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 . $$ n=\square $$

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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: $$ n = 1001 $$ (A) The error in the approximation $$ s \approx s_n $$ is bounded by $$ \frac{1}{n} $$. (B) The smallest value of $$ n $$ such that the upper bound is less than 0.001 is $$ n = 1001 $$.

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