The series is convergent. (A). According to the Remainder Estimate for the Integral Test, the error in the approximation \( s \approx s_{n} \) (where \( s \) is the value of the infinite sum and \( s_{n} \) is the \( n \)-th partial sum) is \( \left|s-s_{n}\right| \leq \square \) (B). Find the smallest value of \( n \) such that this upper bound is less than 0.001 . \( n=\square \)

To solve the problem, we need to follow these steps: 1. **Identify the series and its properties**: We need to know the series in question to apply the Integral Test and find the upper bound for the error. 2. **Apply the Remainder Estimate for the Integral Test**: The error in the approximation \( s \approx s_n \) is given by the formula: \[ |s - s_n| \leq \int_{n}^{\infty} f(x) \, dx \] where \( f(x) \) is the function corresponding to the series. 3. **Find the smallest \( n \)** such that the upper bound is less than 0.001. Since the series is not specified in your question, I will assume a common series for demonstration purposes. Let's consider the series \( \sum_{k=1}^{\infty} \frac{1}{k^p} \) for \( p > 1 \). The function corresponding to this series is \( f(x) = \frac{1}{x^p} \). ### Step A: Calculate the upper bound for the error The integral from \( n \) to \( \infty \) is: \[ \int_{n}^{\infty} \frac{1}{x^p} \, dx \] Calculating this integral: \[ \int_{n}^{\infty} \frac{1}{x^p} \, dx = \left[ \frac{x^{1-p}}{1-p} \right]_{n}^{\infty} = 0 - \frac{n^{1-p}}{1-p} = \frac{n^{1-p}}{p-1} \] Thus, the error bound is: \[ |s - s_n| \leq \frac{n^{1-p}}{p-1} \] ### Step B: Find the smallest \( n \) such that this upper bound is less than 0.001 We need to solve the inequality: \[ \frac{n^{1-p}}{p-1} < 0.001 \] Rearranging gives: \[ n^{1-p} < 0.001(p-1) \] Taking the \( (1-p) \)-th root (noting that \( 1-p < 0 \) for \( p > 1 \)): \[ n > (0.001(p-1))^{\frac{1}{1-p}} \] Now, we need to choose a specific value for \( p \) to find \( n \). Let's assume \( p = 2 \) (which corresponds to the harmonic series of squares). Substituting \( p = 2 \): \[ n > (0.001(2-1))^{\frac{1}{1-2}} = (0.001)^{-1} = 1000 \] Thus, the smallest integer \( n \) satisfying this condition is: \[ n = 1001 \] ### Final Answers (A) The upper bound for the error is: \[ |s - s_n| \leq \frac{n^{1-2}}{2-1} = \frac{1}{n} \] (B) The smallest value of \( n \) such that this upper bound is less than 0.001 is: \[ n = 1001 \]