| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | The limit of the sequence \(\frac{(2n)!!}{(2n+1)!!}\) as \(n\) goes to infinity can be evaluated by simplifying the expression and using the properties of factorials and double factorials. The double factorial \(n!!\) is defined as the product of all the integers from \(1\) to \(n\) that have the same parity (odd or even) as \(n\). Specifically, for even \(n\), \((2n)!! = 2 \cdot 4 \cdot 6 \cdot \ldots \cdot (2n)\), and for odd \(n\), \((2n+1)!! = 1 \cdot 3 \cdot 5 \cdot \ldots \cdot (2n+1)\). Let's rewrite the given expression using these definitions: \[ \frac{(2n)!!}{(2n+1)!!} = \frac{2 \cdot 4 \cdot 6 \cdot \ldots \cdot (2n)}{1 \cdot 3 \cdot 5 \cdot \ldots \cdot (2n+1)} \] This can be further simplified by recognizing that each term in the numerator is twice the corresponding term in the denominator: \[ \frac{(2n)!!}{(2n+1)!!} = \frac{2 \cdot 2 \cdot 2 \cdot \ldots \cdot 2}{1 \cdot 3 \cdot 5 \cdot \ldots \cdot (2n+1)} = \frac{2^n n!}{(2n+1)!!} \] Since \((2n+1)!!\) can also be expressed as a product of odd numbers, it can be rewritten using factorials as follows: \[ (2n+1)!! = \frac{(2n+1)!}{2^n n!} \] Substitute this expression back into the original limit: \[ \lim_{n \to \infty} \frac{(2n)!!}{(2n+1)!!} = \lim_{n \to \infty} \frac{2^n n!}{\frac{(2n+1)!}{2^n n!}} = \lim_{n \to \infty} \frac{(2^n n!)^2}{(2n+1)!} \] We know that \((2n+1)!\) grows much faster than \((2^n n!)^2\) as \(n\) approaches infinity, because \((2n+1)!\) includes all the terms in \((2^n n!)^2\) and many more. Therefore, the limit of the sequence as \(n\) goes to infinity is: \[ \lim_{n \to \infty} \frac{(2n)!!}{(2n+1)!!} = 0 \] Thus, the solution to the problem is that the limit of \(\frac{(2n)!!}{(2n+1)!!}\) as \(n\) goes to infinity is \(0\). \[ \boxed{0} \] |
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| deepseekr1 | To find the limit of \(\frac{(2n)!!}{(2n+1)!!}\) as \(n\) approaches infinity, we start by expressing the double factorials in terms of regular factorials. The double factorial \((2n)!!\) can be written as \(2^n \cdot n!\). Similarly, the double factorial \((2n+1)!!\) can be written as \(\frac{(2n+1)!}{2^n \cdot n!}\). Thus, the given expression becomes: \[ \frac{(2n)!!}{(2n+1)!!} = \frac{2^n \cdot n!}{\frac{(2n+1)!}{2^n \cdot n!}} = \frac{2^{2n} \cdot (n!)^2}{(2n+1)!} \] Using Stirling's approximation \(n! \approx \sqrt{2\pi n} \left(\frac{n}{e}\right)^n\), we approximate the factorials: - \(n! \approx \sqrt{2\pi n} \left(\frac{n}{e}\right)^n\) - \((2n+1)! \approx (2n+1) \cdot \sqrt{4\pi n} \left(\frac{2n}{e}\right)^{2n}\) Substituting these approximations into the expression, we get: \[ \frac{2^{2n} \cdot (\sqrt{2\pi n} \left(\frac{n}{e}\right)^n)^2}{(2n+1) \cdot \sqrt{4\pi n} \left(\frac{2n}{e}\right)^{2n}} \] Simplifying the numerator and denominator: - Numerator: \(2^{2n} \cdot 2\pi n \cdot \left(\frac{n}{e}\right)^{2n}\) - Denominator: \((2n+1) \cdot 2\sqrt{\pi n} \cdot 2^{2n} \cdot \left(\frac{2n}{e}\right)^{2n}\) Canceling common terms and simplifying further, we find: \[ \frac{2\pi n}{(2n+1) \cdot 2\sqrt{\pi n}} = \frac{\sqrt{\pi n}}{2n+1} \] For large \(n\), this expression approximates to \(\frac{\sqrt{\pi}}{2} \cdot \frac{1}{\sqrt{n}}\), which tends to 0 as \(n\) approaches infinity. Additionally, considering the product form \(\prod_{k=1}^n \frac{2k}{2k+1}\) and taking the logarithm, we find the sum of the logarithms diverges to negative infinity, confirming the product tends to 0. Thus, the limit is \(\boxed{0}\). |
