| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | To find the sum of the given series, we can split it into two separate series and find the sum of each series individually. The series is: \[ \sum_{n=1}^{\infty} \left(\frac{1}{e^n} + \frac{1}{n(n+1)}\right) \] Let's break it down into two series: \[ \sum_{n=1}^{\infty} \frac{1}{e^n} \quad \text{and} \quad \sum_{n=1}^{\infty} \frac{1}{n(n+1)} \] ### Solving the First Series: The first series is a geometric series with the first term \(a = \frac{1}{e}\) and the common ratio \(r = \frac{1}{e}\). The sum of an infinite geometric series is given by: \[ S = \frac{a}{1-r} \] Substituting \(a = \frac{1}{e}\) and \(r = \frac{1}{e}\): \[ S_1 = \frac{\frac{1}{e}}{1-\frac{1}{e}} = \frac{1}{e-1} \] ### Solving the Second Series: The second series can be simplified using partial fraction decomposition: \[ \frac{1}{n(n+1)} = \frac{A}{n} + \frac{B}{n+1} \] Solving for \(A\) and \(B\): \[ A(n+1) + Bn = 1 \] Let \(n = 0\), then \(A = 1\). Let \(n = -1\), then \(B = -1\). Thus, \[ \frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1} \] This is a telescoping series. The sum of the series is: \[ S_2 = \lim_{N \to \infty} \sum_{n=1}^{N} \left(\frac{1}{n} - \frac{1}{n+1}\right) = 1 - \lim_{N \to \infty} \frac{1}{N+1} = 1 \] ### Sum of Both Series: The sum of the two series is: \[ S = S_1 + S_2 = \frac{1}{e-1} + 1 = \frac{1}{e-1} + \frac{e-1}{e-1} = \frac{e}{e-1} \] Therefore, the sum of the series \(\sum_{n=1}^{\infty} \left(\frac{1}{e^n} + \frac{1}{n(n+1)}\right)\) is \(\boxed{\frac{e}{e-1}}\). |
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| deepseekr1 | To find the sum of the series \(\sum_{n=1}^{\infty} \left( \frac{1}{e^n} + \frac{1}{n(n+1)} \right)\), we can split it into two separate sums: 1. The geometric series \(\sum_{n=1}^{\infty} \frac{1}{e^n}\). 2. The telescoping series \(\sum_{n=1}^{\infty} \frac{1}{n(n+1)}\). **1. Geometric Series:** The series \(\sum_{n=1}^{\infty} \frac{1}{e^n}\) is a geometric series with the first term \(a = \frac{1}{e}\) and common ratio \(r = \frac{1}{e}\). The sum of a geometric series starting from \(n=1\) is given by: \[ \sum_{n=1}^{\infty} ar^{n-1} = \frac{a}{1 - r} \] However, since our series starts at \(n=1\), we use: \[ \sum_{n=1}^{\infty} \frac{1}{e^n} = \frac{\frac{1}{e}}{1 - \frac{1}{e}} = \frac{\frac{1}{e}}{\frac{e - 1}{e}} = \frac{1}{e - 1} \] **2. Telescoping Series:** The series \(\sum_{n=1}^{\infty} \frac{1}{n(n+1)}\) can be decomposed using partial fractions: \[ \frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1} \] This results in a telescoping series: \[ \sum_{n=1}^{\infty} \left( \frac{1}{n} - \frac{1}{n+1} \right) \] Writing out the first few terms, we see: \[ \left(1 - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{4}\right) + \cdots \] Most terms cancel out, leaving: \[ 1 - \lim_{n \to \infty} \frac{1}{n+1} = 1 - 0 = 1 \] **Combining the Results:** Adding the sums of the two series: \[ \frac{1}{e - 1} + 1 = \frac{1}{e - 1} + \frac{e - 1}{e - 1} = \frac{1 + e - 1}{e - 1} = \frac{e}{e - 1} \] Thus, the sum of the series is \(\boxed{\dfrac{e}{e - 1}}\). |
