| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | To solve the problem, we can start by simplifying the given series. The denominator \( n^2 + 4n + 3 \) can be factored as \( (n+1)(n+3) \). Therefore, the series can be rewritten as: \[ \sum_{n=1}^{\infty} \frac{2}{(n+1)(n+3)} \] Next, we can use partial fraction decomposition to break down the term \( \frac{2}{(n+1)(n+3)} \) into simpler fractions. We can write: \[ \frac{2}{(n+1)(n+3)} = \frac{A}{n+1} + \frac{B}{n+3} \] Multiplying both sides by \( (n+1)(n+3) \), we get: \[ 2 = A(n+3) + B(n+1) \] To find the values of \( A \) and \( B \), we can solve the system of equations obtained by substituting \( n = -1 \) and \( n = -3 \) into the above equation. Doing so, we get: \[ 2 = 2A \Rightarrow A = 1 \] \[ 2 = -2B \Rightarrow B = -1 \] Therefore, we can rewrite the series as: \[ \sum_{n=1}^{\infty} \left(\frac{1}{n+1} - \frac{1}{n+3}\right) \] This series can be simplified further by grouping terms: \[ \sum_{n=1}^{\infty} \left(\frac{1}{n+1} - \frac{1}{n+3}\right) = \left(\frac{1}{2} - \frac{1}{4}\right) + \left(\frac{1}{3} - \frac{1}{5}\right) + \left(\frac{1}{4} - \frac{1}{6}\right) + \left(\frac{1}{5} - \frac{1}{7}\right) + \dots \] Notice that most of the terms cancel out, leaving only the first two terms: \[ \left(\frac{1}{2} - \frac{1}{4}\right) + \left(\frac{1}{3} - \frac{1}{5}\right) + \dots = \frac{1}{2} + \frac{1}{3} = \frac{5}{6} \] Therefore, the sum of the series is: \[ \sum_{n=1}^{\infty} \frac{2}{n^2 + 4n + 3} = \boxed{\frac{5}{6}} \] |
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| deepseekr1 | To find the sum of the series \(\sum_{n=1}^{\infty} \frac{2}{n^2 + 4n + 3}\), we start by factoring the denominator: \[ n^2 + 4n + 3 = (n + 1)(n + 3) \] Next, we use partial fraction decomposition to express the general term: \[ \frac{2}{(n + 1)(n + 3)} = \frac{A}{n + 1} + \frac{B}{n + 3} \] Solving for \(A\) and \(B\), we get: \[ 2 = A(n + 3) + B(n + 1) \] By equating coefficients, we find \(A = 1\) and \(B = -1\). Thus, the decomposition is: \[ \frac{2}{(n + 1)(n + 3)} = \frac{1}{n + 1} - \frac{1}{n + 3} \] The series then becomes a telescoping series: \[ \sum_{n=1}^{\infty} \left( \frac{1}{n + 1} - \frac{1}{n + 3} \right) \] Writing out the first few terms, we observe the cancellation pattern: \[ \left( \frac{1}{2} - \frac{1}{4} \right) + \left( \frac{1}{3} - \frac{1}{5} \right) + \left( \frac{1}{4} - \frac{1}{6} \right) + \left( \frac{1}{5} - \frac{1}{7} \right) + \cdots \] Most terms cancel out, leaving: \[ \frac{1}{2} + \frac{1}{3} - \left( \frac{1}{N+2} + \frac{1}{N+3} \right) \] Taking the limit as \(N \to \infty\), the terms \(\frac{1}{N+2}\) and \(\frac{1}{N+3}\) approach zero, resulting in: \[ \sum_{n=1}^{\infty} \frac{2}{n^2 + 4n + 3} = \frac{1}{2} + \frac{1}{3} = \frac{3}{6} + \frac{2}{6} = \frac{5}{6} \] Thus, the sum of the series is \(\boxed{\dfrac{5}{6}}\). |