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| gpto1 | To find the limit of the ratio \(\frac{(2n)!!}{(2n+1)!!}\) as \(n\) approaches infinity, we'll start by understanding the double factorial notation and expressing the double factorials in terms of more familiar functions. **Definition of Double Factorial:** - For even positive integers: \((2n)!! = 2 \times 4 \times 6 \times \dots \times 2n\) - For odd positive integers: \((2n+1)!! = 1 \times 3 \times 5 \times \dots \times (2n+1)\) **Expressing Double Factorials:** 1. **Even Double Factorial:** \[(2n)!! = 2^n \times n!\] This is because each term in the product \((2 \times 4 \times \dots \times 2n)\) has a factor of 2, and there are \(n\) such factors, leading to \(2^n\). The remaining part is just \(n!\). 2. **Odd Double Factorial:** To express \((2n+1)!!\) in terms of factorial, we use the relationship: \[(2n+1)!! = \frac{(2n+1)!}{2^n \times n!}\] This comes from separating the factorial \((2n+1)!\) into its even and odd components. **Setting Up the Ratio:** Now, our ratio becomes: \[ L = \frac{(2n)!!}{(2n+1)!!} = \frac{2^n \times n!}{\dfrac{(2n+1)!}{2^n \times n!}} = \frac{(2^n \times n!)^2}{(2n+1)!} \] **Using Stirling's Approximation:** For large \(n\), Stirling's approximation provides a good estimate for factorials: \[ \ln n! \approx n\ln n - n + \frac{1}{2}\ln(2\pi n) \] We can now take the natural logarithm of \(L\) to simplify the expression: \[ \ln L = 2n\ln 2 + 2\ln n! - \ln(2n+1)! \] Applying Stirling's approximation: \[ \ln n! \approx n\ln n - n \] \[ \ln(2n+1)! \approx (2n+1)\ln(2n+1) - (2n+1) \] Ignoring the constants (since they become negligible as \(n \to \infty\)): \[ \ln L \approx 2n\ln 2 + 2(n\ln n - n) - [(2n+1)\ln(2n+1) - (2n+1)] \] Simplify the expression: \[ \ln L \approx 2n\ln 2 + 2n\ln n - 2n - [2n\ln 2n + \ln 2n + 1] + 2n + 1 \] We can approximate \(\ln(2n+1) \approx \ln(2n)\) and \((2n+1)\ln(2n+1) \approx 2n\ln(2n)\) for large \(n\). Simplify further: \[ \ln L \approx 2n\ln 2 + 2n\ln n - 2n - [2n\ln 2 + 2n\ln n + \ln 2n + 1] + 2n + 1 \] Most terms cancel out: \[ \ln L \approx -\ln(2n) - 1 \] As \(n\) approaches infinity, \(-\ln(2n)\) dominates, so we get: \[ \ln L \approx -\ln(2n) \] \[ L \approx e^{-\ln(2n)} = \frac{1}{2n} \] Therefore, as \(n\) approaches infinity: \[ L \to \frac{1}{\infty} = 0 \] **Conclusion:** The limit of the ratio \(\frac{(2n)!!}{(2n+1)!!}\) as \(n\) approaches infinity is **0**. **Answer:** 0 |
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| GPT-4o | To find the limit of \(\frac{(2n)!!}{(2n+1)!!}\) as \(n\) approaches infinity, we first need to understand what the double factorial notation means. For an even integer \(2n\): \[ (2n)!! = 2n \cdot (2n-2) \cdot (2n-4) \cdots 2 \] For an odd integer \(2n+1\): \[ (2n+1)!! = (2n+1) \cdot (2n-1) \cdot (2n-3) \cdots 1 \] The double factorial of an even number \( (2n)!! \) is the product of all the even numbers up to \(2n\), and the double factorial of an odd number \( (2n+1)!! \) is the product of all the odd numbers up to \(2n+1\). Now, let's write this fraction out explicitly: \[ \frac{(2n)!!