression seems easier to handle. However, such a recursive formula might still require finding some integrating factor or variable substitution to solve.Alternatively, consider whether there exists some constant k such that letting cₙ = k^n bₙ transforms the equation into a simpler form. However, more observation may be needed.Additionally, I might try to write the recursive formula in a linear form concerning n. For instance, observing the coefficients of n and n-1, I might introduce a new variable, such as letting cₙ = n bₙ, or a similar substitution.Let me try letting cₙ = n bₙ, then the original formula:3n bₙ = 2 + (n -1) bₙ₋₁Substituting cₙ = n bₙ gives the left side as 3n bₙ = 3 cₙ, and the right side as 2 + (n -1) bₙ₋₁ = 2 + (n -1) * (cₙ₋₁)/(n -1) ) When n≥2, since cₙ₋₁ = (n-1) bₙ₋₁, the right side becomes 2 + cₙ₋₁.Thus, the original formula becomes:3 cₙ = 2 + cₙ₋₁That is:cₙ = (1/3) cₙ₋₁ + 2/3This looks like a first-order linear recursive formula, which is very easy to solve!Great! After this variable substitution, the problem has become simpler. Now, I need to confirm whether this substitution is correct.Let me check again:The original formula is 3n bₙ = 2 + (n -1) bₙ₋₁Letting cₙ = n bₙ gives the left side 3n bₙ = 3 cₙOn the right side, when n≥2, (n -1) bₙ₋₁ = (n -1) * [cₙ₋₁ / (n -1)] ) = cₙ₋₁So the right side is 2 + cₙ₋₁Thus, we get 3 cₙ = 2 + cₙ₋₁ → cₙ = (1/3)cₙ₋₁ + 2/3That’s right, this substitution is correct. Now, this recursive formula is a first-order linear non-homogeneous recursive formula, which can be solved using standard methods.First, the corresponding homogeneous recursive formula is cₙ = (1/3)cₙ₋₁, whose general solution is cₙ^(h) = C*(1/3)^{n-1}, where C is a constant.Then, we look for a particular solution. The non-homogeneous term is the constant 2/3, so we assume the particular solution is a constant c_p. Substituting into the equation:c_p = (1/3)c_p + 2/3 → c_p – (1/3)c_p = 2/3 → (2/3)c_p = 2/3 → c_p = 1Thus, the general solution is the homogeneous solution plus the particular solution:cₙ = C*(1/3)^{n-1} + 1Next, we use the initial conditions to determine the constant C. First, the initial condition of the original sequence is a₁ = 3/2, so b₁ = 1/a₁ = 2/3, and c₁ = 1*b₁ = 2/3.Substituting n=1 into the general solution:c₁ = C*(1/3)^{0} + 1 = C*1 + 1 = C + 1But c₁ = 2/3, so:C + 1 = 2/3 → C = 2/3 – 1 = -1/3Therefore, the general term formula for cₙ is:cₙ = – (1/3)*(1/3)^{n-1} + 1 = – (1/3)^n + 1Or written as:cₙ = 1 – (1/3)^nNext, since cₙ = n bₙ, we have bₙ = cₙ / n = [1 – (1/3)^n]/nThen, since bₙ = 1/aₙ, we have aₙ = 1/bₙ = n / [1 – (1/3)^n]However, here I need to check whether it meets the original condition when n=1:When n=1, a₁ = 1 / [1 – (1/3)^1] = 1 / (1 – 1/3) = 1/(2/3) = 3/2, which meets the initial condition given in the problem.Next, I will verify whether the case for n=2 meets the recursive formula:According to the general term formula, a₂ = 2 / [1 – (1/3)^2] = 2 / (1 – 1/9) = 2/(8/9) = 2*(9/8) = 9/4And according to the recursive formula, a₂ = (32a₁)/(2a₁ + 2 -1) = (6*(3/2))/(2*(3/2) +1) = (9)/(3 +1) = 9/4, which is correct.Now verify n=3:The general term gives a₃ = 3 / [1 – (1/3)^3] = 3 / (1 – 1/27) = 3/(26/27) = 81/26According to the recursive formula, a₃ = (33a₂)/(2a₂ +3 -1) = (9*(9/4))/(2*(9/4) +2) = (81/4)/(9/2 +2) = (81/4)/(13/2) = (81/4)*(2/13) = 162/52 = 81/26, which is also correct.So this general term formula is correct.However, it might be further simplified:Noticing that 1 – (1/3)^n = (3^n -1)/3^n, we have aₙ = n / [(3^n -1)/3^n] = n * 3^n / (3^n -1)Thus, the general term can be written as:aₙ = (n * 3^n) / (3^n -1)Now, the problem I need to solve is: Given the sequence $a_n = \frac{n}{1 – 3^{-n}}$, for all positive integers $n$, find the minimum positive integer $\\lambda$ such that the inequality $\prod_{k=1}^n a_k \leq \lambda n!$ holds for all $n$.First, I need to understand the requirements of the problem. The problem provides the general term formula for the sequence and then asks to find the smallest positive integer $\\lambda$ such that for all positive integers $n$, the product $\prod_{k=1}^n a_k$ does not exceed $\lambda n!