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| gpto1 | To find the sum of the infinite series: \[ \sum_{n=1}^{\infty} \frac{2}{n^2 + 4n + 3} \] we can start by simplifying the general term of the series and then using partial fractions to decompose it. Here are the detailed steps: **1. Factor the Denominator** First, factor the quadratic expression in the denominator: \[ n^2 + 4n + 3 = (n + 1)(n + 3) \] **2. Rewrite the General Term** Rewrite the general term of the series using the factored form: \[ \frac{2}{n^2 + 4n + 3} = \frac{2}{(n + 1)(n + 3)} \] **3. Decompose into Partial Fractions** We can express the fraction as the sum of two simpler fractions: \[ \frac{2}{(n + 1)(n + 3)} = \frac{A}{n + 1} + \frac{B}{n + 3} \] To find constants \( A \) and \( B \), multiply both sides by \( (n + 1)(n + 3) \): \[ 2 = A(n + 3) + B(n + 1) \] Simplify and solve for \( A \) and \( B \) by choosing convenient values for \( n \): - Let \( n = -3 \): \[ 2 = A(-3 + 3) + B(-3 + 1) \\ 2 = A(0) + B(-2) \\ 2 = -2B \\ B = -1 \] - Let \( n = -1 \): \[ 2 = A(-1 + 3) + B(-1 + 1) \\ 2 = A(2) + B(0) \\ 2 = 2A \\ A = 1 \] **4. Rewrite the Series** Now, the general term becomes: \[ \frac{2}{(n + 1)(n + 3)} = \frac{1}{n + 1} - \frac{1}{n + 3} \] Therefore, the series is: \[ \sum_{n=1}^{\infty} \left( \frac{1}{n + 1} - \frac{1}{n + 3} \right) \] **5. Compute the Partial Sums** To find the sum of the series, consider the partial sums \( S_N \): \[ S_N = \sum_{n=1}^{N} \left( \frac{1}{n + 1} - \frac{1}{n + 3} \right) \] Split the sum: \[ S_N = \sum_{n=1}^{N} \frac{1}{n + 1} - \sum_{n=1}^{N} \frac{1}{n + 3} \] **6. Simplify the Sums** Express the sums with adjusted indices: - For the first sum, \( k = n + 1 \): \[ \sum_{n=1}^{N} \frac{1}{n + 1} = \sum_{k=2}^{N+1} \frac{1}{k} \] - For the second sum, \( k = n + 3 \): \[ \sum_{n=1}^{N} \frac{1}{n + 3} = \sum_{k=4}^{N+3} \frac{1}{k} \] Now, the partial sum becomes: \[ S_N = \sum_{k=2}^{N+1} \frac{1}{k} - \sum_{k=4}^{N+3} \frac{1}{k} \] **7. Cancel Out Common Terms** Rewrite \( S_N \) by expanding the sums: \[ S_N = \left( \frac{1}{2} + \frac{1}{3} + \sum_{k=4}^{N+1} \frac{1}{k} \right) - \left( \sum_{k=4}^{N+1} \frac{1}{k} + \frac{1}{N+2} + \frac{1}{N+3} \right) \] Notice that the terms \( \sum_{k=4}^{N+1} \frac{1}{k} \) in both sums cancel out. We are left with: \[ S_N = \frac{1}{2} + \frac{1}{3} - \frac{1}{N+2} - \frac{1}{N+3} \] **8. Compute the Limit as \( N \to \infty \)** As \( N \to \infty \), the terms \( \frac{1}{N+2} \) and \( \frac{1}{N+3} \) approach zero: \[ \lim_{N \to \infty} \left( \frac{1}{N+2} + \frac{1}{N+3} \right) = 0 \] Therefore, the sum of the series is: \[ S = \lim_{N \to \infty} S_N = \frac{1}{2} + \frac{1}{3} = \frac{5}{6} \] **Answer:** \(\displaystyle \sum_{n=1}^{\infty} \frac{2}{n^2 + 4n + 3} = \frac{5}{6}\) |