}{(2n+1)!!} = \frac{2n \cdot (2n-2) \cdot (2n-4) \cdots 2}{(2n+1) \cdot (2n-1) \cdot (2n-3) \cdots 1} \] We can approximate the ratio by considering the leading terms for large \(n\). Note that for large \(n\), the dominant terms in the double factorials are \(2n\) and \(2n+1\), respectively. Let's look at the ratio of consecutive terms: \[ \frac{(2n) \cdot (2n-2) \cdot (2n-4) \cdots 2}{(2n+1) \cdot (2n-1) \cdot (2n-3) \cdots 1} \] Each term in the numerator is approximately half the corresponding term in the denominator for large \(n\). More precisely, each term in the numerator is of the form \(2k\) and each term in the denominator is of the form \(2k+1\), where \(k\) ranges from \(n\) down to 1. We can analyze this by considering the product of the ratios of consecutive terms: \[ \prod_{k=1}^{n} \frac{2k}{2k+1} \] As \(n\) goes to infinity, this product tends to zero because each fraction \(\frac{2k}{2k+1}\) is slightly less than 1 and the product of a large number of such fractions will tend to zero. To be more rigorous, we can use the Wallis product formula or Stirling's approximation to show that the limit is zero: Using Stirling's approximation \(n! \approx \sqrt{2 \pi n} \left( \frac{n}{e} \right)^n\), we can write: \[ (2n)!! \approx \sqrt{\pi n} \left( \frac{2n}{e} \right)^n \] \[ (2n+1)!! \approx \sqrt{\pi n} \left( \frac{2n+1}{e} \right)^n \] Using these approximations, we can write the ratio as: \[ \frac{(2n)!!}{(2n+1)!!} \approx \frac{\sqrt{\pi n} \left( \frac{2n}{e} \right)^n}{\sqrt{\pi n} \left( \frac{2n+1}{e} \right)^n} = \left( \frac{2n}{2n+1} \right)^n \] As \(n\) goes to infinity: \[ \left( \frac{2n}{2n+1} \right)^n = \left( 1 - \frac{1}{2n+1} \right)^n \approx \frac{1}{e^{1/2}} = e^{-1/2} \] Thus: \[ \left( \frac{2n}{2n+1} \right)^n \rightarrow 0 \text{ as } n \rightarrow \infty \] Therefore, the limit of \(\frac{(2n)!!}{(2n+1)!!}\) as \(n\) approaches infinity is: \[ \boxed{0} \] |
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| llama3.1-70B-Instruct | A nice problem! The double factorial notation is defined as: $$n!! = \begin{cases} n(n-2)(n-4)\cdots(2\text{ or }1) & \text{if } n \text{ is even} \\ n(n-2)(n-4)\cdots(3\text{ or }1) & \text{if } n \text{ is odd} \end{cases}$$ Now, let's analyze the given expression: $$\frac{(2n)!!}{(2n+1)!!}$$ We can rewrite the numerator as: $$(2n)!! = (2n)(2n-2)(2n-4)\cdots(4)(2)$$ And the denominator as: $$(2n+1)!! = (2n+1)(2n-1)(2n-3)\cdots(3)(1)$$ Now, we can see that the numerator and denominator have many common factors. In fact, we can cancel out all the factors except for the last two: $$\frac{(2n)!!}{(2n+1)!!} = \frac{(2n)(2n-2)}{(2n+1)(2n-1)}$$ Simplifying further, we get: $$\frac{(2n)!!}{(2n+1)!!} = \frac{2n(2n-2)}{(2n+1)(2n-1)} = \frac{4n^2 - 4n}{4n^2 - 1}$$ Now, we can take the limit as $n$ goes to infinity: $$\lim_{n\to\infty} \frac{4n^2 - 4n}{4n^2 - 1} = \lim_{n\to\infty} \frac{4 - \frac{4}{n}}{4 - \frac{1}{n^2}} = \frac{4}{4} = 1$$ Therefore, the limit of $(2n)!!/(2n+1)!!$ as $n$ goes to infinity is $\boxed{1}$. |
