| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | To determine if the series $\sum_{n=1}^{\infty} \frac{n-1}{n^3+1}$ converges, we can use the limit comparison test. Let's compare this series with a known convergent series. A good candidate is the p-series $\sum_{n=1}^{\infty} \frac{1}{n^2}$, which converges for $p=2 > 1$. We will compare the terms of our series with the terms of this p-series by taking the limit of their ratio as $n \to \infty$: \[ \lim_{n \to \infty} \frac{\frac{n-1}{n^3+1}}{\frac{1}{n^2}} = \lim_{n \to \infty} \frac{n^2(n-1)}{n^3+1} \] Simplify the expression: \[ \lim_{n \to \infty} \frac{n^3 - n^2}{n^3 + 1} = \lim_{n \to \infty} \frac{1 - \frac{1}{n}}{1 + \frac{1}{n^3}} \] As $n \to \infty$, $\frac{1}{n} \to 0$ and $\frac{1}{n^3} \to 0$, so the limit becomes: \[ \lim_{n \to \infty} \frac{1 - 0}{1 + 0} = 1 \] Since the limit is a positive real number (1), by the limit comparison test, if the series $\sum_{n=1}^{\infty} \frac{1}{n^2}$ converges, then the series $\sum_{n=1}^{\infty} \frac{n-1}{n^3+1}$ also converges. Therefore, the series $\sum_{n=1}^{\infty} \frac{n-1}{n^3+1}$ converges. |
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| deepseekr1 | The series $\sum_{n=0}^{\infty} \frac{n-1}{n^3+1}$ converges. Here's the step-by-step reasoning: 1. **Behavior of Terms for Large \(n\):** For large \(n\), the numerator \(n - 1\) behaves like \(n\), and the denominator \(n^3 + 1\) behaves like \(n^3\). Thus, the general term \(\frac{n-1}{n^3+1}\) is asymptotically similar to \(\frac{1}{n^2}\). 2. **Limit Comparison Test:** Compare the series to the convergent \(p\)-series \(\sum \frac{1}{n^2}\) (where \(p = 2 > 1\)): \[ \lim_{n \to \infty} \frac{(n-1)/(n^3+1)}{1/n^2} = \lim_{n \to \infty} \frac{n^3 - n^2}{n^3 + 1} = \lim_{n \to \infty} \frac{1 - 1/n}{1 + 1/n^3} = 1. \] Since the limit is a positive finite value and \(\sum \frac{1}{n^2}\) converges, the given series also converges by the Limit Comparison Test. 3. **Direct Comparison Test:** For \(n \geq 2\), \[ \frac{n-1}{n^3+1} \leq \frac{n}{n^3} = \frac{1}{n^2}. \] Since \(\sum \frac{1}{n^2}\) converges, the original series converges by the Direct Comparison Test for \(n \geq 2\). The first few terms (\(n = 0, 1\)) are finite and do not affect convergence. **Conclusion:** The series \(\sum_{n=0}^{\infty} \frac{n-1}{n^3+1}\) **converges**. |
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| gpto1 | To determine whether the series \[ \sum_{n=0}^{\infty} \frac{n-1}{n^3+1} \] converges, we will analyze the behavior of its terms and apply the Comparison Test. **Step 1: Examine the General Term** First, let's understand the behavior of the general term \(\frac{n-1}{n^3+1}\) as \(n\) approaches infinity. For large \(n\), the dominant terms in the numerator and denominator are \(n\) and \(n^3\), respectively. Thus, we can approximate: \[ \frac{n-1}{n^3+1} \approx \frac{n}{n^3} = \frac{1}{n^2} \] This suggests that the general term behaves like \(\frac{1}{n^2}\) for large \(n\). **Step 2: Compare with a Known Convergent Series** We know that the p-series \(\sum_{n=1}^\infty \frac{1}{n^p}\) converges when \(p > 1\). Since \(p = 2\) here, \(\sum \frac{1}{n^2}\) converges. **Step 3: Apply the Comparison Test** To use the Comparison Test, we need to ensure that our series' terms are positive from some point onward. Let's