$. In other words, this $\\lambda$ needs to satisfy $\prod_{k=1}^n a_k \leq \lambda n!$ for all $n \geq 1$, and $\\lambda$ is the smallest positive integer that meets this condition.First, I should calculate the product $\prod_{k=1}^n a_k$, and then compare it with $n!$ to find their ratio, thus determining the minimum value of $\\lambda$.First, write out the expression for $a_k$: $$ a_k = \frac{k}{1 – 3^{-k}} = \frac{k \cdot 3^k}{3^k – 1} $$ So, the product $\prod_{k=1}^n a_k$ can be expressed as: $$ \prod_{k=1}^n \frac{k \cdot 3^k}{3^k – 1} = \left( \prod_{k=1}^n k \right) \cdot \prod_{k=1}^n \frac{3^k}{3^k – 1} $$ Note that $\prod_{k=1}^n k = n!$, so the product simplifies to: $$ n! \cdot \prod_{k=1}^n \frac{3^k}{3^k – 1} $$ Therefore, the original inequality $\prod_{k=1}^n a_k \leq \lambda n!$ can be transformed into: $$ n! \cdot \prod_{k=1}^n \frac{3^k}{3^k – 1} \leq \lambda n! $$ Dividing both sides by $n!$, we get: $$ \prod_{k=1}^n \frac{3^k}{3^k – 1} \leq \lambda $$ Thus, the problem reduces to finding the smallest positive integer $\\lambda$ such that for all $n \geq 1$, we have: $$ \prod_{k=1}^n \frac{3^k}{3^k – 1} \leq \lambda $$ Next, I need to calculate the product $\prod_{k=1}^n \frac{3^k}{3^k – 1}$ and find its maximum value, because $\\lambda$ needs to satisfy this product for all $n$, so the minimum value of $\\lambda$ is the maximum value of this product across all $n$ rounded up to the nearest integer.First, I need to analyze how this product changes with $n$. As $n$ increases, does the product increase or decrease? Or is there an extremum?First, consider each term $\frac{3^k}{3^k – 1} = 1 + \frac{1}{3^k – 1}$, clearly each term is greater than 1, so the product is increasing. However, as $k$ increases, $\frac{3^k}{3^k – 1}$ approaches 1, so the entire product may approach some limit value rather than increasing indefinitely. Therefore, this product may be a monotonically increasing sequence with an upper bound, thus having a limit, and the maximum value may be this limit value.Therefore, I need to calculate the limit of this product as $n$ approaches infinity, and then determine whether this limit is indeed the maximum value, or whether there exists some finite $n$ at which the product reaches its maximum value.However, it is also possible that as $n$ increases, the product, while increasing, approaches some finite value, so the maximum term may be reached at some finite $n$, or at the limit. Further analysis is needed.First, calculate the values for the first few terms to see the trend:When n=1: $$ \prod_{k=1}^1 \frac{3^1}{3^1 – 1} = \frac{3}{3 – 1} = \frac{3}{2} = 1.5 $$When n=2: $$ \frac{3}{2} \times \frac{9}{9 – 1} = \frac{3}{2} \times \frac{9}{8} = \frac{27}{16} \approx 1.6875 $$When n=3: $$ \frac{27}{16} \times \frac{27}{27 – 1} = \frac{27}{16} \times \frac{27}{26} \approx 1.6875 \times 1.03846 \approx 1.7538 $$When n=4: $$ \frac{27}{16} \times \frac{27}{26} \times \frac{81}{80} \approx 1.7538 \times 1.0125 \approx 1.775 $$When n=5: $$ 1.775 \times (243/(243-1)) = 1.775 \times 243/242 \approx 1.775 \times 1.00413 \approx 1.7823 $$When n=6: $$ 1.7823 \times (729/728) \approx 1.7823 \times 1.00137 \approx 1.7847 $$When n=7: $$ 1.7847 \times (2187/2186) \approx 1.7847 \times 1.000457 \approx 1.7855 $$When n=8: $$ 1.7855 \times (6561/6560) \approx 1.7855 \times 1.000152 \approx 1.7857 $$When n=9: $$ 1.7857 \times (19683/19682) \approx 1.7857 \times 1.0000508 \approx 1.7858 $$When n=10: $$ 1.7858 \times (59049/59048) \approx 1.7858 \times 1.0000169 \approx 1.7858 $$It seems that as n increases, the product gradually approaches some value, approximately around 1.7858, and then increases very slowly, almost not changing anymore. Therefore, the maximum value of this product should be the limit value as n approaches infinity, and all finite n corresponding to the product are less than this value, as the product is monotonically increasing and converges to this