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| llama3.1-405B-Instruct | A nice problem! To evaluate the limit, we can use the fact that $(2n)!! = 2^n \cdot n!$ and $(2n+1)!! = (2n+1) \cdot (2n-1) \cdot (2n-3) \cdots 3 \cdot 1 = (2n+1) \cdot (2n-1)!!$. So, we can rewrite the expression as: $$\frac{(2n)!!}{(2n+1)!!} = \frac{2^n \cdot n!}{(2n+1) \cdot (2n-1)!!}$$ Now, we can use the fact that $(2n-1)!! = (2n-1) \cdot (2n-3) \cdots 3 \cdot 1 = (2n-1) \cdot (2n-3)!!$ to rewrite the expression as: $$\frac{(2n)!!}{(2n+1)!!} = \frac{2^n \cdot n!}{(2n+1) \cdot (2n-1) \cdot (2n-3)!!}$$ Notice that $(2n-3)!!$ is a product of $n-1$ odd integers, each of which is less than or equal to $2n-3$. Therefore, we have: $$(2n-3)!! \leq (2n-3)^{n-1}$$ Using this inequality, we can bound the expression: $$\frac{(2n)!!}{(2n+1)!!} \leq \frac{2^n \cdot n!}{(2n+1) \cdot (2n-1) \cdot (2n-3)^{n-1}}$$ Now, we can use Stirling's approximation for $n!$: $$n! \approx \sqrt{2\pi n} \left(\frac{n}{e}\right)^n$$ Substituting this into the expression, we get: $$\frac{(2n)!!}{(2n+1)!!} \leq \frac{2^n \cdot \sqrt{2\pi n} \left(\frac{n}{e}\right)^n}{(2n+1) \cdot (2n-1) \cdot (2n-3)^{n-1}}$$ Simplifying and taking the limit as $n \to \infty$, we get: $$\lim_{n \to \infty} \frac{(2n)!!}{(2n+1)!!} \leq \lim_{n \to \infty} \frac{2^n \cdot \sqrt{2\pi n} \left(\frac{n}{e}\right)^n}{(2n+1) \cdot (2n-1) \cdot (2n-3)^{n-1}} = 0$$ Therefore, we have: $$\lim_{n \to \infty} \frac{(2n)!!}{(2n+1)!!} = 0$$ So, the limit is 0. |
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| DeepSeek-Coder-V2 | To solve the problem of finding the limit of \(\frac{(2n)!!}{(2n+1)!!}\) as \(n\) goes to infinity, we will follow a step-by-step approach. First, let's recall the definitions of double factorials: - \((2n)!!\) is the product of all even numbers up to \(2n\), i.e., \((2n)!! = 2 \cdot 4 \cdot 6 \cdots (2n)\). - \((2n+1)!!\) is the product of all odd numbers up to \(2n+1\), i.e., \((2n+1)!! = 1 \cdot 3 \cdot 5 \cdots (2n+1)\). We can express these products in a more convenient form: \[ (2n)!! = 2^n \cdot n! \] \[ (2n+1)!! = \frac{(2n+1)!}{(2n)!!} = \frac{(2n+1)!}{2^n \cdot n!} \] Now, we substitute these expressions into the given limit: \[ \lim_{n \to \infty} \frac{(2n)!!}{(2n+1)!!} = \lim_{n \to \infty} \frac{2^n \cdot n!}{\frac{(2n+1)!}{2^n \cdot n!}} = \lim_{n \to \infty} \frac{2^{2n} \cdot (n!)^2}{(2n+1)!} \] Next, we simplify the expression inside the limit: \[ \lim_{n \to \infty} \frac{2^{2n} \cdot (n!)^2}{(2n+1)!} \] To understand the behavior of this limit, we can use Stirling's approximation for factorials, which states: \[ n! \approx \sqrt{2\pi n} \left(\frac{n}{e}\right)^n \] Applying Stirling's approximation to \(n!\) and \((2n+1)!\): \[ n! \approx \sqrt{2\pi n} \left(\frac{n}{e}\right)^n \] \[ (2n+1)! \approx \sqrt{2\pi (2n+1)} \left(\frac{2n+1}{e}\right)^{2n+1} \] Substituting these approximations into our limit expression: \[ \lim_{n \to \infty} \frac{2^{2n} \cdot (\sqrt{2\pi n} \left(\frac{n}{e}\right)^n)^2}{\sqrt{2\pi (2n+1)} \left(\frac{2n+1}{e}\right)^{2n+1}} \] Simplifying further: \[ \lim_{n \to \infty} \frac{2^{2n} \cdot 2\pi n \left(\frac{n}{e}\right)^{2n}}{\sqrt{2\pi (2n+1)} \left(\frac{2n+1}{e}\right)^{2n+1}} \] As \(n\) approaches infinity, the dominant terms in the numerator and denominator are: \[ \lim_{n \to \infty} \frac{2^{2n} \cdot n^{2n}}{ (2n+1)^{2n+1}} \] Since \(2n+1 \approx 2n\) for large \(n\), we can approximate: \[ \lim_{n \to \infty} \frac{2^{2n} \cdot n^{2n}}{ (2n)^{2n+1}} = \lim_{n \to \infty} \frac{2^{2n} \cdot n^{2n}}{ 2^{2n} \cdot n^{2n} \cdot 2n} = \lim_{n \to \infty} \frac{1}{2n} = 0 \] Thus, the limit is: \[ \boxed{0} \] |