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| GPT-4o | To find the sum of the series \(\sum_{n=1}^{\infty} \frac{2}{n^2 + 4n + 3}\), we start by simplifying the general term of the series. First, we factor the denominator \(n^2 + 4n + 3\): \[ n^2 + 4n + 3 = (n+1)(n+3) \] Thus, the series becomes: \[ \sum_{n=1}^{\infty} \frac{2}{(n+1)(n+3)} \] Next, we use partial fraction decomposition to rewrite the fraction: \[ \frac{2}{(n+1)(n+3)} = \frac{A}{n+1} + \frac{B}{n+3} \] To find \(A\) and \(B\), we solve the equation: \[ 2 = A(n+3) + B(n+1) \] Expanding and equating coefficients: \[ 2 = An + 3A + Bn + B \] \[ 2 = (A+B)n + (3A+B) \] Equating the coefficients of \(n\) and the constant term, we get: \[ A + B = 0 \quad \text{and} \quad 3A + B = 2 \] Solving these equations, we first find \(B\) from \(A + B = 0\): \[ B = -A \] Substituting \(B = -A\) into \(3A + B = 2\): \[ 3A - A = 2 \] \[ 2A = 2 \] \[ A = 1 \] Then, substituting \(A = 1\) back into \(B = -A\): \[ B = -1 \] Thus, the partial fractions are: \[ \frac{2}{(n+1)(n+3)} = \frac{1}{n+1} - \frac{1}{n+3} \] Hence, the series can be written as: \[ \sum_{n=1}^{\infty} \left( \frac{1}{n+1} - \frac{1}{n+3} \right) \] This is a telescoping series. To see how it telescopes, we write out the first few terms: \[ \left( \frac{1}{2} - \frac{1}{4} \right) + \left( \frac{1}{3} - \frac{1}{5} \right) + \left( \frac{1}{4} - \frac{1}{6} \right) + \left( \frac{1}{5} - \frac{1}{7} \right) + \cdots \] Most terms cancel out: \[ \left( \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \cdots \right) - \left( \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \cdots \right) \] After cancellation, we are left with: \[ \sum_{n=1}^{\infty} \left( \frac{1}{n+1} - \frac{1}{n+3} \right) = \left( \frac{1}{2} + \frac{1}{3} \right) \] Therefore, the sum of the series is: \[ \frac{1}{2} + \frac{1}{3} = \frac{3}{6} + \frac{2}{6} = \frac{5}{6} \] Thus, the sum of the series \(\sum_{n=1}^{\infty} \frac{2}{n^2 + 4n + 3}\) is: \[ \boxed{\frac{5}{6}} \] |
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| llama3.1-70B-Instruct | The given series can be written as: $$\sum_{n=1}^{\infty} \frac{2}{n^2 + 4n + 3} = \sum_{n=1}^{\infty} \frac{2}{(n+1)(n+3)}$$ Using partial fraction decomposition, we can write: $$\frac{2}{(n+1)(n+3)} = \frac{A}{n+1} + \frac{B}{n+3}$$ Solving for A and B, we get: $$A = 1, B = -1$$ So, the series becomes: $$\sum_{n=1}^{\infty} \frac{2}{n^2 + 4n + 3} = \sum_{n=1}^{\infty} \left(\frac{1}{n+1} - \frac{1}{n+3}\right)$$ This is a telescoping series, and we can write: $$\sum_{n=1}^{\infty} \left(\frac{1}{n+1} - \frac{1}{n+3}\right) = \left(\frac{1}{2} - \frac{1}{4}\right) + \left(\frac{1}{3} - \frac{1}{5}\right) + \left(\frac{1}{4} - \frac{1}{6}\right) + \cdots$$ The terms cancel out, leaving only the first two terms: $$\sum_{n=1}^{\infty} \frac{2}{n^2 + 4n + 3} = \frac{1}{2} + \frac{1}{3} = \frac{5}{6}$$ Therefore, the sum of the given series is $\frac{5}{6}$. |
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| llama3.1-405B-Instruct | ## Step 1: Decompose the given series into partial fractions We start by decomposing $\frac{2}{n^2 + 4n + 3}$ into partial fractions. The denominator can be factored as $(n+1)(n+3)$. ## Step 2: Write the partial fraction decomposition form The partial fraction decomposition will have the form $\frac{A}{n+1} + \frac{B}{n+3}$. ## Step 3: Find the values of A and B Multiplying both sides by $(n+1)(n+3)$ to clear the denominators gives $2 = A(n+3) + B(n+1)$. Setting $n=-1$ yields $2 = 2A$, so $A = 1$. Setting $n=-3$ yields $2 = -2B$, so $B = -1$. ## Step 4: Rewrite the series using the partial fraction decomposition Thus, the series becomes $\sum_{n=1}^{\infty} \left( \frac{1}{n+1} - \frac{1}{n+3} \right)$. ## Step 5: Expand the series to observe the telescoping pattern Expanding the first few terms of the series gives $\left( \frac{1}{2} - \frac{1}{4} \right) + \left( \frac{1}{3} - \frac{1}{5} \right) + \left( \frac{1}{4} - \frac{1}{6} \right) + \left( \frac{1}{5} - \frac{1}{7} \right) + \cdots$. ## Step 6: Observe the cancellation pattern Notice that the $\frac{1}{4}$ term in the first parenthesis cancels with the $-\frac{1}{4}$ term in the third parenthesis, the $\frac{1}{5}$ term in the second parenthesis cancels with the $-\frac{1}{5}$ term in the fourth parenthesis, and so on. ## Step 7: Sum the remaining terms After cancellation, only $\frac{1}{2}$ and $\frac{1}{3}$ remain, so the sum of the series is $\frac{1}{2} + \frac{1}{3}$. ## Step 8: Calculate the sum $\frac{1}{2} + \frac{1}{3} = \frac{3}{6} + \frac{2}{6} = \frac{5}{6}$. The final answer is: $\boxed{\frac{5}{6}}$ |
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| DeepSeek-Coder-V2 | To solve the problem of finding the sum of \(\sum_{n=1}^{\infty} \frac{2}{n^2 + 4n + 3}\), we will follow these steps: 1. **Factor the denominator**: \[ n^2 + 4n + 3 = (n+1)(n+3) \] 2. **Rewrite the fraction using partial fractions**: \[ \frac{2}{(n+1)(n+3)} = \frac{A}{n+1} + \frac{B}{n+3} \] To find \(A\) and \(B\), we solve: \[ 2 = A(n+3) + B(n+1) \] Setting \(n = -1\): \[ 2 = A(2) \implies A = 1 \] Setting \(n = -3\): \[ 2 = B(-2) \implies B = -1 \] Thus, we have: \[ \frac{2}{(n+1)(n+3)} = \frac{1}{n+1} - \frac{1}{n+3} \] 3. **Rewrite the series**: \[ \sum_{n=1}^{\infty} \frac{2}{n^2 + 4n + 3} = \sum_{n=1}^{\infty} \left( \frac{1}{n+1} - \frac{1}{n+3} \right) \] 4. **Simplify the series using telescoping**: \[ \sum_{n=1}^{\infty} \left( \frac{1}{n+1} - \frac{1}{n+3} \right) \] Notice that this series is telescoping. The terms will cancel out in a pattern: \[ \left( \frac{1}{2} - \frac{1}{4} \right) + \left( \frac{1}{3} - \frac{1}{5} \right) + \left( \frac{1}{4} - \frac{1}{6} \right) + \left( \frac{1}{5} - \frac{1}{7} \right) + \cdots \] Most terms cancel out, leaving: \[ \frac{1}{2} + \frac{1}{3} \] 5. **Sum the remaining terms**: \[ \frac{1}{2} + \frac{1}{3} = \frac{3}{6} + \frac{2}{6} = \frac{5}{6} \] Thus, the sum of the series is: \[ \boxed{\frac{5}{6}} \] |
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| Qwen2-72B-Instruct | To solve the problem, we can start by simplifying the given series. The denominator $n^2 + 4n + 3$ can be factored as $(n+1)(n+3)$. Therefore, the series can be rewritten as: $$\sum_{n=1}^{\infty} \frac{2}{(n+1)(n+3)}$$ Next, we can use partial fraction decomposition to break down the term $\frac{2}{(n+1)(n+3)}$ into simpler fractions. We can write: $$\frac{2}{(n+1)(n+3)} = \frac{A}{n+1} + \frac{B}{n+3}$$ Multiplying both sides by $(n+1)(n+3)$, we get: $$2 = A(n+3) + B(n+1)$$ To find the values of A and B, we can solve