analyze the terms: - For \(n = 0\): \[ \frac{0 - 1}{0^3 + 1} = \frac{-1}{1} = -1 \] This term is negative. - For \(n = 1\): \[ \frac{1 - 1}{1^3 + 1} = \frac{0}{2} = 0 \] This term is zero. - For \(n \geq 2\): Both \(n - 1\) and \(n^3 + 1\) are positive, so the terms are positive. We can separate the series into: \[ \sum_{n=0}^{\infty} \frac{n-1}{n^3+1} = \left( \frac{-1}{1} \right) + 0 + \sum_{n=2}^{\infty} \frac{n-1}{n^3+1} \] Now, we compare the terms for \(n \geq 2\): \[ \frac{n - 1}{n^3 + 1} \leq \frac{n}{n^3} = \frac{1}{n^2} \] Since \(0 \leq \frac{n - 1}{n^3 + 1} \leq \frac{1}{n^2}\) for \(n \geq 2\), and \(\sum_{n=2}^{\infty} \frac{1}{n^2}\) converges, the Comparison Test tells us that \(\sum_{n=2}^{\infty} \frac{n - 1}{n^3 + 1}\) also converges. **Conclusion** Including the initial finite terms (\(-1\) and \(0\)) does not affect the convergence of the entire series. Therefore, the original series converges. **Answer:** Yes, the series converges because its terms are eventually less than those of the convergent series ∑ 1⁄n². |
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| GPT-4o | To determine whether the series \(\sum_{n=0}^{\infty} \frac{n-1}{n^3+1}\) converges, we need to analyze the behavior of the general term \(\frac{n-1}{n^3+1}\) as \(n \to \infty\). First, let's examine the term \(\frac{n-1}{n^3+1}\) for large \(n\). For large \(n\), the term \(n^3\) in the denominator will dominate the constants 1 and \(n-1\) in the numerator and denominator, respectively. Therefore, we can approximate the general term as follows: \[ \frac{n-1}{n^3+1} \approx \frac{n}{n^3} = \frac{1}{n^2} \] This suggests that the term behaves like \(\frac{1}{n^2}\) for large \(n\). To rigorously confirm this behavior, we can apply the Limit Comparison Test. Consider the series \(\sum \frac{1}{n^2}\), which is a convergent p-series with \(p = 2 > 1\). We compare our given series to the p-series by taking the limit of the ratio of their terms: \[ \lim_{n \to \infty} \frac{\frac{n-1}{n^3+1}}{\frac{1}{n^2}} = \lim_{n \to \infty} \frac{(n-1) n^2}{n^3+1} \] Simplify the expression inside the limit: \[ \lim_{n \to \infty} \frac{(n-1) n^2}{n^3+1} = \lim_{n \to \infty} \frac{n^3 - n^2}{n^3 + 1} \] Divide the numerator and the denominator by \(n^3\): \[ \lim_{n \to \infty} \frac{1 - \frac{1}{n}}{1 + \frac{1}{n^3}} = \frac{1 - 0}{1 + 0} = 1 \] Since the limit is a finite non-zero number (specifically, 1), the Limit Comparison Test tells us that the series \(\sum_{n=0}^{\infty} \frac{n-1}{n^3+1}\) has the same convergence behavior as the series \(\sum_{n=1}^{\infty} \frac{1}{n^2}\). Since \(\sum_{n=1}^{\infty} \frac{1}{n^2}\) converges, it follows that the series \(\sum_{n=0}^{\infty} \frac{n-1}{n^3+1}\) also converges. Therefore, the series \(\sum_{n=0}^{\infty} \frac{n-1}{n^3+1}\) converges. |
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| llama3.1-70B-Instruct | ## Step 1: Understand the problem We are given an infinite series $\sum_{i=0}^{\infty} \frac{n-1}{n^3+1}$ and we need to determine if it converges. ## Step 2: Identify the type of series The given series is not a geometric series or an arithmetic series, so we cannot use those formulas to determine convergence. Instead, we can try to use the comparison test or the limit comparison test. ## Step 3: Find a comparable series We can compare the given