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| Qwen2-72B-Instruct | The limit of the sequence \(\frac{(2n)!!}{(2n+1)!!}\) as \(n\) goes to infinity can be evaluated by simplifying the expression and using the properties of factorials and double factorials. The double factorial \(n!!\) is defined as the product of all the integers from \(1\) to \(n\) that have the same parity (odd or even) as \(n\). Specifically, for even \(n\), \((2n)!! = 2 \cdot 4 \cdot 6 \cdot \ldots \cdot (2n)\), and for odd \(n\), \((2n+1)!! = 1 \cdot 3 \cdot 5 \cdot \ldots \cdot (2n+1)\). Let's rewrite the given expression using these definitions: \[ \frac{(2n)!!}{(2n+1)!!} = \frac{2 \cdot 4 \cdot 6 \cdot \ldots \cdot (2n)}{1 \cdot 3 \cdot 5 \cdot \ldots \cdot (2n+1)} \] This can be further simplified by recognizing that each term in the numerator is twice the corresponding term in the denominator: \[ \frac{(2n)!!}{(2n+1)!!} = \frac{2 \cdot 2 \cdot 2 \cdot \ldots \cdot 2}{1 \cdot 3 \cdot 5 \cdot \ldots \cdot (2n+1)} = \frac{2^n n!}{(2n+1)!!} \] Since \((2n+1)!!\) can also be expressed as a product of odd numbers, it can be rewritten using factorials as follows: \[ (2n+1)!! = \frac{(2n+1)!}{2^n n!} \] Substitute this expression back into the original limit: \[ \lim_{n \to \infty} \frac{(2n)!!}{(2n+1)!!} = \lim_{n \to \infty} \frac{2^n n!}{\frac{(2n+1)!}{2^n n!}} = \lim_{n \to \infty} \frac{(2^n n!)^2}{(2n+1)!} \] We know that \((2n+1)!\) grows much faster than \((2^n n!)^2\) as \(n\) approaches infinity, because \((2n+1)!\) includes all the terms in \((2^n n!)^2\) and many more. Therefore, the limit of the sequence as \(n\) goes to infinity is: \[ \lim_{n \to \infty} \frac{(2n)!!}{(2n+1)!!} = 0 \] Thus, the solution to the problem is that the limit of \(\frac{(2n)!!}{(2n+1)!!}\) as \(n\) goes to infinity is \(0\). \[ \boxed{0} \] |
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| baichuan4 | To find the limit of the expression $(\frac{(2n)!!}{(2n+1)!!})$ as $n$ goes to infinity, we can analyze the ratio of the double factorials involved. First, let's recall the definition of a double factorial for an even number: $$(2n)!! = 2n \cdot (2n-2) \cdot (2n-4) \cdots 4 \cdot 2$$ And for an odd number, it is defined as: $$(2n+1)!! = (2n+1) \cdot (2n-1) \cdot (2n-3) \cdots 3 \cdot 1$$ Now, consider the ratio of these two expressions: $$\frac{(2n)!!}{(2n+1)!!} = \frac{2n \cdot (2n-2) \cdot (2n-4) \cdots 4 \cdot 2}{(2n+1) \cdot (2n-1) \cdot (2n-3) \cdots 3 \cdot 1}$$ As $n$ goes to infinity, the ratio of consecutive terms in the numerator and denominator will approach 1: $$\lim_{n \to \infty} \frac{2n}{2n+1} = 1, \quad \lim_{n \to \infty} \frac{2n-2}{2n-1} = 1, \quad \text{and so on.}$$ However, because the denominator always has one more term than the numerator, the overall ratio will be less than 1. Specifically, each ratio of consecutive terms will contribute a factor slightly less than 1, and there will be $n$ such factors. Therefore, the limit of the entire expression will be 0 as $n$ goes to infinity: $$\lim_{n \to \infty} \frac{(2n)!!}{(2n+1)!!} = 0$$ This result makes intuitive sense because the numerator is the product of $n$ even numbers, while the denominator is the product of $n$ odd numbers, which are all slightly larger than the corresponding even numbers. As $n$ grows large, the difference between the numerator and denominator becomes significant, driving the overall ratio towards zero. |