the system of equations obtained by substituting n = -1 and n = -3 into the above equation. Doing so, we get: $$2 = 2A \Rightarrow A = 1$$ $$2 = -2B \Rightarrow B = -1$$ Therefore, we can rewrite the series as: $$\sum_{n=1}^{\infty} \left(\frac{1}{n+1} - \frac{1}{n+3}\right)$$ This series can be simplified further by grouping terms: $$\sum_{n=1}^{\infty} \left(\frac{1}{n+1} - \frac{1}{n+3}\right) = \left(\frac{1}{2} - \frac{1}{4}\right) + \left(\frac{1}{3} - \frac{1}{5}\right) + \left(\frac{1}{4} - \frac{1}{6}\right) + \left(\frac{1}{5} - \frac{1}{7}\right) + \dots$$ Notice that most of the terms cancel out, leaving only the first two terms: $$\left(\frac{1}{2} - \frac{1}{4}\right) + \left(\frac{1}{3} - \frac{1}{5}\right) + \dots = \frac{1}{2} + \frac{1}{3} = \frac{5}{6}$$ Therefore, the sum of the series is: $$\sum_{n=1}^{\infty} \frac{2}{n^2 + 4n + 3} = \boxed{\frac{5}{6}}$$ |
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| baichuan4 | To find the sum of the series $\sum_{n=1}^{\infty} \frac{2}{n^2 + 4n + 3}$, we can begin by factoring the denominator and then using partial fractions to break it down into simpler fractions. The denominator $n^2 + 4n + 3$ can be factored as $(n+1)(n+3)$. Now, let's use partial fractions to break down the fraction: $$\frac{2}{n^2 + 4n + 3} = \frac{A}{n+1} + \frac{B}{n+3}$$ To find $A$ and $B$, we multiply both sides by $(n+1)(n+3)$: $$2 = A(n+3) + B(n+1)$$ Now, we can solve for $A$ and $B$ by setting $n$ to values that make one of the variables vanish. For instance, setting $n = -1$ gives us $2 = B(0)$, so $B = 2$. Similarly, setting $n = -3$ gives us $2 = A(0)$, so $A = 2$. Thus, we have: $$\frac{2}{n^2 + 4n + 3} = \frac{2}{n+1} + \frac{-2}{n+3}$$ Now, we can write the original series as: $$\sum_{n=1}^{\infty} \left(\frac{2}{n+1} - \frac{2}{n+3}\right)$$ This series is now in the form of a telescoping series, where the terms cancel out except for the first few. To find the sum, we can write out the first few terms: $$\left(\frac{2}{2} - \frac{2}{4}\right) + \left(\frac{2}{3} - \frac{2}{5}\right) + \left(\frac{2}{4} - \frac{2}{6}\right) + \ldots$$ We can see that all the terms cancel out except for $\frac{2}{2}$ and $-\frac{2}{5}$: $$\frac{2}{2} - \frac{2}{5} = 1 - \frac{2}{5} = \frac{3}{5}$$ Therefore, the sum of the series $\sum_{n=1}^{\infty} \frac{2}{n^2 + 4n + 3}$ is $\frac{3}{5}$. |
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| doubao | We can factor the denominator: $n^2 + 4n + 3 = (n+1)(n+3)$. So the series can be rewritten as: $\sum_{n=1}^{\infty} \frac{2}{(n+1)(n+3)} = \sum_{n=1}^{\infty} \left(\frac{1}{n+1} - \frac{1}{n+3}\right)$. Now let's write out the first few terms: $\left(\frac{1}{2} - \frac{1}{4}\right) + \left(\frac{1}{3} - \frac{1}{5}\right) + \left(\frac{1}{4} - \frac{1}{6}\right) + \left(\frac{1}{5} - \frac{1}{7}\right) + \cdots$. Notice that many terms will cancel out. In fact, all terms except for the first $\frac{1}{2}$ and the second term $\frac{1}{3}$ will cancel. So the sum is $\frac{1}{2} + \frac{1}{3} = \frac{5}{6}$. |