limit.Thus, the maximum value of the product should be the limit value when n approaches infinity, approximately 1.7858, and the maximum value of all finite n is less than this limit value. Therefore, the minimum positive integer $\\lambda$ should be the ceiling of this limit value, which is 2, because 1.7858 is approximately 1.7858, and rounding up gives 2, while 1.7858 is less than 2, so the minimum positive integer $\\lambda$ is 2.However, I need to verify whether this conclusion is correct, whether there exists some n such that the product exceeds 2, or whether the limit value is indeed less than 2, thus making the minimum value of $\\lambda$ equal to 2.However, based on previous calculations, when n=10, the product is approximately 1.7858, and as n continues to increase, the product will get closer to the limit value, but the increment becomes smaller and smaller, so the limit value should not exceed 2, thus is $\\lambda=2$ sufficient?Or, perhaps my calculations are incorrect, or the limit value is actually higher?Let me calculate this limit value more precisely, or find its expression, or estimate whether it exceeds 2.Another method is to consider transforming the product into exponential form, or finding some series expansion.However, a simpler method might be to use logarithmic transformation to convert the product into a sum:Let P(n) = \prod_{k=1}^n (3^k)/(3^k – 1)Taking the natural logarithm gives:ln P(n) = \sum_{k=1}^n [ln(3^k) – ln(3^k – 1)] = \sum_{k=1}^n [k ln3 – ln(3^k – 1)]However, this might not be easy to sum, but perhaps can be estimated.Or, noting that 3^k – 1 = 3^k(1 – 3^{-k}), we have:ln P(n) = \sum_{k=1}^n [ln(3^k) – ln(3^k(1 – 3^{-k}))] = \sum_{k=1}^n [ln(3^k) – ln(3^k) – ln(1 – 3^{-k})] = -\sum_{k=1}^n ln(1 – 3^{-k})Thus, ln P(n) = -\sum_{k=1}^n ln(1 – 3^{-k})As n approaches infinity, this sum becomes:ln P(∞) = -\sum_{k=1}^∞ ln(1 – 3^{-k})Does this series have a known sum?According to mathematical knowledge, this series may be related to the q-Pochhammer symbol, that is:For |q| < 1, we have:∏_{k=1}^∞ (1 – q^k) = (q; q)∞In this case, q = 1/3, so:∏_{k=1}^∞ (1 – (1/3)^k) = (1/3; 1/3)∞Therefore, P(∞) = 1 / ∏_{k=1}^∞ (1 – 3^{-k}) = 1 / (1/3; 1/3)∞However, what is the specific value of this value? It may need to be calculated using numerical methods.According to numerical calculations of the q-Pochhammer symbol, when q=1/3, (1/3; 1/3)_∞ ≈ 0.5601265771… (referencing some mathematical software or tables)Thus, P(∞) = 1 / 0.5601265771 ≈ 1.7858, which is consistent with previous calculations.Therefore, the limit value of this infinite product is approximately 1.7858, so the maximum product value is this limit value, and all finite n corresponding to the product are less than this value, as the product is monotonically increasing and converges to this limit.Thus, $\\lambda$ needs to satisfy $\\lambda ≥ P(n)$ for all n, and the maximum value of P(n) is P(∞)≈1.7858, so the minimum positive integer $\\lambda$ is 2, because 1.7858 is less than 2, and if we take $\\lambda=2$, then for all n, the product P(n) ≤ 2, and if we take $\\lambda=1$, then when n is large enough, the product will exceed 1, for example, when n=1, the product is 1.5, which already exceeds 1, so $\\lambda=1$ clearly does not satisfy, but the problem requires the minimum positive integer $\\lambda$ such that for all n, the product ≤ λ n!, and based on previous analysis, when $\\lambda=2$, the product P(n) ≤2, and the limit of the product is approximately 1.7858, so $\\lambda=2$ is the minimum positive integer that satisfies the condition.However, to ensure everything is correct, I need to check this conclusion again.First, confirm that as n→∞, P(n) approaches approximately 1.7858, and all finite n corresponding to P(n) are less than this value, so the maximum P(n) is the limit value when n approaches infinity, approximately 1.7858, thus the minimum positive integer $\\lambda$ is 2, because 1.7858 < 2, and when $\\lambda=2$, for all n, P(n) ≤2 holds, while $\\lambda=1$ does not hold, so the answer is 2.
