| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | To solve the problem, we need to evaluate the infinite series $\sum_{n=1}^{\infty} (\cos(1/n^2) - \cos(1/(n+1)^2))$. Let's first simplify the expression inside the sum. Using the trigonometric identity for the cosine of a difference, we have: $\cos(1/n^2) - \cos(1/(n+1)^2) = -2\sin\left(\frac{\frac{1}{n^2} + \frac{1}{(n+1)^2}}{2}\right)\sin\left(\frac{\frac{1}{n^2} - \frac{1}{(n+1)^2}}{2}\right)$ Simplifying the arguments of the sine functions, we get: $-2\sin\left(\frac{\frac{1}{n^2} + \frac{1}{(n+1)^2}}{2}\right)\sin\left(\frac{\frac{1}{n^2} - \frac{1}{(n+1)^2}}{2}\right) = -2\sin\left(\frac{1}{2}\left(\frac{1}{n^2} + \frac{1}{(n+1)^2}\right)\right)\sin\left(\frac{1}{2}\left(\frac{1}{n^2} - \frac{1}{(n+1)^2}\right)\right)$ Let's denote $a_n = \frac{1}{2}\left(\frac{1}{n^2} + \frac{1}{(n+1)^2}\right)$ and $b_n = \frac{1}{2}\left(\frac{1}{n^2} - \frac{1}{(n+1)^2}\right)$. Now, we can rewrite the sum as: $\sum_{n=1}^{\infty} (\cos(1/n^2) - \cos(1/(n+1)^2)) = \sum_{n=1}^{\infty} -2\sin(a_n)\sin(b_n)$ To evaluate this sum, we can use the product-to-sum identity for sine: $\sin(a)\sin(b) = \frac{1}{2}(\cos(a-b) - \cos(a+b))$ Applying this identity to our sum, we get: $\sum_{n=1}^{\infty} -2\sin(a_n)\sin(b_n) = \sum_{n=1}^{\infty} -\cos(a_n - b_n) + \cos(a_n + b_n)$ Now, let's simplify the expressions for $a_n - b_n$ and $a_n + b_n$: $a_n - b_n = \frac{1}{2}\left(\frac{1}{n^2} + \frac{1}{(n+1)^2}\right) - \frac{1}{2}\left(\frac{1}{n^2} - \frac{1}{(n+1)^2}\right) = \frac{1}{(n+1)^2}$ $a_n + b_n = \frac{1}{2}\left(\frac{1}{n^2} + \frac{1}{(n+1)^2}\right) + \frac{1}{2}\left(\frac{1}{n^2} - \frac{1}{(n+1)^2}\right) = \frac{1}{n^2}$ Substituting these expressions back into the sum, we get: $\sum_{n=1}^{\infty} -\cos(a_n - b_n) + \cos(a_n + b_n) = \sum_{n=1}^{\infty} -\cos\left(\frac{1}{(n+1)^2}\right) + \cos\left(\frac{1}{n^2}\right)$ Now, we can see that the terms in the sum telescope, meaning that each term cancels out with the next term except for the first and last terms. Therefore, the sum simplifies to: $\sum_{n=1}^{\infty} -\cos\left(\frac{1}{(n+1)^2}\right) + \cos\left(\frac{1}{n^2}\right) = -\cos(0) + \cos(1) = \boxed{\cos(1) - 1}$ |
-0.459 |
| deepseekr1 | The given problem is to find the sum of the infinite series \(\sum_{n=1}^{\infty} \left( \cos\left(\frac{1}{n^2}\right) - \cos\left(\frac{1}{(n+1)^2}\right) \right)\). First, we observe that this is a telescoping series. Writing out the first few terms, we see: \[ \left( \cos\left(\frac{1}{1^2}\right) - \cos\left(\frac{1}{2^2}\right) \right) + \left( \cos\left(\frac{1}{2^2}\right) - \cos\left(\frac{1}{3^2}\right) \right) + \left( \cos\left(\frac{1}{3^2}\right) - \cos\left(\frac{1}{4^2}\right) \right) + \cdots \] Each term \(-\cos\left(\frac{1}{(n+1)^2}\right)\) cancels with the subsequent term \(\cos\left(\frac{1}{(n+1)^2}\right)\), leaving only the first term \(\cos(1)\) and the limit of the last term as \(n\) approaches infinity. The partial sum \(S_N\) up to \(N\) terms is: \[ S_N = \cos(1) - \cos\left(\frac{1}{(N+1)^2}\right) \] Taking the limit as \(N\) approaches infinity, we need to evaluate: \[ \lim_{N \to \infty} \left( \cos(1) - \cos\left(\frac{1}{(N+1)^2}\right) \right) \] As \(N\) becomes very large, \(\frac{1}{(N+1)^2}\) approaches 0, and \(\cos\left(\frac{1}{(N+1)^2}\right)\) approaches \(\cos(0) = 1\). Therefore, the limit is: \[ \cos(1) - 1 \] Thus, the sum of the series is \(\boxed{\cos 1 - 1}\). |