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| doubao | We can rewrite the expression as: \[ \frac{(2n)!!}{(2n+1)!!} = \frac{(2n)(2n-2)\cdots(2)(1)}{(2n+1)(2n-1)\cdots(3)(1)} = \frac{2^n n!}{(2n+1)!!} \] Now, consider the ratio of consecutive terms: \[ \frac{a_{n+1}}{a_n} = \frac{\frac{2^{n+1} (n+1)!}{(2n+3)!!}}{\frac{2^n n!}{(2n+1)!!}} = \frac{2(n+1)}{2n+3} \] As n goes to infinity, this ratio approaches 2/3 < 1. By the ratio test, the series converges, and the limit of the expression is 0. |
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| minimax | The double factorial notation, denoted by "!!", is defined for odd and even integers. For even integers, $(2n)!!$ is the product of all even integers from 2 up to $2n$, and for odd integers, $(2n+1)!!$ is the product of all odd integers from 1 up to $2n+1$. The limit we are looking at is: $$\lim_{n \to \infty} \frac{(2n)!!}{(2n+1)!!}$$ To evaluate this limit, we can use the Wallis product for $\pi$, which is given by: $$\frac{\pi}{2} = \prod_{n=1}^{\infty} \frac{(2n)^2}{(2n-1)(2n+1)}$$ This product can be rewritten as: $$\frac{\pi}{2} = \prod_{n=1}^{\infty} \frac{4n^2}{(2n-1)(2n+1)} = \prod_{n=1}^{\infty} \frac{2n}{2n-1} \cdot \frac{2n}{2n+1}$$ Notice that the terms in the product resemble the terms in our limit. If we take the reciprocal of the Wallis product, we get: $$\frac{2}{\pi} = \prod_{n=1}^{\infty} \frac{2n-1}{2n} \cdot \frac{2n+1}{2n}$$ Now, if we look at the terms in our limit, we can see that they are the terms in the reciprocal Wallis product: $$\frac{(2n)!!}{(2n+1)!!} = \frac{2 \cdot 4 \cdot 6 \cdot \ldots \cdot 2n}{1 \cdot 3 \cdot 5 \cdot \ldots \cdot (2n+1)} = \frac{2}{1} \cdot \frac{2}{3} \cdot \frac{4}{3} \cdot \frac{4}{5} \cdot \frac{6}{5} \cdot \frac{6}{7} \cdot \ldots \cdot \frac{2n}{2n-1} \cdot \frac{2n}{2n+1}$$ This is essentially the reciprocal Wallis product, but with each pair of terms $(2k)/(2k-1)$ and $(2k)/(2k+1)$ replaced by a single term $(2k)/(2k+1)$. Therefore, as $n$ goes to infinity, the limit of our expression is the reciprocal of the Wallis product for $\pi$: $$\lim_{n \to \infty} \frac{(2n)!!}{(2n+1)!!} = \frac{2}{\pi}$$ So the limit of the given expression as $n$ approaches infinity is $\frac{2}{\pi}$. |
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| kimi | To find the limit of the expression \(\frac{(2n)!!}{(2n+1)!!}\) as \(n\) goes to infinity, we can start by expressing the double factorials in terms of factorials and simplify the expression. The double factorial \((2n)!!\) is the product of all even numbers from 2 to \(2n\), and \((2n+1)!!\) is the product of all odd numbers from 1 to \(2n+1\). We can write these as: \[ (2n)!! = 2 \cdot 4 \cdot 6 \cdot \ldots \cdot (2n) \] \[ (2n+1)!! = 1 \cdot 3 \cdot 5 \cdot \ldots \cdot (2n+1) \] Now, let's express the factorials in terms of regular factorials: \[ (2n)!! = \frac{(2n)!}{2^n \cdot n!} \] \[ (2n+1)!! = \frac{(2n+1)!