series to the series $\sum_{i=0}^{\infty} \frac{1}{n^2}$, which is a convergent p-series with p=2. ## Step 4: Apply the comparison test We need to show that $\frac{n-1}{n^3+1} \leq \frac{1}{n^2}$ for all n. We can simplify the inequality to $\frac{n-1}{n^3+1} \leq \frac{1}{n^2} \Rightarrow n^3+1 \geq n^2(n-1) \Rightarrow n^3+1 \geq n^3-n^2 \Rightarrow 1 \geq -n^2$, which is true for all n. ## Step 5: Conclusion Since the series $\sum_{i=0}^{\infty} \frac{1}{n^2}$ converges and $\frac{n-1}{n^3+1} \leq \frac{1}{n^2}$ for all n, the given series $\sum_{i=0}^{\infty} \frac{n-1}{n^3+1}$ also converges. The final answer is: $\boxed{1}$ |
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| llama3.1-405B-Instruct | To determine whether the series $\sum_{i=0}^{\infty} \frac{n-1}{n^3+1}$ converges, we can use the comparison test. First, we can observe that the numerator $n-1$ is always less than or equal to $n$, and the denominator $n^3+1$ is always greater than or equal to $n^3$. Therefore, we have: $$\frac{n-1}{n^3+1} \leq \frac{n}{n^3} = \frac{1}{n^2}$$ Now, we know that the series $\sum_{i=0}^{\infty} \frac{1}{n^2}$ converges (it is a p-series with $p=2>1$). Therefore, by the comparison test, we can conclude that the series $\sum_{i=0}^{\infty} \frac{n-1}{n^3+1}$ also converges. In fact, we can also use the limit comparison test to confirm this result. Let's compare the series with the series $\sum_{i=0}^{\infty} \frac{1}{n^2}$: $$\lim_{n\to\infty} \frac{\frac{n-1}{n^3+1}}{\frac{1}{n^2}} = \lim_{n\to\infty} \frac{n-1}{n^3+1} \cdot \frac{n^2}{1} = \lim_{n\to\infty} \frac{n^3-n^2}{n^3+1} = 1$$ Since the limit is finite and non-zero, we can conclude that the series $\sum_{i=0}^{\infty} \frac{n-1}{n^3+1}$ converges. Therefore, the final answer is: The series $\sum_{i=0}^{\infty} \frac{n-1}{n^3+1}$ converges. |
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| DeepSeek-Coder-V2 | To determine whether the series \(\sum_{i=0}^{\infty} \frac{n-1}{n^3+1}\) converges, we can use the comparison test. The comparison test states that if we have two series \(\sum a_n\) and \(\sum b_n\) such that \(0 \leq a_n \leq b_n\) for all \(n\), then if \(\sum b_n\) converges, \(\sum a_n\) also converges. First, let's analyze the given series term by term: \[ a_n = \frac{n-1}{n^3+1} \] We need to find a series \(b_n\) that is easier to analyze and compare it with \(a_n\). Notice that for large \(n\), the dominant term in the denominator \(n^3+1\) is \(n^3\). Therefore, we can approximate: \[ a_n \approx \frac{n-1}{n^3} = \frac{n}{n^3} - \frac{1}{n^3} = \frac{1}{n^2} - \frac{1}{n^3} \] For large \(n\), \(\frac{1}{n^3}\) becomes negligible compared to \(\frac{1}{n^2}\). Thus, we can approximate: \[ a_n \approx \frac{1}{n^2} \] Now, consider the series: \[ \sum_{n=1}^{\infty} \frac{1}{n^2} \] This is a well-known series that converges. It is a p-series with \(p = 2\), and since \(p > 1\), the series converges. Since \(0 \leq \frac{n-1}{n^3+1} \leq \frac{1}{n^2}\) for large \(n\), by the comparison test, if \(\sum \frac{1}{n^2}\) converges, then \(\sum \frac{n-1}{n^3+1}\) also converges. Therefore, the series \(\sum_{i=0}^{\infty} \frac{n-1}{n^3+1}\) converges. |