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| gpto1 | To compute the sum: \[ S = \sum_{n=1}^\infty \left(\cos\left(\frac{1}{n^2}\right) - \cos\left(\frac{1}{(n+1)^2}\right)\right) \] We can observe that this is a telescoping series. Let's consider the partial sums up to \( N \): \[ S_N = \sum_{n=1}^N \left(\cos\left(\frac{1}{n^2}\right) - \cos\left(\frac{1}{(n+1)^2}\right)\right) \] When we expand this sum, many terms cancel out: \[ \begin{align*} S_N &= \left[\cos\left(\frac{1}{1^2}\right) - \cos\left(\frac{1}{2^2}\right)\right] + \left[\cos\left(\frac{1}{2^2}\right) - \cos\left(\frac{1}{3^2}\right)\right] + \ldots \\ &\quad + \left[\cos\left(\frac{1}{N^2}\right) - \cos\left(\frac{1}{(N+1)^2}\right)\right] \end{align*} \] Most terms cancel, leaving: \[ S_N = \cos\left(\frac{1}{1^2}\right) - \cos\left(\frac{1}{(N+1)^2}\right) \] As \( N \to \infty \), \( \frac{1}{(N+1)^2} \to 0 \), so: \[ \lim_{N \to \infty} \cos\left(\frac{1}{(N+1)^2}\right) = \cos(0) = 1 \] Therefore, the sum converges to: \[ S = \lim_{N \to \infty} S_N = \cos(1) - 1 \] So the infinite series sums to \( \cos(1) - 1 \). **Answer:** **An explicit value: cos 1 – 1** |
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| GPT-4o | To find the sum of the series \(\sum_{n=1}^{\infty} (\cos(1/n^2) - \cos(1/(n+1)^2))\), we observe that this is a telescoping series. Let's first rewrite the series for clarity: \[ \sum_{n=1}^{\infty} \left(\cos\left(\frac{1}{n^2}\right) - \cos\left(\frac{1}{(n+1)^2}\right)\right) \] A telescoping series is one in which most terms cancel out with other terms in the series. To see this cancellation more clearly, let's write out the first few terms of the series: \[ \begin{aligned} &\left(\cos\left(\frac{1}{1^2}\right) - \cos\left(\frac{1}{2^2}\right)\right) + \left(\cos\left(\frac{1}{2^2}\right) - \cos\left(\frac{1}{3^2}\right)\right) + \left(\cos\left(\frac{1}{3^2}\right) - \cos\left(\frac{1}{4^2}\right)\right) + \cdots \end{aligned} \] Notice that each \(\cos\left(\frac{1}{n^2}\right)\) term for \(n \geq 2\) cancels with the \(- \cos\left(\frac{1}{(n+1)^2}\right)\) term from the previous term in the series. Thus, the series simplifies significantly because of this cancellation. Specifically, all intermediate terms cancel out, leaving us with only the first term of the first fraction and the last term of the last fraction: \[ \cos\left(\frac{1}{1^2}\right) - \lim_{n \to \infty} \cos\left(\frac{1}{(n+1)^2}\right) \] We know from the properties of the cosine function that \(\cos(x)\) approaches \(\cos(0) = 1\) as \(x\) approaches 0. Therefore, as \(n\) approaches infinity, \( \frac{1}{(n+1)^2} \) approaches 0, and thus \(\cos\left(\frac{1}{(n+1)^2}\right) \) approaches \(\cos(0) = 1\). Hence, the series sum simplifies to: \[ \cos(1) - 1 \] So, the sum of the series \(\sum_{n=1}^{\infty} (\cos(1/n^2) - \cos(1/(n+1)^2))\) is: \[ \boxed{\cos(1) - 1} \] |
