Math  /  Calculus

QuestionFind the sum of the convergent series by using a well-known function. (Round your answer to four decimal places.) n=0(1)n192n+1(2n+1)\sum_{n=0}^{\infty}(-1)^{n} \frac{1}{9^{2 n+1}(2 n+1)}

Studdy Solution
Calculate the sum of the series using the arctangent function:
arctan(19)\arctan\left(\frac{1}{9}\right)
Use a calculator to find the value:
arctan(19)0.110657\arctan\left(\frac{1}{9}\right) \approx 0.110657
Round the result to four decimal places:
0.1107\boxed{0.1107}

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