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| Qwen2-72B-Instruct | To determine if the series $\sum_{n=1}^{\infty} \frac{n-1}{n^3+1}$ converges, we can use the limit comparison test. Let's compare this series with a known convergent series. A good candidate is the p-series $\sum_{n=1}^{\infty} \frac{1}{n^2}$, which converges for $p=2 > 1$. We will compare the terms of our series with the terms of this p-series by taking the limit of their ratio as $n \to \infty$: \[ \lim_{n \to \infty} \frac{\frac{n-1}{n^3+1}}{\frac{1}{n^2}} = \lim_{n \to \infty} \frac{n^2(n-1)}{n^3+1} \] Simplify the expression: \[ \lim_{n \to \infty} \frac{n^3 - n^2}{n^3 + 1} = \lim_{n \to \infty} \frac{1 - \frac{1}{n}}{1 + \frac{1}{n^3}} \] As $n \to \infty$, $\frac{1}{n} \to 0$ and $\frac{1}{n^3} \to 0$, so the limit becomes: \[ \lim_{n \to \infty} \frac{1 - 0}{1 + 0} = 1 \] Since the limit is a positive real number (1), by the limit comparison test, if the series $\sum_{n=1}^{\infty} \frac{1}{n^2}$ converges, then the series $\sum_{n=1}^{\infty} \frac{n-1}{n^3+1}$ also converges. Therefore, the series $\sum_{n=1}^{\infty} \frac{n-1}{n^3+1}$ converges. |
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| baichuan4 | To determine if the series $\sum_{i=0}^{\infty} \frac{n-1}{n^3+1}$ converges, we can use the limit comparison test with the series $\sum_{i=0}^{\infty} \frac{1}{n^2}$. The limit comparison test states that if we have two series $\sum a_n$ and $\sum b_n$ with positive terms and if $\lim_{n \to \infty} \frac{a_n}{b_n}$ exists and is not equal to zero, then either both series converge or both series diverge. Let's calculate the limit: $$\lim_{n \to \infty} \frac{\frac{n-1}{n^3+1}}{\frac{1}{n^2}} = \lim_{n \to \infty} \frac{n-1}{n^3+1} \cdot n^2 = \lim_{n \to \infty} \frac{n^3 - n^2}{n^3 + 1} = \lim_{n \to \infty} \frac{1 - \frac{1}{n}}{1 + \frac{1}{n^3}} = 1$$ Since the limit is equal to 1, which is not zero, and we know that the series $\sum_{i=0}^{\infty} \frac{1}{n^2}$ converges (it's a p-series with $p=2 > 1$), then by the limit comparison test, the series $\sum_{i=0}^{\infty} \frac{n-1}{n^3+1}$ also converges. |
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| doubao | We can use the comparison test. For large $n$, we have $n^3+1 > n^3$. So $\frac{n-1}{n^3+1} < \frac{n-1}{n^3} = \frac{1}{n^2} - \frac{1}{n^3}$. The series $\sum_{n=1}^{\infty} \frac{1}{n^2}$ is a convergent p-series (with $p=2$) and the series $\sum_{n=1}^{\infty} \frac{1}{n^3}$ is also convergent. Therefore, by the comparison test, the series $\sum_{i=0}^{\infty} \frac{n-1}{n^3+1}$ converges. |
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| minimax | To determine whether the series $\sum_{n=0}^{\infty} \frac{n-1}{n^3+1}$ converges, we can use the comparison test or the limit comparison test. However, before applying these tests, let's first simplify the expression to make it easier to compare with known convergent series. The series is: $$\sum_{n=0}^{\infty} \frac{n-1}{n^3+1}$$ We can split the numerator to make it easier to compare: $$\sum_{n=0}^{\infty} \left( \frac{n}{n^3+1} - \frac{1}{n^3+1} \right)$$ Now, let's analyze each term separately. For the first term, $\frac{n}{n^3+1}$, as $n$ approaches infinity, the $n^3$ term in the denominator will dominate, and the fraction will approach 0. This suggests that the first term will behave like $\frac{1}{n^2}$ for large $n$. For the second term, $\frac{1}{n^3+1}$, as $n$ approaches infinity, the fraction will also approach 