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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

STEP 1

1. The series given is an alternating series.
2. The series resembles the Taylor series expansion for a known function.
3. We will use the arctangent function, which has a known series expansion, to find the sum.

STEP 2

1. Identify the known function whose series expansion matches the given series.
2. Use the known function to find the sum of the series.
3. Calculate the sum to four decimal places.

STEP 3

Recognize that the series
∑n=0∞(−1)n192n+1(2n+1)\sum_{n=0}^{\infty}(-1)^{n} \frac{1}{9^{2n+1}(2n+1)}
resembles the Taylor series expansion for the arctangent function:
arctan⁡(x)=∑n=0∞(−1)nx2n+12n+1\arctan(x) = \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{2n+1}
for ∣x∣≤1|x| \leq 1.

STEP 4

Match the given series to the arctangent series. Notice that x=19x = \frac{1}{9} in the given series because:
(19)2n+1=192n+1\left(\frac{1}{9}\right)^{2n+1} = \frac{1}{9^{2n+1}}
Thus, the series is:
arctan⁡(19)\arctan\left(\frac{1}{9}\right)

STEP 5

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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