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| minimax | To find the sum of the given infinite series, we can start by performing partial fraction decomposition on the general term of the series. The general term is: $$\frac{2}{n^2 + 4n + 3}$$ We want to express this as a sum of simpler fractions: $$\frac{2}{n^2 + 4n + 3} = \frac{A}{n+1} + \frac{B}{n+3}$$ To find the constants A and B, we multiply both sides by the denominator on the left side: $$2 = A(n+3) + B(n+1)$$ Now, we can solve for A and B by plugging in values for n that simplify the equation. Let's plug in n = -1 and n = -3: For n = -1: $$2 = A(-1+3) + B(-1+1)$$ $$2 = 2A$$ $$A = 1$$ For n = -3: $$2 = A(-3+3) + B(-3+1)$$ $$2 = -2B$$ $$B = -1$$ Now we have: $$\frac{2}{n^2 + 4n + 3} = \frac{1}{n+1} - \frac{1}{n+3}$$ The series can now be written as: $$\sum_{n=1}^{\infty} \left(\frac{1}{n+1} - \frac{1}{n+3}\right)$$ This is a telescoping series, where many terms cancel out when we write out the first few terms: $$\left(\frac{1}{2} - \frac{1}{4}\right) + \left(\frac{1}{3} - \frac{1}{5}\right) + \left(\frac{1}{4} - \frac{1}{6}\right) + \left(\frac{1}{5} - \frac{1}{7}\right) + \ldots$$ Notice that the negative term in one bracket cancels with the positive term in the next bracket. This cancellation continues indefinitely, leaving only the first few terms of the series: $$\frac{1}{2} - \lim_{n \to \infty} \frac{1}{n+3}$$ Since the limit of $\frac{1}{n+3}$ as $n$ approaches infinity is 0, the sum of the series is simply: $$\sum_{n=1}^{\infty} \frac{2}{n^2 + 4n + 3} = \frac{1}{2}$$ Therefore, the sum of the given infinite series is $\frac{1}{2}$. |
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| kimi | To find the sum of the series $\sum_{n=1}^{\infty} \frac{2}{n^2 + 4n + 3}$, we can start by simplifying the general term of the series. First, observe that the denominator can be factored: $$n^2 + 4n + 3 = (n + 1)(n + 3).$$ Now, the term becomes: $$\frac{2}{n^2 + 4n + 3} = \frac{2}{(n + 1)(n + 3)}.$$ We can use partial fraction decomposition to break this fraction into simpler parts: $$\frac{2}{(n + 1)(n + 3)} = \frac{A}{n + 1} + \frac{B}{n + 3}.$$ Multiplying both sides by the denominator $((n + 1)(n + 3))$ gives us: $$2 = A(n + 3) + B(n + 1).$$ To find $A$ and $B$, we can set up a system of equations by choosing values for $n$ that will eliminate one of the variables: 1. Let $n = -1$: $$2 = A(-1 + 3) \Rightarrow 2 = 2A \Rightarrow A = 1.$$ 2. Let $n = -3$: $$2 = B(-3 + 1) \Rightarrow 2 = -2B \Rightarrow B = -1.$$ So, the partial fractions are: $$\frac{2}{(n + 1)(n + 3)} = \frac{1}{n + 1} - \frac{1}{n + 3}.$$ Now, we can rewrite the series as: $$\sum_{n=1}^{\infty} \left(\frac{1}{n + 1} - \frac{1}{n + 3}\right).$$ This is a telescoping series, where most terms will cancel out. Writing out the first few terms, we get: $$\left(\frac{1}{2} - \frac{1}{4}\right) + \left(\frac{1}{3} - \frac{1}{5}\right) + \left(\frac{1}{4} - \frac{1}{6}\right) + \cdots.$$ Notice that the $-\frac{1}{4}$ from the first term cancels with the $\frac{1}{4}$ from the third term, and so on. This pattern continues indefinitely, and all terms cancel except for the first few terms on the left side. So, the sum of the series is: $$\left(\frac{1}{2} + \frac{1}{3}\right) - \lim_{n \to \infty} \frac{1}{n + 3} = \frac{1}{2} + \frac{1}{3} - 0 = \frac{3}{6} + \frac{2}{6} = \frac{5}{6}.$$ Therefore, the sum of the series is $\boxed{\frac{5}{6}}$. |