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| gpto1 | To find the sum: \[ S = \sum_{n=1}^{\infty} \left( \frac{1}{e^n} + \frac{1}{n(n+1)} \right) \] We can split this into two separate sums: \[ S = \left( \sum_{n=1}^\infty \frac{1}{e^n} \right) + \left( \sum_{n=1}^\infty \frac{1}{n(n+1)} \right) \] **First Sum:** \[ S_1 = \sum_{n=1}^\infty \frac{1}{e^n} = \frac{1}{e} + \frac{1}{e^2} + \frac{1}{e^3} + \dots \] This is a geometric series with first term \( a = \frac{1}{e} \) and common ratio \( r = \frac{1}{e} \) (since \( e > 1 \), \( |r| < 1 \)), so it converges. The sum of an infinite geometric series is: \[ S = \frac{a}{1 - r} \] Applying this, we get: \[ S_1 = \frac{\frac{1}{e}}{1 - \frac{1}{e}} = \frac{1}{e - 1} \] **Second Sum:** \[ S_2 = \sum_{n=1}^\infty \frac{1}{n(n+1)} \] We can decompose the fraction using partial fractions: \[ \frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1} \] Thus, \[ S_2 = \sum_{n=1}^\infty \left( \frac{1}{n} - \frac{1}{n+1} \right) \] This series telescopes: \[ S_2 = \left( 1 - \frac{1}{2} \right) + \left( \frac{1}{2} - \frac{1}{3} \right) + \left( \frac{1}{3} - \frac{1}{4} \right) + \dots \] Most terms cancel out, leaving: \[ S_2 = \lim_{N \to \infty} \left( 1 - \frac{1}{N+1} \right) = 1 \] **Final Sum:** \[ S = S_1 + S_2 = \frac{1}{e - 1} + 1 \] **Answer: An explicit sum: S = (1 ⁄ (e – 1)) + 1; that is, S = 1 + 1⁄(e – 1) |
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| GPT-4o | To solve the given problem, we need to find the sum of the infinite series: \[ \sum_{n=1}^{\infty} \left( \frac{1}{e^n} + \frac{1}{n(n+1)} \right) \] We can split this sum into two separate series: \[ \sum_{n=1}^{\infty} \frac{1}{e^n} + \sum_{n=1}^{\infty} \frac{1}{n(n+1)} \] ### 1. Evaluating \(\sum_{n=1}^{\infty} \frac{1}{e^n}\) This is a geometric series where the common ratio \( r = \frac{1}{e} \). The sum of an infinite geometric series with first term \( a \) and common ratio \( r \) (where \( |r| < 1 \)) is given by: \[ \sum_{n=0}^{\infty} ar^n = \frac{a}{1 - r} \] In our case, the series starts from \( n = 1 \), so we make the adjustment: \[ \sum_{n=1}^{\infty} \frac{1}{e^n} = \sum_{n=1}^{\infty} \left(\frac{1}{e}\right)^n = \frac{\frac{1}{e}}{1 - \frac{1}{e}} = \frac{\frac{1}{e}}{\frac{e-1}{e}} = \frac{1}{e-1} \] ### 2. Evaluating \(\sum_{n=1}^{\infty} \frac{1}{n(n+1)}\) We can simplify \(\frac{1}{n(n+1)}\) using partial fractions: \[ \frac{1}{n(n+1)} = \frac{A}{n} + \frac{B}{n+1} \] Solving for \(A\) and \(B\): \[ 1 = A(n+1) + Bn \] Setting \(n = 0\): \[ 1 = A(0+1) \implies A = 1 \] Setting \(n = -1\): \[ 1 = B(-1) \implies B = -1 \] Thus, \[ \frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1} \] The sum of the series becomes a telescoping series: \[ \sum_{n=1}^{\infty} \left( \frac{1}{n} - \frac{1}{n+1} \right) \] Writing out the first few terms: \[ \left( \frac{1}{1} - \frac{1}{2} \right) + \left( \frac{1}{2} - \frac{1}{3} \right) + \left( \frac{1}{3} - \frac{1}{4} \right) + \cdots \] Most terms cancel out, leaving: \[ 1 - \lim_{n \to \infty} \frac{1}{n+1} = 1 - 0 = 1 \] ### Combining the Results Adding the two sums together: \[ \sum_{n=1}^{\infty} \left( \frac{1}{e^n} + \frac{1}{n(n+1)} \right) = \frac{1}{e-1} + 1 \] Thus, the sum of the given series is: \[ \boxed{\frac{1}{e-1} + 1} \] |