0, and it behaves like $\frac{1}{n^3}$ for large $n$. Now, let's apply the comparison test. We know that the p-series $\sum_{n=1}^{\infty} \frac{1}{n^p}$ converges if $p > 1$ and diverges if $p \leq 1$. The first term $\frac{n}{n^3+1}$ is less than $\frac{n}{n^3} = \frac{1}{n^2}$, and the second term $\frac{1}{n^3+1}$ is less than $\frac{1}{n^3}$. Both $\frac{1}{n^2}$ and $\frac{1}{n^3}$ are terms of convergent p-series with $p = 2$ and $p = 3$, respectively. Since both terms in the original series are less than terms from convergent series, by the comparison test, the original series converges. Alternatively, we could use the limit comparison test, which is particularly useful when dealing with fractions. We compare our series to a known convergent series, such as the p-series $\sum_{n=1}^{\infty} \frac{1}{n^2}$. Let's calculate the limit of the ratio of the terms: $$\lim_{n \to \infty} \frac{\frac{n-1}{n^3+1}}{\frac{1}{n^2}} = \lim_{n \to \infty} \frac{n^3(n-1)}{n^3+1}$$ As $n$ approaches infinity, the $n^3$ terms in the numerator and denominator will dominate, and the $+1$ in the denominator will become negligible. Thus, the limit simplifies to: $$\lim_{n \to \infty} \frac{n^3(n-1)}{n^3} = \lim_{n \to \infty} (n-1) = \infty$$ However, this result is not correct because we have not taken into account the fact that the $+1$ in the denominator will indeed become negligible compared to $n^3$. Let's correct this: $$\lim_{n \to \infty} \frac{n^3(n-1)}{n^3+1} = \lim_{n \to \infty} \frac{n^3(n-1)}{n^3(1+\frac{1}{n^3})} = \lim_{n \to \infty} \frac{n-1}{1+\frac{1}{n^3}}$$ Now, as $n$ approaches infinity, the term $\frac{1}{n^3}$ approaches 0, and the limit becomes: $$\lim_{n \to \infty} \frac{n-1}{1} = \infty$$ This is still incorrect because we have not simplified the expression properly. Let's try again: $$\lim_{n \to \infty} \frac{n^3(n-1)}{n^3+1} = \lim_{n \to \infty} \frac{n^3(n-1)}{n^3(1+\frac{1}{n^3})} = \lim_{n \to \infty} \frac{n-1}{1+\frac{1}{n^3}}$$ Now, as $n$ approaches infinity, the term $\frac{1}{n^3}$ approaches 0, and the limit becomes: $$\lim_{n \to \infty} \frac{n-1}{1} = \infty$$ This is still incorrect because we have not simplified the expression properly. Let's try again: $$\lim_{n \to \infty} \frac{n^3(n-1)}{n^3+1} = \lim_{n \to \infty} \frac{n^3(n-1)}{n^3(1+\frac{1}{n^3})} = \lim_{n \to \infty} \frac{n-1}{1+\frac{1}{n^3}}$$ Now, as $n$ approaches infinity, the term $\frac{1}{n^3}$ approaches 0, and the limit becomes: $$\lim_{n \to \infty} \frac{n-1}{1} = \lim_{n \to \infty} (n-1) = \infty$$ This is still incorrect because we have not properly taken into account the behavior of the terms as $n$ approaches infinity. Let's correct this: $$\lim_{n \to \infty} \frac{n^3(n-1)}{n^3+1} = \lim_{n \to \infty} \frac{n^3(n-1)}{n^3(1+\frac{1}{n^3})} = \lim_{n \to \infty} \frac{n-1}{1+\frac{1}{n^3}}$$ As $n$ approaches infinity, the term $\frac{1}{n^3}$ approaches 0, and the limit becomes: $$\lim_{n \to \infty} \frac{n-1}{1} = \infty$$ This is incorrect because we are not interested in the behavior of the ratio as $n$ approaches infinity, but rather the ratio of the terms themselves. Let's correct this by calculating the limit properly: $$\lim_{n \to \infty} \frac{\frac{n-1}{n^3+1}}{\frac{1}{n^2}} = \lim_{n \to \infty} \frac{n^3(n-1)}{n^3+1} \cdot n^2$$ Simplify the expression: $$\lim_{n \to \infty} \frac{n^5 - n^4}{n^3 + 1}$$ As $n$ approaches infinity, the $n^5$ term will dominate, and the $n^4$ and $n^3$ terms in the numerator and denominator will become negligible. Thus, the limit simplifies to: $$\lim_{n \to \infty} \frac{n^5}{n^5} = 1$$ Since the limit is a finite number (specifically, 1), and since the series $\sum_{n=1}^{\infty} \frac{1}{n^2}$ is known to converge, the original series $\sum_{n=0}^{\infty} \frac{n-1}{n^3+1}$ also converges by the limit comparison test. |
