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Let f be a positive, continuous, and decreasing function for x ≥ 1, such that an...

Let f be a positive, continuous, and decreasing function for x ≥ 1, such that an = f(n). If the series ∞ an n = 1 converges to S, then the remainder RN = S − SN is bounded by 0 ≤ RN ≤ ∞ N f(x) dx. Use the result above to approximate the sum of the convergent series using the indicated number of terms. (Round your answers to four decimal places.) ∞ n = 1 1 n2 + 1 , twelve terms 0.9969 Correct: Your answer is correct. Include an estimate of the maximum error for your approximation. 0.06379 Incorrect: Your answer is incorrect. Need Help? Read It

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Answer #1

1)

\\S_n\approx S_{12}=\sum _{n=1}^{12}\:\frac{1}{n^2+1}=\frac{1}{2}+\frac{1}{5}+\frac{1}{10}+\frac{1}{17}+\frac{1}{26}+\frac{1}{37}+\frac{1}{50}+\frac{1}{65}+\frac{1}{82}+\frac{1}{101}+\frac{1}{122}+\frac{1}{145}=\frac{1492837748579}{1497500823325}=0.996886\approx0.9969

Answer: 0.9969

2)

\\Here,\:0\le R_N\le \int _N^{\infty }\:f\left(x\right)\:dx \\\because N=12 \\\:0\le R_{12}\le \int _{12}^{\infty }\frac{1}{x^2+1}\:dx \\Consider, \int _{12}^{\infty }\frac{1}{x^2+1}\:dx=\left[tan^{-1}\:\:\left(x\right)\right]^{\infty }_{12}=\frac{\pi \:}{2}-\arctan \:\left(12\right)=0.08314123188\approx0.0831

Answer: 0.0831

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