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| llama3.1-70B-Instruct | Step 1: Analyze the problem We are given an infinite series, which is a sum of two separate series. The first series is a geometric series with the common ratio 1/e, and the second series is a telescoping series. ### Step 2: Outline the solution strategy To find the sum of the given series, we will: 1. Find the sum of the geometric series. 2. Find the sum of the telescoping series. 3. Add the sums of the two series. ### Step 3: Find the sum of the geometric series The sum of an infinite geometric series with first term a and common ratio r is given by: S = a / (1 - r) In this case, a = 1/e and r = 1/e. ```python import math # Define the variables a = 1 / math.e r = 1 / math.e # Calculate the sum of the geometric series sum_geometric = a / (1 - r) ``` ### Step 4: Find the sum of the telescoping series The sum of the telescoping series can be found by noticing that each term cancels out all but the first and last terms. ```python # Calculate the sum of the telescoping series sum_telescoping = 1 ``` ### Step 5: Add the sums of the two series Now, we add the sums of the geometric series and the telescoping series. ```python # Calculate the final sum final_sum = sum_geometric + sum_telescoping ``` ### Combine the code into a function and execute it using iPython Here's the complete code: |
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| llama3.1-405B-Instruct | ## Step 1: Break down the series into two separate series The given series can be broken down into two separate series: $\sum_{n=1}^{\infty} 1/e^n$ and $\sum_{n=1}^{\infty} 1/(n*(n+1))$. ## Step 2: Evaluate the first series The first series is a geometric series with first term $1/e$ and common ratio $1/e$. The sum of an infinite geometric series is given by $a/(1-r)$, where $a$ is the first term and $r$ is the common ratio. Therefore, the sum of the first series is $(1/e)/(1-1/e) = 1/(e-1)$. ## Step 3: Evaluate the second series The second series can be evaluated using partial fractions. We can rewrite $1/(n*(n+1))$ as $1/n - 1/(n+1)$. Then, the series becomes $\sum_{n=1}^{\infty} (1/n - 1/(n+1))$. This is a telescoping series, and its sum is given by $1 - 1/2 + 1/2 - 1/3 + ... = 1$. ## Step 4: Add the sums of the two series The sum of the original series is the sum of the two separate series: $1/(e-1) + 1$. The final answer is: $\boxed{1/(e-1) + 1}$ |