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| llama3.1-70B-Instruct | The given series is a telescoping series, which means that most of the terms in the series cancel each other out, leaving only a few terms. We can rewrite the series as: $$\sum_{n=1}^{\infty} \left(\frac{1}{n^2} - \frac{1}{(n+1)^2}\right)$$ Now, let's expand the first few terms of the series: $$\left(\frac{1}{1^2} - \frac{1}{2^2}\right) + \left(\frac{1}{2^2} - \frac{1}{3^2}\right) + \left(\frac{1}{3^2} - \frac{1}{4^2}\right) + \cdots$$ As we can see, the terms start to cancel each other out: $$\frac{1}{1^2} - \frac{1}{2^2} + \frac{1}{2^2} - \frac{1}{3^2} + \frac{1}{3^2} - \frac{1}{4^2} + \cdots$$ The only term that remains is the first term, which is $\frac{1}{1^2} = 1$. Therefore, the sum of the series is: $$\sum_{n=1}^{\infty} \left(\frac{1}{n^2} - \frac{1}{(n+1)^2}\right) = 1$$ |
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| llama3.1-405B-Instruct | ## Step 1: Understand the problem The problem asks for the sum of an infinite series, which is the difference between the cost of 1/n^2 and the cost of 1/(n+1)^2 for each n from 1 to infinity. ## Step 2: Identify the pattern The cost function is not explicitly defined, but we can assume it is a decreasing function, meaning that cost(1/n^2) > cost(1/(n+1)^2) for all n. ## Step 3: Simplify the series We can rewrite the series as: $\sum_{n=1}^{\infty} (cost(1/n^2) - cost(1/(n+1)^2)) = \sum_{n=1}^{\infty} cost(1/n^2) - \sum_{n=1}^{\infty} cost(1/(n+1)^2)$ ## Step 4: Notice the telescoping series The second sum can be rewritten as: $\sum_{n=1}^{\infty} cost(1/(n+1)^2) = \sum_{n=2}^{\infty} cost(1/n^2)$ ## Step 5: Cancel out terms Now we can cancel out the terms in the two sums: $\sum_{n=1}^{\infty} cost(1/n^2) - \sum_{n=2}^{\infty} cost(1/n^2) = cost(1/1^2)$ ## Step 6: Evaluate the remaining term The only remaining term is cost(1/1^2), which is simply cost(1). The final answer is: $\boxed{1}$ |
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| DeepSeek-Coder-V2 | To solve the problem of finding the sum of \(\sum_{n=1}^{\infty} (cost(1/n^2) - cost(1/(n+1)^2))\), we need to understand the behavior of the function \(cost(x)\) and how it interacts with the series. Given the problem, it seems that \(cost(x)\) is a placeholder for some function that we need to define or assume certain properties for. For simplicity, let's assume \(cost(x)\) is a continuous function. ### Step-by-Step Solution: 1. **Understanding the Series:** The series is given by: \[ \sum_{n=1}^{\infty} (cost(1/n^2) - cost(1/(n+1)^2)) \] 2. **Telescoping Series:** Notice that the series is a telescoping series. In a telescoping series, most terms cancel out, leaving only the first and the last terms. 3. **Cancellation of Terms:** Let's write out the first few terms to see the cancellation: \[ (cost(1) - cost(1/4)) + (cost(1/4) - cost(1/9)) + (cost(1/9) - cost(1/16)) + \cdots \] Here, we see that each \(cost(1/(n+1)^2)\) term cancels with the \(cost(1/n^2)\) term of the next term in the series. 4. **Remaining Terms:** After all the cancellations, the only terms that remain are the first term's positive part and the last term's negative part. However, since the series goes to infinity, the last term's negative part effectively disappears. 5. **Sum of the Series:** Therefore, the sum of the series simplifies to: \[ cost(1) \] ### Conclusion: The sum of the series \(\sum_{n=1}^{\infty} (cost(1/n^2) - cost(1/(n+1)^2))\) is: \[ \boxed{cost(1)} \] |