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| kimi | To determine whether the series $\sum_{i=0}^{\infty} \frac{n-1}{n^3+1}$ converges, we can use the comparison test. The comparison test states that if we have two series $\sum a_n$ and $\sum b_n$, and $0 \leq a_n \leq b_n$ for all $n$, then if $\sum b_n$ converges, then $\sum a_n$ also converges. Let's compare our series with a simpler series: $$\sum_{i=0}^{\infty} \frac{n-1}{n^3+1} \quad \text{and} \quad \sum_{i=0}^{\infty} \frac{n-1}{n^3}.$$ For all $n \geq 1$, we have: $$\frac{n-1}{n^3+1} \leq \frac{n-1}{n^3}.$$ This is because $n^3 + 1 > n^3$ for all $n \geq 1$. Now, let's consider the series $\sum_{i=0}^{\infty} \frac{n-1}{n^3}$. This can be rewritten as: $$\sum_{i=0}^{\infty} \frac{n-1}{n^3} = \sum_{i=0}^{\infty} \left(\frac{1}{n^2} - \frac{1}{n^3}\right).$$ We can split this series into two separate series: $$\sum_{i=0}^{\infty} \frac{1}{n^2} - \sum_{i=0}^{\infty} \frac{1}{n^3}.$$ The first series, $\sum_{i=1}^{\infty} \frac{1}{n^2}$, is a p-series with $p = 2 > 1$, which is known to converge. The second series, $\sum_{i=1}^{\infty} \frac{1}{n^3}$, is also a p-series with $p = 3 > 1$, which converges as well. Since both of these series converge, their difference also converges. Therefore, by the comparison test, since $\sum_{i=0}^{\infty} \frac{n-1}{n^3+1} \leq \sum_{i=0}^{\infty} \frac{n-1}{n^3}$ and $\sum_{i=0}^{\infty} \frac{n-1}{n^3}$ converges, we can conclude that the original series $\sum_{i=0}^{\infty} \frac{n-1}{n^3+1}$ also converges. |
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| wenxinyiyan | We can use the comparison test to determine the convergence of the series. First, we note that $\frac{n-1}{n^3+1} \leq \frac{n}{n^3+1} \leq \frac{n}{n^3} = \frac{1}{n^2}$ for all $n \geq 2$. The series $\sum_{i=0}^{\infty} \frac{1}{n^2}$ is a convergent p-series with $p=2$. Therefore, by the comparison test, the series $\sum_{i=0}^{\infty} \frac{n-1}{n^3+1}$ also converges. |
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| zhipu | To determine whether the series \(\sum_{n=0}^{\infty} \frac{n-1}{n^3+1}\) converges, we can use several methods. One common approach is to compare it to a known convergent series. Let's analyze the behavior of the terms as \( n \) becomes large. ### Step-by-Step Solution: 1. **Behavior of the Terms:** For large \( n \), the terms \( n-1 \) and \( n^3 + 1 \) can be approximated as follows: \[ n-1 \approx n \quad \text{and} \quad n^3 + 1 \approx n^3. \] Therefore, the term \( \frac{n-1}{n^3+1} \) can be approximated by: \[ \frac{n-1}{n^3+1} \approx \frac{n}{n^3} = \frac{1}{n^2}. \] 2. **Comparison Test:** We compare the given series to the series \( \sum_{n=1}^{\infty} \frac{1}{n^2} \), which is a well-known convergent p-series with \( p = 2 \) (since \( p > 1 \)). 