}{2^n \cdot (n+1)! \] Substituting these into the original expression, we get: \[ \frac{(2n)!!}{(2n+1)!!} = \frac{\frac{(2n)!}{2^n \cdot n!}}{\frac{(2n+1)!}{2^n \cdot (n+1)!}} = \frac{(2n)! \cdot (n+1)!}{2^n \cdot n! \cdot (2n+1)!} \] Now, we can cancel out \((2n)!\) in the numerator and denominator: \[ = \frac{(n+1)!}{2^n \cdot n! \cdot (2n+1)} \] Next, we can express \((n+1)!\) in terms of \(n!\): \[ = \frac{(n+1) \cdot n!}{2^n \cdot n! \cdot (2n+1)} \] The \(n!\) terms cancel out: \[ = \frac{n+1}{2^n \cdot (2n+1)} \] As \(n\) goes to infinity, \(2^n\) grows much faster than \(n+1\) and \(2n+1\), so the fraction approaches 0. Therefore, the limit is: \[ \lim_{n \to \infty} \frac{(2n)!!}{(2n+1)!!} = 0 \] |
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| wenxinyiyan | 为了找到极限 lim n→∞ (2n)!!/(2n+1)!!，我们可以使用以下步骤： 首先，我们注意到 (2n)!! = 2 × 4 × 6 × ... × (2n)，而 (2n+1)!! = 1 × 3 × 5 × ... × (2n+1)。 我们可以将 (2n)!! 分解为 2^n × (1 × 2 × 3 × ... × n)，即 (2n)!! = 2^n × n!。 类似地，(2n+1)!! 可以分解为 (1 × 3 × 5 × ... × (2n-1)) × (2n+1)，即 (2n+1)!! = (2n-1)!! × (2n+1)。 现在，我们可以将原极限重写为 lim n→∞ (2^n × n!)/((2n-1)!! × (2n+1))。 接下来，我们注意到 (2n-1)!! = 1 × 3 × 5 × ... × (2n-1) = (2n)!/(2^n × n!)，因此我们可以将极限进一步简化为 lim n→∞ (2^n × n!)/((2n)!/(2^n × n!) × (2n+1))。 这可以进一步简化为 lim n→∞ (2^n × n!)^2/((2n)! × (2n+1))。 现在，我们使用斯特林公式 n! ≈ sqrt(2πn) × (n/e)^n，将极限重写为 lim n→∞ (2^n × sqrt(2πn) × (n/e)^n)^2/(sqrt(4πn) × (2n/e)^(2n) × (2n+1))。 这可以进一步简化为 lim n→∞ (2^n × sqrt(2πn) × n^n/e^n)^2/(sqrt(4πn) × (2n)^(2n)/e^(2n) × (2n+1))。 现在，我们注意到分子和分母中的 e^(2n) 和 (2n)^(2n) 可以相互抵消，因此极限可以进一步简化为 lim n→∞ (2^n × sqrt(2πn) × n^n)^2/(sqrt(4πn) × (2n+1))。 这可以进一步简化为 lim n→∞ (2^n × n^n)^2/(2n+1)。 最后，我们注意到分子中的 (2^n × n^n)^2 = 2^(2n) × n^(2n) 与分母中的 (2n+1) 相比，随着 n 趋于无穷大，分子的增长速度远快于分母，因此极限为无穷大。 因此，lim n→∞ (2n)!!/(2n+1)!! = ∞。 |
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| zhipu | To find the limit of \(\frac{(2n)!!}{(2n+1)!!}\) as \(n\) approaches infinity, we need to understand the behavior of the double factorials in the numerator and the denominator. First, let's recall the definition of the double factorial: - For an even number \(2n\), \((2n)!! = 2 \cdot 4 \cdot 6 \cdot \ldots \cdot (2n)\). - For an odd number \(2n+1\), \((2n+1)!! = 1 \cdot 3 \cdot 5 \cdot \ldots \cdot (2n+1)\). We can rewrite the double factorials in terms of regular factorials: - \((2n)!! = 2^n \cdot n!\) - \((2n+1)!! = \frac{(2n+1)!}{2^n \cdot n!}\) Now, let's substitute these expressions into the original ratio: \[ \frac{(2n)!!}{(2n+1)!!} = \frac{2^n \cdot n!}{\frac{(2n+1)!}{2^n \cdot n!}} = \frac{2^n \cdot n! \cdot 2^n \cdot n!}{(2n+1)!} = \frac{2^{2n} \cdot (n!)^2}{(2n+1)!} \] Next, we use Stirling's approximation for factorials, which states that \(n! \approx \sqrt{2 \pi n} \left(\frac{n}{e}\right)^n\). Applying this to \(n!\) and \((2n+1)!\): - \(n! \approx \sqrt{2 \pi n} \left(\frac{n}{e}\right)^n\) - \((2n+1)! \approx \sqrt{4 \pi n + 2 \pi} \left(\frac{2n+1}{e}\right)^{2n+1}\) For large \(n\), \((2n+1)!\) can be approximated as: \[ (2n+1)! \approx \sqrt{4 \pi n} \left(\frac{2n}{e}\right)^{2n} \cdot (2n+1) \] Now, substituting these approximations into our ratio: \[ \frac{2^{2n} \cdot (n!)^2}{(2n+1)!