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| DeepSeek-Coder-V2 | To solve the problem of finding the sum of \(\sum_{n=1}^{\infty} \left(\frac{1}{e^n} + \frac{1}{n(n+1)}\right)\), we will break it down into two separate sums and solve each one individually. First, let's consider the sum \(\sum_{n=1}^{\infty} \frac{1}{e^n}\): \[ \sum_{n=1}^{\infty} \frac{1}{e^n} \] This is a geometric series with the first term \(a = \frac{1}{e}\) and common ratio \(r = \frac{1}{e}\). The sum of an infinite geometric series is given by: \[ S = \frac{a}{1 - r} \] Substituting \(a = \frac{1}{e}\) and \(r = \frac{1}{e}\): \[ S = \frac{\frac{1}{e}}{1 - \frac{1}{e}} = \frac{\frac{1}{e}}{\frac{e-1}{e}} = \frac{1}{e-1} \] So, the sum of the series \(\sum_{n=1}^{\infty} \frac{1}{e^n}\) is \(\frac{1}{e-1}\). Next, let's consider the sum \(\sum_{n=1}^{\infty} \frac{1}{n(n+1)}\): \[ \sum_{n=1}^{\infty} \frac{1}{n(n+1)} \] We can use partial fraction decomposition to simplify this term: \[ \frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1} \] Thus, the series becomes: \[ \sum_{n=1}^{\infty} \left(\frac{1}{n} - \frac{1}{n+1}\right) \] This is a telescoping series, where most terms cancel out. Writing out the first few terms: \[ \left(1 - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{4}\right) + \cdots \] All intermediate terms cancel, leaving: \[ 1 - \lim_{n \to \infty} \frac{1}{n+1} = 1 - 0 = 1 \] So, the sum of the series \(\sum_{n=1}^{\infty} \frac{1}{n(n+1)}\) is 1. Combining the results of the two sums, we get: \[ \sum_{n=1}^{\infty} \left(\frac{1}{e^n} + \frac{1}{n(n+1)}\right) = \frac{1}{e-1} + 1 \] Therefore, the final answer is: \[ \boxed{\frac{1}{e-1} + 1} \] |
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| Qwen2-72B-Instruct | To find the sum of the given series, we can split it into two separate series and find the sum of each series individually. The series is: \[ \sum_{n=1}^{\infty} \left(\frac{1}{e^n} + \frac{1}{n(n+1)}\right) \] Let's break it down into two series: \[ \sum_{n=1}^{\infty} \frac{1}{e^n} \quad \text{and} \quad \sum_{n=1}^{\infty} \frac{1}{n(n+1)} \] ### Solving the First Series: The first series is a geometric series with the first term $a = \frac{1}{e}$ and the common ratio $r = \frac{1}{e}$. The sum of an infinite geometric series is given by: \[ S = \frac{a}{1-r} \] Substituting $a = \frac{1}{e}$ and $r = \frac{1}{e}$: \[ S_1 = \frac{\frac{1}{e}}{1-\frac{1}{e}} = \frac{1}{e-1} \] ### Solving the Second Series: The second series can be simplified using partial fraction decomposition: \[ \frac{1}{n(n+1)} = \frac{A}{n} + \frac{B}{n+1} \] Solving for $A$ and $B$: \[ A(n+1) + Bn = 1 \] Let $n = 0$, then $A = 1$. Let $n = -1$, then $B = -1$. Thus, \[ \frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1} \] This is a telescoping series. The sum of the series is: \[ S_2 = \lim_{N \to \infty} \sum_{n=1}^{N} \left(\frac{1}{n} - \frac{1}{n+1}\right) = 1 - \lim_{N \to \infty} \frac{1}{N+1} = 1 \] ### Sum of Both Series: The sum of the two series is: \[ S = S_1 + S_2 = \frac{1}{e-1} + 1 = \frac{1}{e-1} + \frac{e-1}{e-1} = \frac{e}{e-1} \] Therefore, the sum of the series $\sum_{n=1}^{\infty} \left(\frac{1}{e^n} + \frac{1}{n(n+1)}\right)$ is $\boxed{\frac{e}{e-1}}$. |