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| wenxinyiyan | 首先，我们对给定的级数进行裂项： $\frac{2}{n^2 + 4n + 3} = \frac{1}{n + 1} - \frac{1}{n + 3}$ 这样，原级数可以表示为： $\sum_{n=1}^{\infty} (\frac{1}{n + 1} - \frac{1}{n + 3})$ 接下来，我们对前n项进行求和： $S_n = \frac{1}{2} - \frac{1}{4} + \frac{1}{3} - \frac{1}{5} + \frac{1}{4} - \frac{1}{6} + \ldots + \frac{1}{n} - \frac{1}{n + 2} + \frac{1}{n + 1} - \frac{1}{n + 3}$ 整理上述求和结果，我们得到： $S_n = \frac{1}{2} + \frac{1}{3} - \frac{1}{n + 2} - \frac{1}{n + 3}$ 最后，我们求极限： $\lim_{n \to \infty} S_n = \lim_{n \to \infty} (\frac{1}{2} + \frac{1}{3} - \frac{1}{n + 2} - \frac{1}{n + 3}) = \frac{5}{6}$ 因此，原级数的和为$\frac{5}{6}$。 |
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| zhipu | To find the sum of the infinite series \(\sum_{n=1}^{\infty} \frac{2}{n^2 + 4n + 3}\), we start by simplifying the expression inside the summation. First, notice that the denominator \(n^2 + 4n + 3\) can be factored: \[ n^2 + 4n + 3 = (n + 1)(n + 3) \] So the given series can be rewritten as: \[ \sum_{n=1}^{\infty} \frac{2}{(n + 1)(n + 3)} \] Next, we use partial fraction decomposition to break down the fraction: \[ \frac{2}{(n + 1)(n + 3)} = \frac{A}{n + 1} + \frac{B}{n + 3} \] Multiplying both sides by \((n + 1)(n + 3)\) gives: \[ 2 = A(n + 3) + B(n + 1) \] To find \(A\) and \(B\), we set up the following system of equations by equating coefficients: 1. For the \(n\) term: \[ A + B = 0 \] 2. For the constant term: \[ 3A + B = 2 \] From the first equation, \(A + B = 0\), we get: \[ B = -A \] Substituting \(B = -A\) into the second equation: \[ 3A - A = 2 \] \[ 2A = 2 \] \[ A = 1 \] Then, substituting \(A = 1\) back into \(B = -A\): \[ B = -1 \] Thus, the partial fraction decomposition is: \[ \frac{2}{(n + 1)(n + 3)} = \frac{1}{n + 1} - \frac{1}{n + 3} \] Now, we can rewrite the series: \[ \sum_{n=1}^{\infty} \left( \frac{1}{n + 1} - \frac{1}{n + 3} \right) \] This series is telescoping, meaning many terms will cancel out as we sum them. Let's write out the first few terms to see the pattern: \[ \left( \frac{1}{2} - \frac{1}{4} \right) + \left( \frac{1}{3} - \frac{1}{5} \right) + \left( \frac{1}{4} - \frac{1}{6} \right) + \left( \frac{1}{5} - \frac{1}{7} \right) + \cdots \] Notice that each term \(\frac{1}{n+3}\) will cancel with the term \(\frac{1}{n+1}\) three terms later. The remaining terms are the first few terms that do not get canceled: \[ \frac{1}{2} + \frac{1}{3} + \frac{1}{4} \] However, since the series is infinite, we need to account for all the terms that do not get canceled. The effective remaining terms are: \[ \frac{1}{2} + \frac{1}{3} \] Summing these gives: \[ \frac{1}{2} + \frac{1}{3} = \frac{3}{6} + \frac{2}{6} = \frac{5}{6} \] Since each term in the original series was multiplied by 2, we need to multiply the sum by 2: \[ 2 \left( \frac{5}{6} \right) = \frac{10}{6} = \frac{5}{3} \] Therefore, the sum of the series \(\sum_{n=1}^{\infty} \frac{2}{n^2 + 4n + 3}\) is: \[ \boxed{\frac{5}{3}} \] |