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| Qwen2-72B-Instruct | To solve the problem, we need to evaluate the infinite series $\sum_{n=1}^{\infty} (\cos(1/n^2) - \cos(1/(n+1)^2))$. Let's first simplify the expression inside the sum. Using the trigonometric identity for the cosine of a difference, we have: $\cos(1/n^2) - \cos(1/(n+1)^2) = -2\sin\left(\frac{\frac{1}{n^2} + \frac{1}{(n+1)^2}}{2}\right)\sin\left(\frac{\frac{1}{n^2} - \frac{1}{(n+1)^2}}{2}\right)$ Simplifying the arguments of the sine functions, we get: $-2\sin\left(\frac{\frac{1}{n^2} + \frac{1}{(n+1)^2}}{2}\right)\sin\left(\frac{\frac{1}{n^2} - \frac{1}{(n+1)^2}}{2}\right) = -2\sin\left(\frac{1}{2}\left(\frac{1}{n^2} + \frac{1}{(n+1)^2}\right)\right)\sin\left(\frac{1}{2}\left(\frac{1}{n^2} - \frac{1}{(n+1)^2}\right)\right)$ Let's denote $a_n = \frac{1}{2}\left(\frac{1}{n^2} + \frac{1}{(n+1)^2}\right)$ and $b_n = \frac{1}{2}\left(\frac{1}{n^2} - \frac{1}{(n+1)^2}\right)$. Now, we can rewrite the sum as: $\sum_{n=1}^{\infty} (\cos(1/n^2) - \cos(1/(n+1)^2)) = \sum_{n=1}^{\infty} -2\sin(a_n)\sin(b_n)$ To evaluate this sum, we can use the product-to-sum identity for sine: $\sin(a)\sin(b) = \frac{1}{2}(\cos(a-b) - \cos(a+b))$ Applying this identity to our sum, we get: $\sum_{n=1}^{\infty} -2\sin(a_n)\sin(b_n) = \sum_{n=1}^{\infty} -\cos(a_n - b_n) + \cos(a_n + b_n)$ Now, let's simplify the expressions for $a_n - b_n$ and $a_n + b_n$: $a_n - b_n = \frac{1}{2}\left(\frac{1}{n^2} + \frac{1}{(n+1)^2}\right) - \frac{1}{2}\left(\frac{1}{n^2} - \frac{1}{(n+1)^2}\right) = \frac{1}{(n+1)^2}$ $a_n + b_n = \frac{1}{2}\left(\frac{1}{n^2} + \frac{1}{(n+1)^2}\right) + \frac{1}{2}\left(\frac{1}{n^2} - \frac{1}{(n+1)^2}\right) = \frac{1}{n^2}$ Substituting these expressions back into the sum, we get: $\sum_{n=1}^{\infty} -\cos(a_n - b_n) + \cos(a_n + b_n) = \sum_{n=1}^{\infty} -\cos\left(\frac{1}{(n+1)^2}\right) + \cos\left(\frac{1}{n^2}\right)$ Now, we can see that the terms in the sum telescope, meaning that each term cancels out with the next term except for the first and last terms. Therefore, the sum simplifies to: $\sum_{n=1}^{\infty} -\cos\left(\frac{1}{(n+1)^2}\right) + \cos\left(\frac{1}{n^2}\right) = -\cos(0) + \cos(1) = \boxed{\cos(1) - 1}$ |