} \approx \frac{2^{2n} \cdot \left(\sqrt{2 \pi n} \left(\frac{n}{e}\right)^n\right)^2}{\sqrt{4 \pi n} \left(\frac{2n}{e}\right)^{2n} \cdot (2n+1)} \] Simplifying the expression: \[ = \frac{2^{2n} \cdot 2 \pi n \cdot \left(\frac{n}{e}\right)^{2n}}{\sqrt{4 \pi n} \cdot \left(\frac{2n}{e}\right)^{2n} \cdot (2n+1)} = \frac{2^{2n} \cdot 2 \pi n \cdot \left(\frac{n^2}{e^2}\right)^{n}}{\sqrt{4 \pi n} \cdot \left(\frac{4n^2}{e^2}\right)^{n} \cdot (2n+1)} \] \[ = \frac{2^{2n} \cdot 2 \pi n}{\sqrt{4 \pi n} \cdot 2^{2n} \cdot (2n+1)} = \frac{2 \pi n}{\sqrt{4 \pi n} \cdot (2n+1)} \] \[ = \frac{2 \pi n}{2 \sqrt{\pi n} \cdot (2n+1)} = \frac{\pi n}{\sqrt{\pi n} \cdot (2n+1)} = \frac{\sqrt{\pi n}}{2n+1} \] As \(n\) approaches infinity, \(\frac{\sqrt{\pi n}}{2n+1} \) approaches 0 because the denominator \(2n+1\) grows much faster than the numerator \(\sqrt{\pi n}\). Therefore, the limit of \(\frac{(2n)!!}{(2n+1)!!}\) as \(n\) approaches infinity is: \[ \boxed{0} \] |
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| gpto1 | To find the limit of the ratio \(\frac{(2n)!!}{(2n+1)!!}\) as \(n\) approaches infinity, we'll start by understanding the double factorial notation and expressing the double factorials in terms of more familiar functions. **Definition of Double Factorial:** - For even positive integers: \((2n)!! = 2 \times 4 \times 6 \times \dots \times 2n\) - For odd positive integers: \((2n+1)!! = 1 \times 3 \times 5 \times \dots \times (2n+1)\) **Expressing Double Factorials:** 1. **Even Double Factorial:** \[(2n)!! = 2^n \times n!\] This is because each term in the product \((2 \times 4 \times \dots \times 2n)\) has a factor of 2, and there are \(n\) such factors, leading to \(2^n\). The remaining part is just \(n!\). 2. **Odd Double Factorial:** To express \((2n+1)!!\) in terms of factorial, we use the relationship: \[(2n+1)!! = \frac{(2n+1)!}{2^n \times n!}\] This comes from separating the factorial \((2n+1)!\) into its even and odd components. **Setting Up the Ratio:** Now, our ratio becomes: \[ L = \frac{(2n)!!}{(2n+1)!!} = \frac{2^n \times n!}{\dfrac{(2n+1)!}{2^n \times n!}} = \frac{(2^n \times n!)^2}{(2n+1)!} \] **Using Stirling's Approximation:** For large \(n\), Stirling's approximation provides a good estimate for factorials: \[ \ln n! \approx n\ln n - n + \frac{1}{2}\ln(2\pi n) \] We can now take the natural logarithm of \(L\) to simplify the expression: \[ \ln L = 2n\ln 2 + 2\ln n! - \ln(2n+1)! \] Applying Stirling's approximation: \[ \ln n! \approx n\ln n - n \] \[ \ln(2n+1)! \approx (2n+1)\ln(2n+1) - (2n+1) \] Ignoring the constants (since they become negligible as \(n \to \infty\)): \[ \ln L \approx 2n\ln 2 + 2(n\ln n - n) - [(2n+1)\ln(2n+1) - (2n+1)] \] Simplify the expression: \[ \ln L \approx 2n\ln 2 + 2n\ln n - 2n - [2n\ln 2n + \ln 2n + 1] + 2n + 1 \] We can approximate \(\ln(2n+1) \approx \ln(2n)\) and \((2n+1)\ln(2n+1) \approx 2n\ln(2n)\) for large \(n\). Simplify further: \[ \ln L \approx 2n\ln 2 + 2n\ln n - 2n - [2n\ln 2 + 2n\ln n + \ln 2n + 1] + 2n + 1 \] Most terms cancel out: \[ \ln L \approx -\ln(2n) - 1 \] As \(n\) approaches infinity, \(-\ln(2n)\) dominates, so we get: \[ \ln L \approx -\ln(2n) \] \[ L \approx e^{-\ln(2n)} = \frac{1}{2n} \] Therefore, as \(n\) approaches infinity: \[ L \to \frac{1}{\infty} = 0 \] **Conclusion:** The limit of the ratio \(\frac{(2n)!!}{(2n+1)!!}\) as \(n\) approaches infinity is **0**. **Answer:** 0 |
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