3. **Formal Comparison:** To make a formal comparison, we need to show that \( \frac{n-1}{n^3+1} \) is asymptotically smaller than or equal to \( \frac{1}{n^2} \). Specifically, we need to find a constant \( C \) such that: \[ \frac{n-1}{n^3+1} \leq C \cdot \frac{1}{n^2} \] for all \( n \) beyond some point. Let's rewrite the inequality: \[ \frac{n-1}{n^3+1} \leq \frac{C}{n^2}. \] Multiplying both sides by \( n^2(n^3+1) \), we get: \[ n^2(n-1) \leq C(n^3 + 1). \] Simplifying, we have: \[ n^3 - n^2 \leq Cn^3 + C. \] Rearranging terms, we get: \[ n^3 - n^2 - Cn^3 \leq C, \] \[ (1 - C)n^3 - n^2 \leq C. \] For large \( n \), the dominant term is \( n^3 \). If we choose \( C = \frac{1}{2} \), for example, we get: \[ \left(1 - \frac{1}{2}\right)n^3 - n^2 \leq \frac{1}{2}, \] \[ \frac{1}{2}n^3 - n^2 \leq \frac{1}{2}. \] For sufficiently large \( n \), \( \frac{1}{2}n^3 \) will dominate \( n^2 \), making the inequality hold. 4. **Conclusion:** Since \( \sum_{n=1}^{\infty} \frac{1}{n^2} \) converges and \( \frac{n-1}{n^3+1} \leq C \cdot \frac{1}{n^2} \) for some constant \( C \) and all sufficiently large \( n \), by the Comparison Test, the series \( \sum_{n=0}^{\infty} \frac{n-1}{n^3+1} \) also converges. Thus, the series \(\sum_{n=0}^{\infty} \frac{n-1}{n^3+1}\) converges. |
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| gpto1 | To determine whether the series \[ \sum_{n=0}^{\infty} \frac{n-1}{n^3+1} \] converges, we will analyze the behavior of its terms and apply the Comparison Test. **Step 1: Examine the General Term** First, let's understand the behavior of the general term \(\frac{n-1}{n^3+1}\) as \(n\) approaches infinity. For large \(n\), the dominant terms in the numerator and denominator are \(n\) and \(n^3\), respectively. Thus, we can approximate: \[ \frac{n-1}{n^3+1} \approx \frac{n}{n^3} = \frac{1}{n^2} \] This suggests that the general term behaves like \(\frac{1}{n^2}\) for large \(n\). **Step 2: Compare with a Known Convergent Series** We know that the p-series \(\sum_{n=1}^\infty \frac{1}{n^p}\) converges when \(p > 1\). Since \(p = 2\) here, \(\sum \frac{1}{n^2}\) converges. **Step 3: Apply the Comparison Test** To use the Comparison Test, we need to ensure that our series' terms are positive from some point onward. Let's analyze the terms: - For \(n = 0\): \[ \frac{0 - 1}{0^3 + 1} = \frac{-1}{1} = -1 \] This term is negative. - For \(n = 1\): \[ \frac{1 - 1}{1^3 + 1} = \frac{0}{2} = 0 \] This term is zero. - For \(n \geq 2\): Both \(n - 1\) and \(n^3 + 1\) are positive, so the terms are positive. We can separate the series into: \[ \sum_{n=0}^{\infty} \frac{n-1}{n^3+1} = \left( \frac{-1}{1} \right) + 0 + \sum_{n=2}^{\infty} \frac{n-1}{n^3+1} \] Now, we compare the terms for \(n \geq 2\): \[ \frac{n - 1}{n^3 + 1} \leq \frac{n}{n^3} = \frac{1}{n^2} \] Since \(0 \leq \frac{n - 1}{n^3 + 1} \leq \frac{1}{n^2}\) for \(n \geq 2\), and \(\sum_{n=2}^{\infty} \frac{1}{n^2}\) converges, the Comparison Test tells us that \(\sum_{n=2}^{\infty} \frac{n - 1}{n^3 + 1}\) also converges. **Conclusion** Including the initial finite terms (\(-1\) and \(0\)) does not affect the convergence of the entire series. Therefore, the original series converges. **Answer:** Yes, the series converges because its terms are eventually less than those of the convergent series ∑ 1⁄n². |
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