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| baichuan4 | To find the sum of the series $\sum_{n=1}^{\infty} (\cos(1/n^2) - \cos(1/(n+1)^2))$, we can observe that it is an infinite telescoping series. This means that most of the terms will cancel out when we group them together, leaving us with just the first term minus the limit of the last term as $n$ goes to infinity. Let's write out the first few terms to see how they telescope: $\sum_{n=1}^{\infty} (\cos(1/n^2) - \cos(1/(n+1)^2)) = (\cos(1) - \cos(1/4)) + (\cos(1/4) - \cos(1/9)) + (\cos(1/9) - \cos(1/16)) + \ldots$ As you can see, all the intermediate terms like $\cos(1/4)$ and $\cos(1/9)$ will cancel out because they appear in both positive and negative forms in consecutive pairs. The only terms that do not cancel are the first term $\cos(1)$ and the last term $-\cos(1/(n+1)^2)$. So, the sum of the series simplifies to: $\sum_{n=1}^{\infty} (\cos(1/n^2) - \cos(1/(n+1)^2)) = \cos(1) - \lim_{n \to \infty} \cos(1/(n+1)^2)$ Now, we need to evaluate the limit: $\lim_{n \to \infty} \cos(1/(n+1)^2)$ As $n$ approaches infinity, $1/(n+1)^2$ approaches $0$, and we know that $\cos(0) = 1$. Therefore: $\lim_{n \to \infty} \cos(1/(n+1)^2) = 1$ Substituting this back into our sum, we get: $\sum_{n=1}^{\infty} (\cos(1/n^2) - \cos(1/(n+1)^2)) = \cos(1) - 1$ Since $\cos(1)$ is approximately $0.540302$, we have: $\sum_{n=1}^{\infty} (\cos(1/n^2) - \cos(1/(n+1)^2)) \approx 0.540302 - 1 = -0.459698$ So the sum of the series is approximately $-0.459698$. |
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| doubao | This is a telescoping series. Let's expand the terms: \[ \begin{align*} &\sum_{n=1}^{\infty} (\cos(\frac{1}{n^2}) - \cos(\frac{1}{(n+1)^2}))\\ =&(\cos(1) - \cos(\frac{1}{4})) + (\cos(\frac{1}{4}) - \cos(\frac{1}{9})) + (\cos(\frac{1}{9}) - \cos(\frac{1}{16})) + \cdots \end{align*} \] We can see that all the middle terms cancel out, leaving only the first term (cos(1)) and the last term (which approaches 0 as n goes to infinity). So the sum is simply cos(1). |
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| minimax | The given problem involves an infinite series with a telescoping nature. The term "cost" is not standard mathematical notation, so I will assume it is a typographical error or a placeholder for some function. However, since the problem doesn't define what "cost" is, I will proceed with the assumption that it is a function that can be subtracted and that the series converges. The series given is: $$\sum_{n=1}^{\infty} (f(1/n^2) - f(1/(n+1)^2))$$ This is a telescoping series, where each term cancels out a part of the subsequent term. To see this, let's write out the first few terms explicitly: $$f(1/1^2) - f(1/2^2) + f(1/2^2) - f(1/3^2) + f(1/3^2) - f(1/4^2) + \ldots$$ Notice that $f(1/2^2)$ cancels out in the first and second terms, $f(1/3^2)$ cancels out in the second and third terms, and so on. This pattern continues for all terms in the series. As a result, all intermediate terms cancel out, and we are left with only the first term of the first function evaluated at $1/1^2$ and the last term of the last function evaluated at $1/\infty^2$ (which is zero if $f$ is continuous at zero): $$\sum_{n=1}^{\infty} (f(1/n^2) - f(1/(n+1)^2)) = f(1/1^2) - \lim_{n \to \infty} f(1/(n+1)^2)$$ If we assume that $f$ is continuous and that $f(0) = 0$, then as $n$ approaches infinity, $1/(n+1)^2$ approaches zero, and the limit of $f(1/(n+1)^2)$ as $n$ approaches infinity is $f(0)$, which is zero. Therefore, the sum of the series is simply: $$f(1) - 0 = f(1)$$ Without the definition of the function "cost" or "f", we cannot provide a numerical value for the sum. If you can provide the definition of the function "cost" or "f", I can give you a more specific answer. |
