Question

For the convergent alternating series ∑[infinity]ₖ₌₀ (-1)ᵏ/(3k+2)³, evaluate the nth partial sum for n=2. Then find an upper bound for the error S-Sₙ.

Answer 1

The nth partial sum of the series for n=2 is calculated by evaluating the first three terms and summing them, which results in S2 = 1/8 - 1/125 + 1/512. To estimate the error, we use the magnitude of the first omitted term, leading to an upper bound for the error of 1/1331.

Step-by-step explanation:

The question asks to evaluate the nth partial sum for n=2 of the convergent alternating series ∑[infinity]ₖ₀ (-1)ʳ/(3k+2)³ and then to find an upper bound for the error S-Sₙ. To find the nth partial sum for n=2, we sum the first three terms (k=0, 1, 2) of the series:

• When k=0: (-1)°/(3(0)+2)³ = 1/2³ = 1/8
• When k=1: (-1)¹/(3(1)+2)³ = -1/5³ = -1/125
• When k=2: (-1)²/(3(2)+2)³ = 1/8³ = 1/512

Summing these values gives the partial sum S₂ = 1/8 - 1/125 + 1/512.

To find an upper bound for the error S-Sₙ, we use the fact that the error in estimating the sum by the nth partial sum in an alternating series is less than the absolute value of the first omitted term. Hence, for n=2, the next term (k=3) provides the upper bound for the error:

Upper bound for error = |(-1)³/(3(3)+2)³| = |(-1)¹/(11)³| = 1/1331.