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| baichuan4 | To find the sum of the series $\sum_{n=1}^{\infty} \left(\frac{1}{e^n} + \frac{1}{n(n+1)}\right)$, we can split it into two separate series: 1. $\sum_{n=1}^{\infty} \frac{1}{e^n}$ 2. $\sum_{n=1}^{\infty} \frac{1}{n(n+1)}$ We will evaluate each of these series separately and then add their sums together. ### Series 1: Geometric Series The first series is a geometric series with the common ratio $1/e < 1$. For a geometric series with the first term $a$ and common ratio $r$, where $|r| < 1$, the sum to infinity is given by: $$\sum_{n=1}^{\infty} ar^{n-1} = \frac{a}{1-r}$$ In our case, $a = 1/e$ and $r = 1/e$, so the sum of the first series is: $$\sum_{n=1}^{\infty} \frac{1}{e^n} = \frac{1/e}{1 - 1/e} = \frac{1}{e - 1}$$ ### Series 2: Telescoping Series The second series can be simplified by recognizing it as a telescoping series: $$\sum_{n=1}^{\infty} \frac{1}{n(n+1)} = \sum_{n=1}^{\infty} \left(\frac{1}{n} - \frac{1}{n+1}\right)$$ When you expand this out, most of the terms will cancel out except for the first and last terms: $$\left(1 - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{4}\right) + \ldots = 1 - \lim_{n \to \infty} \frac{1}{n+1}$$ The limit of the last term goes to 0 as $n$ approaches infinity, so the sum of the second series is: $$\sum_{n=1}^{\infty} \frac{1}{n(n+1)} = 1$$ ### Total Sum Now, adding the sums of both series together, we get: $$\sum_{n=1}^{\infty} \left(\frac{1}{e^n} + \frac{1}{n(n+1)}\right) = \frac{1}{e - 1} + 1 = \frac{e}{e - 1}$$ Therefore, the sum of the given series is $\frac{e}{e - 1}$. |
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| doubao | Let's consider the two parts of the series separately. For the series $\sum_{n=1}^{\infty} \frac{1}{e^n}$, this is a geometric series with a common ratio of $\frac{1}{e}$. The sum of a geometric series with first term $a$ and common ratio $r$ (where $|r|<1$) is $\frac{a}{1-r}$. Here, $a=\frac{1}{e}$ and $r=\frac{1}{e}$, so the sum is $\frac{\frac{1}{e}}{1-\frac{1}{e}}=\frac{1}{e-1}$. For the series $\sum_{n=1}^{\infty} \frac{1}{n(n+1)}$, we can use partial fraction decomposition. We have $\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}$. So the sum telescopes: \[ \begin{align*} &\sum_{n=1}^{\infty} \left(\frac{1}{n}-\frac{1}{n+1}\right)\\ =&(1-\frac{1}{2})+(\frac{1}{2}-\frac{1}{3})+(\frac{1}{3}-\frac{1}{4})+\cdots\\ =&1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\cdots\\ =&1 \end{align*} \] Adding the two sums together, we get $\frac{1}{e-1}+1=\frac{1}{e-1}+\frac{e-1}{e-1}=\frac{e}{e-1}$. |
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| minimax | To find the sum of the series $\sum_{n=1}^{\infty} (1/e^n + 1/(n*(n+1)))$, we can split the series into two separate series and find the sum of each one separately. The first series is a geometric series with the common ratio $1/e$: $$\sum_{n=1}^{\infty} \frac{1}{e^n} = \frac{1/e}{1 - (1/e)} = \frac{1/e}{(e-1)/e} = \frac{1}{e-1}$$ The second series is a telescoping series: $$\sum_{n=1}^{\infty} \frac{1}{n(n+1)}$$ We can rewrite the terms of this series using partial fraction decomposition: $$\frac{1}{n(n+1)} = \frac{A}{n} + \frac{B}{n+1}$$ To find the values of A and B, we solve the equation: $$1 = A(n+1) + Bn$$ Setting $n = 0$, we get $A = 1$. Setting $n = -1$, we get $B = -1$. So the decomposition is: $$\frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}$$ Now, let's look at the series: $$\sum_{n=1}^{\infty} \left(\frac{1}{n} - \frac{1}{n+1}\right)$$ This is a telescoping series, where each term cancels out the subsequent term except for the first part of the first term and the last part of the last term. When we write out the first few terms, we see: $$\left(\frac{1}{1} - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{4}\right) + \ldots$$ All the terms in the parentheses cancel except for the first term of the first pair, which is $1/1$. Since the series is infinite, there is no "last term," and thus the series sums to just $1$. Now we can add the sums of the two series together: $$\sum_{n=1}^{\infty} (1/e^n + 1/(n*(n+1))) = \frac{1}{e-1} + 1$$ Therefore, the sum of the given series is $\frac{1}{e-1} + 1$. |
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| kimi | To find the sum of the infinite series \(\sum_{n=1}^{\infty} \left(\frac{1}{e^n} + \frac{1}{n(n+1)}\right)\), we can break it down into two separate series and then add them together. First, let's consider the series \(\sum_{n=1}^{\infty} \frac{1}{e^n}\). This is a geometric series with the first term \(a = \frac{1}{e}\) and common ratio \(r = \frac{1}{e}\). The sum \(S\) of an infinite geometric series is given by \(S = \frac{a}{1 - r}\), provided that \(|r| < 1\). In this case, \(r = \frac{1}{e} < 1\), so the sum is: \[S_1 = \frac{\frac{1}{e}}{1 - \frac{1}{e}} = \frac{1}{e - 1}\] Next, let's consider the series \(\sum_{n=1}^{\infty} \frac{1}{n(n+1)}\). This series can be simplified using partial fractions. We have: \[\frac{1}{n(n+1)} = \frac{A}{n} + \frac{B}{n+1}\] By solving for \(A\) and \(B\), we find that \(A = 1\) and \(B = -1\), so: \[\frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}\] Now, when we sum this series, many terms will cancel out: \[S_2 = \left(1 - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{4}\right) + \cdots\] All the intermediate terms cancel out, leaving us with: \[S_2 = 1\] Finally, we add the two sums together: \[S = S_1 + S_2 = \frac{1}{e - 1} + 1\] To simplify this further, we can combine the terms over a common denominator: \[S = \frac{1}{e - 1} + \frac{e - 1}{e - 1} = \frac{e}{e - 1}\] So the sum of the infinite series is: \[S = \frac{e}{e - 1}\] |
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| wenxinyiyan | 首先，我们将给定的级数拆分为两个部分：$\sum_{n=1}^{\infty} 1/e^n$ 和 $\sum_{n=1}^{\infty} 1/(n*(n+1))$。 第一部分是一个几何级数，其首项为$1/e$，公比也为$1/e$。根据几何级数的求和公式，该部分的和为： $S_1 = \frac{1/e}{1-1/e} = \frac{1}{e-1}$ 第二部分可以进一步拆分为两个级数的差：$\sum_{n=1}^{\infty} 1/n - \sum_{n=1}^{\infty} 1/(n+1)$。观察这两个级数，我们可以发现它们都是调和级数，但是第二个级数的每一项都比第一个级数对应的项少1。因此，该部分的和为： $S_2 = (1 + 1/2 + 1/3 + ...) - (1/2 + 1/3 + 1/4 + ...) = 1$ 最后，我们将两部分的和相加，得到： $S = S_1 + S_2 = \frac{1}{e-1} + 1 = \frac{e}{e-1}$ |