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| kimi | The given problem is to find the sum of the series: $$\sum_{n=1}^{\infty} \left(\cos\left(\frac{1}{n^2}\right) - \cos\left(\frac{1}{(n+1)^2}\right)\right)$$ This is a telescoping series, where each term cancels out part of the previous term. Let's write out the first few terms to see how this works: $$\begin{align*} &\left(\cos\left(\frac{1}{1^2}\right) - \cos\left(\frac{1}{2^2}\right)\right) + \\ &\left(\cos\left(\frac{1}{2^2}\right) - \cos\left(\frac{1}{3^2}\right)\right) + \\ &\left(\cos\left(\frac{1}{3^2}\right) - \cos\left(\frac{1}{4^2}\right)\right) + \ldots \end{align*}$$ Notice that each $-\cos\left(\frac{1}{(n+1)^2}\right)$ cancels with the $\cos\left(\frac{1}{n^2}\right)$ of the next term. This leaves us with: $$\cos\left(\frac{1}{1^2}\right) - \lim_{n \to \infty} \cos\left(\frac{1}{(n+1)^2}\right)$$ As $n$ approaches infinity, $\frac{1}{(n+1)^2}$ approaches 0, and $\cos(x)$ approaches 1 when $x$ approaches 0. Therefore, the limit of the cosine terms as $n$ approaches infinity is 1. So the sum of the series is: $$\cos(1) - 1$$ This is the final answer. |
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| wenxinyiyan | 首先，我们观察这个级数，它是一个交错级数，形式为 $\sum_{n=1}^{\infty} (a_n - a_{n+1})$，其中 $a_n = cost(1/n^2)$。 为了求这个级数的和，我们可以先对 $cost(x)$ 进行泰勒展开，得到： $cost(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots$ 然后，我们将 $1/n^2$ 和 $1/(n+1)^2$ 分别代入上述泰勒展开式中，得到： $cost(1/n^2) = 1 - \frac{1}{2!n^4} + \frac{1}{4!n^8} - \frac{1}{6!n^{12}} + \cdots$ $cost(1/(n+1)^2) = 1 - \frac{1}{2!(n+1)^4} + \frac{1}{4!(n+1)^8} - \frac{1}{6!(n+1)^{12}} + \cdots$ 接下来，我们将 $cost(1/n^2) - cost(1/(n+1)^2)$ 展开，得到： $cost(1/n^2) - cost(1/(n+1)^2) = \frac{1}{2!n^4} - \frac{1}{2!(n+1)^4} + \frac{1}{4!n^8} - \frac{1}{4!(n+1)^8} + \cdots$ 现在，我们将上述级数按照 $n$ 的幂次进行整理，得到： $\sum_{n=1}^{\infty} (cost(1/n^2) - cost(1/(n+1)^2)) = \left( \frac{1}{2!1^4} - \frac{1}{2!2^4} \right) + \left( \frac{1}{2!2^4} - \frac{1}{2!3^4} \right) + \cdots + \left( \frac{1}{4!1^8} - \frac{1}{4!2^8} \right) + \left( \frac{1}{4!2^8} - \frac{1}{4!3^8} \right) + \cdots$ 观察上述级数，我们发现它是一个交错级数，且每一项的绝对值都在逐渐减小。因此，根据交错级数的性质，我们可以知道这个级数是收敛的。 最后，我们注意到每一项中都有一个正项和一个负项相抵消，只剩下第一个正项和最后一个负项。因此，这个级数的和为： $\sum_{n=1}^{\infty} (cost(1/n^2) - cost(1/(n+1)^2)) = \frac{1}{2!1^4} - \lim_{n \to \infty} \frac{1}{2!n^4} = \frac{1}{2} - 0 = \frac{1}{2}$ 所以，这个级数的和为 $\frac{1}{2}$。 |
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| zhipu | To find the sum of the infinite series \(\sum_{n=1}^{\infty} \left(\cos\left(\frac{1}{n^2}\right) - \cos\left(\frac{1}{(n+1)^2}\right)\right)\), we can use the telescoping series technique. A telescoping series is a series where most of the terms cancel out when you add them together, leaving only a few terms remaining. This often happens when each term in the series can be written as the difference of two terms. Let's examine the given series term by term: \[ \sum_{n=1}^{\infty} \left(\cos\left(\frac{1}{n^2}\right) - \cos\left(\frac{1}{(n+1)^2}\right)\right) \] Write out the first few terms to see the pattern: \[ \begin{align*} &\left(\cos\left(\frac{1}{1^2}\right) - \cos\left(\frac{1}{2^2}\right)\right) + \\ &\left(\cos\left(\frac{1}{2^2}\right) - \cos\left(\frac{1}{3^2}\right)\right) + \\ &\left(\cos\left(\frac{1}{3^2}\right) - \cos\left(\frac{1}{4^2}\right)\right) + \\ &\left(\cos\left(\frac{1}{4^2}\right) - \cos\left(\frac{1}{5^2}\right)\right) + \\ &\cdots \end{align*} \] Notice that each term \(\cos\left(\frac{1}{n^2}\right)\) cancels with the next term \(-\cos\left(\frac{1}{n^2}\right)\) in the series. This is the hallmark of a telescoping series. So, when we sum the series, all intermediate terms cancel out, leaving only the first term of the first pair and the last term of the last pair: \[ \cos\left(\frac{1}{1^2}\right) - \lim_{n \to \infty} \cos\left(\frac{1}{(n+1)^2}\right) \] Evaluate the limit as \(n\) approaches infinity: \[ \lim_{n \to \infty} \cos\left(\frac{1}{(n+1)^2}\right) \] As \(n\) approaches infinity, \(\frac{1}{(n+1)^2}\) approaches 0. The cosine function is continuous, so: \[ \cos\left(\lim_{n \to \infty} \frac{1}{(n+1)^2}\right) = \cos(0) = 1 \] Thus, the sum of the series is: \[ \cos\left(\frac{1}{1^2}\right) - 1 = \cos(1) - 1 \] Therefore, the sum of the series \(\sum_{n=1}^{\infty} \left(\cos\left(\frac{1}{n^2}\right) - \cos\left(\frac{1}{(n+1)^2}\right)\right)\) is: \[ \boxed{\cos(1) - 1} \] |
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| gpto1 | To compute the sum: \[ S = \sum_{n=1}^\infty \left(\cos\left(\frac{1}{n^2}\right) - \cos\left(\frac{1}{(n+1)^2}\right)\right) \] We can observe that this is a telescoping series. Let's consider the partial sums up to \( N \): \[ S_N = \sum_{n=1}^N \left(\cos\left(\frac{1}{n^2}\right) - \cos\left(\frac{1}{(n+1)^2}\right)\right) \] When we expand this sum, many terms cancel out: \[ \begin{align*} S_N &= \left[\cos\left(\frac{1}{1^2}\right) - \cos\left(\frac{1}{2^2}\right)\right] + \left[\cos\left(\frac{1}{2^2}\right) - \cos\left(\frac{1}{3^2}\right)\right] + \ldots \\ &\quad + \left[\cos\left(\frac{1}{N^2}\right) - \cos\left(\frac{1}{(N+1)^2}\right)\right] \end{align*} \] Most terms cancel, leaving: \[ S_N = \cos\left(\frac{1}{1^2}\right) - \cos\left(\frac{1}{(N+1)^2}\right) \] As \( N \to \infty \), \( \frac{1}{(N+1)^2} \to 0 \), so: \[ \lim_{N \to \infty} \cos\left(\frac{1}{(N+1)^2}\right) = \cos(0) = 1 \] Therefore, the sum converges to: \[ S = \lim_{N \to \infty} S_N = \cos(1) - 1 \] So the infinite series sums to \( \cos(1) - 1 \). **Answer:** **An explicit value: cos 1 – 1** |
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| deepseekv3 | Alright, I have this infinite series to sum up: \[ \sum_{n=1}^{\infty} \left( \cos\left(\frac{1}{n^2}\right) - \cos\left(\frac{1}{(n+1)^2}\right) \right) \] At first glance, it looks a bit intimidating, but I think I can break it down step by step. Let's see. ### Understanding the Series First, I need to understand what's inside the summation. It's the difference between two cosine terms: \[ \cos\left(\frac{1}{n^2}\right) - \cos\left(\frac{1}{(n+1)^2}\right) \] So, for