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| gpto1 | To find the sum of the infinite series: \[ \sum_{n=1}^{\infty} \frac{2}{n^2 + 4n + 3} \] we can start by simplifying the general term of the series and then using partial fractions to decompose it. Here are the detailed steps: **1. Factor the Denominator** First, factor the quadratic expression in the denominator: \[ n^2 + 4n + 3 = (n + 1)(n + 3) \] **2. Rewrite the General Term** Rewrite the general term of the series using the factored form: \[ \frac{2}{n^2 + 4n + 3} = \frac{2}{(n + 1)(n + 3)} \] **3. Decompose into Partial Fractions** We can express the fraction as the sum of two simpler fractions: \[ \frac{2}{(n + 1)(n + 3)} = \frac{A}{n + 1} + \frac{B}{n + 3} \] To find constants \( A \) and \( B \), multiply both sides by \( (n + 1)(n + 3) \): \[ 2 = A(n + 3) + B(n + 1) \] Simplify and solve for \( A \) and \( B \) by choosing convenient values for \( n \): - Let \( n = -3 \): \[ 2 = A(-3 + 3) + B(-3 + 1) \\ 2 = A(0) + B(-2) \\ 2 = -2B \\ B = -1 \] - Let \( n = -1 \): \[ 2 = A(-1 + 3) + B(-1 + 1) \\ 2 = A(2) + B(0) \\ 2 = 2A \\ A = 1 \] **4. Rewrite the Series** Now, the general term becomes: \[ \frac{2}{(n + 1)(n + 3)} = \frac{1}{n + 1} - \frac{1}{n + 3} \] Therefore, the series is: \[ \sum_{n=1}^{\infty} \left( \frac{1}{n + 1} - \frac{1}{n + 3} \right) \] **5. Compute the Partial Sums** To find the sum of the series, consider the partial sums \( S_N \): \[ S_N = \sum_{n=1}^{N} \left( \frac{1}{n + 1} - \frac{1}{n + 3} \right) \] Split the sum: \[ S_N = \sum_{n=1}^{N} \frac{1}{n + 1} - \sum_{n=1}^{N} \frac{1}{n + 3} \] **6. Simplify the Sums** Express the sums with adjusted indices: - For the first sum, \( k = n + 1 \): \[ \sum_{n=1}^{N} \frac{1}{n + 1} = \sum_{k=2}^{N+1} \frac{1}{k} \] - For the second sum, \( k = n + 3 \): \[ \sum_{n=1}^{N} \frac{1}{n + 3} = \sum_{k=4}^{N+3} \frac{1}{k} \] Now, the partial sum becomes: \[ S_N = \sum_{k=2}^{N+1} \frac{1}{k} - \sum_{k=4}^{N+3} \frac{1}{k} \] **7. Cancel Out Common Terms** Rewrite \( S_N \) by expanding the sums: \[ S_N = \left( \frac{1}{2} + \frac{1}{3} + \sum_{k=4}^{N+1} \frac{1}{k} \right) - \left( \sum_{k=4}^{N+1} \frac{1}{k} + \frac{1}{N+2} + \frac{1}{N+3} \right) \] Notice that the terms \( \sum_{k=4}^{N+1} \frac{1}{k} \) in both sums cancel out. We are left with: \[ S_N = \frac{1}{2} + \frac{1}{3} - \frac{1}{N+2} - \frac{1}{N+3} \] **8. Compute the Limit as \( N \to \infty \)** As \( N \to \infty \), the terms \( \frac{1}{N+2} \) and \( \frac{1}{N+3} \) approach zero: \[ \lim_{N \to \infty} \left( \frac{1}{N+2} + \frac{1}{N+3} \right) = 0 \] Therefore, the sum of the series is: \[ S = \lim_{N \to \infty} S_N = \frac{1}{2} + \frac{1}{3} = \frac{5}{6} \] **Answer:** \(\displaystyle \sum_{n=1}^{\infty} \frac{2}{n^2 + 4n + 3} = \frac{5}{6}\) |
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