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| zhipu | To find the sum of the series \(\sum_{n=1}^{\infty} \left( \frac{1}{e^n} + \frac{1}{n(n+1)} \right)\), we can split the series into two separate series and sum them individually: \[ \sum_{n=1}^{\infty} \left( \frac{1}{e^n} + \frac{1}{n(n+1)} \right) = \sum_{n=1}^{\infty} \frac{1}{e^n} + \sum_{n=1}^{\infty} \frac{1}{n(n+1)} \] ### Sum of the First Series: \(\sum_{n=1}^{\infty} \frac{1}{e^n}\) This is a geometric series with the first term \(a = \frac{1}{e}\) and common ratio \(r = \frac{1}{e}\). The sum of an infinite geometric series is given by: \[ \sum_{n=1}^{\infty} ar^{n-1} = \frac{a}{1 - r} \] For our series, \(a = \frac{1}{e}\) and \(r = \frac{1}{e}\), so: \[ \sum_{n=1}^{\infty} \frac{1}{e^n} = \frac{\frac{1}{e}}{1 - \frac{1}{e}} = \frac{\frac{1}{e}}{\frac{e-1}{e}} = \frac{1}{e-1} \] ### Sum of the Second Series: \(\sum_{n=1}^{\infty} \frac{1}{n(n+1)}\) This series can be summed using partial fraction decomposition. We can write: \[ \frac{1}{n(n+1)} = \frac{A}{n} + \frac{B}{n+1} \] Solving for \(A\) and \(B\): \[ 1 = A(n+1) + Bn \implies 1 = An + A + Bn \implies 1 = (A + B)n + A \] Comparing coefficients, we get: \[ A + B = 0 \quad \text{and} \quad A = 1 \implies B = -1 \] Thus: \[ \frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1} \] Now, the series becomes a telescoping series: \[ \sum_{n=1}^{\infty} \left( \frac{1}{n} - \frac{1}{n+1} \right) \] Writing out the first few terms, we see: \[ \left( \frac{1}{1} - \frac{1}{2} \right) + \left( \frac{1}{2} - \frac{1}{3} \right) + \left( \frac{1}{3} - \frac{1}{4} \right) + \cdots \] All intermediate terms cancel out, leaving: \[ 1 - \frac{1}{\infty} = 1 - 0 = 1 \] ### Combining the Results Now, we combine the sums of the two series: \[ \sum_{n=1}^{\infty} \left( \frac{1}{e^n} + \frac{1}{n(n+1)} \right) = \frac{1}{e-1} + 1 \] Thus, the sum of the given series is: \[ \boxed{\frac{1}{e-1} + 1} \] |
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| gpto1 | To find the sum: \[ S = \sum_{n=1}^{\infty} \left( \frac{1}{e^n} + \frac{1}{n(n+1)} \right) \] We can split this into two separate sums: \[ S = \left( \sum_{n=1}^\infty \frac{1}{e^n} \right) + \left( \sum_{n=1}^\infty \frac{1}{n(n+1)} \right) \] **First Sum:** \[ S_1 = \sum_{n=1}^\infty \frac{1}{e^n} = \frac{1}{e} + \frac{1}{e^2} + \frac{1}{e^3} + \dots \] This is a geometric series with first term \( a = \frac{1}{e} \) and common ratio \( r = \frac{1}{e} \) (since \( e > 1 \), \( |r| < 1 \)), so it converges. The sum of an infinite geometric series is: \[ S = \frac{a}{1 - r} \] Applying this, we get: \[ S_1 = \frac{\frac{1}{e}}{1 - \frac{1}{e}} = \frac{1}{e - 1} \] **Second Sum:** \[ S_2 = \sum_{n=1}^\infty \frac{1}{n(n+1)} \] We can decompose the fraction using partial fractions: \[ \frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1} \] Thus, \[ S_2 = \sum_{n=1}^\infty \left( \frac{1}{n} - \frac{1}{n+1} \right) \] This series telescopes: \[ S_2 = \left( 1 - \frac{1}{2} \right) + \left( \frac{1}{2} - \frac{1}{3} \right) + \left( \frac{1}{3} - \frac{1}{4} \right) + \dots \] Most terms cancel out, leaving: \[ S_2 = \lim_{N \to \infty} \left( 1 - \frac{1}{N+1} \right) = 1 \] **Final Sum:** \[ S = S_1 + S_2 = \frac{1}{e - 1} + 1 \] **Answer: An explicit sum: S = (1 ⁄ (e – 1)) + 1; that is, S = 1 + 1⁄(e – 1) |
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