each term in the series, we're taking the cosine of \( \frac{1}{n^2} \) and subtracting the cosine of \( \frac{1}{(n+1)^2} \). ### Writing Out the First Few Terms Maybe if I write out the first few terms, I can see if there's a pattern or if it telescopes. Let's compute the first few terms: 1. For \( n = 1 \): \[ \cos\left(\frac{1}{1^2}\right) - \cos\left(\frac{1}{(1+1)^2}\right) = \cos(1) - \cos\left(\frac{1}{4}\right) \] 2. For \( n = 2 \): \[ \cos\left(\frac{1}{2^2}\right) - \cos\left(\frac{1}{(2+1)^2}\right) = \cos\left(\frac{1}{4}\right) - \cos\left(\frac{1}{9}\right) \] 3. For \( n = 3 \): \[ \cos\left(\frac{1}{3^2}\right) - \cos\left(\frac{1}{(3+1)^2}\right) = \cos\left(\frac{1}{9}\right) - \cos\left(\frac{1}{16}\right) \] And so on. ### Observing the Pattern Looking at these terms, I notice that there's a cancellation happening: \[ \left( \cos(1) - \cos\left(\frac{1}{4}\right) \right) + \left( \cos\left(\frac{1}{4}\right) - \cos\left(\frac{1}{9}\right) \right) + \left( \cos\left(\frac{1}{9}\right) - \cos\left(\frac{1}{16}\right) \right) + \cdots \] Each \( \cos\left(\frac{1}{n^2}\right) \) term cancels with the \( -\cos\left(\frac{1}{n^2}\right) \) from the next term. This is called a telescoping series. ### Simplifying the Series In a telescoping series, most terms cancel out, and we're left with only the first and last terms. However, since this is an infinite series, the "last" term approaches a limit. Let's write the partial sum \( S_N \) up to \( N \): \[ S_N = \sum_{n=1}^{N} \left( \cos\left(\frac{1}{n^2}\right) - \cos\left(\frac{1}{(n+1)^2}\right) \right) \] Expanding this: \[ S_N = \left( \cos(1) - \cos\left(\frac{1}{4}\right) \right) + \left( \cos\left(\frac{1}{4}\right) - \cos\left(\frac{1}{9}\right) \right) + \cdots + \left( \cos\left(\frac{1}{N^2}\right) - \cos\left(\frac{1}{(N+1)^2}\right) \right) \] After cancellation, we get: \[ S_N = \cos(1) - \cos\left(\frac{1}{(N+1)^2}\right) \] ### Taking the Limit as \( N \) Approaches Infinity Now, to find the sum of the infinite series, we take the limit as \( N \) approaches infinity: \[ \lim_{N \to \infty} S_N = \lim_{N \to \infty} \left( \cos(1) - \cos\left(\frac{1}{(N+1)^2}\right) \right) \] As \( N \) becomes very large, \( \frac{1}{(N+1)^2} \) approaches 0. So, we need to find: \[ \cos(1) - \cos(0) \] We know that \( \cos(0) = 1 \), so: \[ \cos(1) - 1 \] ### Final Answer Therefore, the sum of the infinite series is: \[ \cos(1) - 1 \] Expressed concisely: \[ \boxed{\cos(1) - 1} \] ### Verification To ensure this is correct, let's verify the steps: 1. **Telescoping Pattern**: The series is indeed telescoping, as each \( \cos\left(\frac{1}{n^2}\right) \) cancels with the \( -\cos\left(\frac{1}{n^2}\right) \) from the next term. 2. **Partial Sum**: The partial sum \( S_N = \cos(1) - \cos\left(\frac{1}{(N+1)^2}\right) \) correctly represents the cancellation. 3. **Limit Calculation**: As \( N \to \infty \), \( \frac{1}{(N+1)^2} \to 0 \), and \( \cos(0) = 1 \), leading to the final result \( \cos(1) - 1 \). Everything seems to check out. \boxed{\cos(1) - 1} |
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