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Question
Find a bound for the error when approximating the series
\[
\sum_{n=1}^{\infty} \frac{6(-1)^{n+1}}{4 n^{3}}
\]
by the partial sum $S_{2}$.
Round your answer to six decimal plajees if necessary.

Step 1 ：The given series is an alternating series. The error in approximating the sum of an alternating series by a partial sum is less than or equal to the absolute value of the next term.

Step 2 ：In this case, we are approximating the series by the partial sum \(S_2\), so the error is less than or equal to the absolute value of the third term.

Step 3 ：Calculate the third term: \(\frac{6(-1)^{3+1}}{4 \cdot 3^{3}} = \frac{1}{18}\)

Step 4 ：The absolute value of the third term is \(\left|\frac{1}{18}\right| = \frac{1}{18}\)

Step 5 ：Rounding the absolute value of the third term to six decimal places gives 0.055556.

Step 6 ：Final Answer: The bound for the error when approximating the series by the partial sum \(S_2\) is \(\boxed{0.055556}\).