Approximate the sum of the series by using the first six terms. $\sum _ { n = 1 } ^ { \infty } \frac { ( - 1 ) ^ { n + 1 } 2 } { n ^ { 3 } }$
A) $1.799

Question

Approximate the sum of the series by using the first six terms. $\sum _ { n = 1 } ^ { \infty } \frac { ( - 1 ) ^ { n + 1 } 2 } { n ^ { 3 } }$
A) $1.799 < S < 1.805$
B) $1.794 < S < 1.806$
C) $1.794 < S < 1.805$
D) $1.792 < S < 1.792$
E)$1.698 < S < 1.902$

Quizplus · Accepted Answer

To approximate the sum of the series using the first six terms, we calculate each term and sum them up. The series is alternating, and its nth term is given by $\frac { ( - 1 ) ^ { n + 1 } 2 } { n ^ { 3 } }$. Calculating the first six terms:1st term: $ \frac{2}{1^3} = 2 $2nd term: $ -\frac{2}{2^3} = -\frac{1}{4} = -0.25 $3rd term: $ \frac{2}{3^3} = \frac{2}{27} \approx 0.074 $4th term: $ -\frac{2}{4^3} = -\frac{1}{32} = -0.0625 $5th term: $ \frac{2}{5^3} = \frac{2}{125} = 0.016 $6th term: $ -\frac{2}{6^3} = -\frac{1}{108} \approx -0.0185 $Sum of the first six terms: $2 - 0.25 + 0.074 - 0.0625 + 0.016 - 0.0185 \approx 1.759 $Since the series is alternating and decreasing in absolute value, the error in using the first six terms is less than the absolute value of the seventh term, which is $ \frac{2}{7^3} \approx 0.0041 $. Therefore, the actual sum is within 0.0041 of our approximation, making option C, $1.794 < S < 1.805$, the closest range that could contain the true sum based on our approximation and considering the error margin.

Approximate the Sum of the Series by Using the First ∑n=1∞(−1)n+12n3\sum _ { n = 1 } ^ { \infty } \frac { ( - 1 ) ^ { n + 1 } 2 } { n ^ { 3 } }∑n=1∞​n3(−1)n+12​

Approximate the Sum of the Series by Using the First $\sum _ { n = 1 } ^ { \infty } \frac { ( - 1 ) ^ { n + 1 } 2 } { n ^ { 3 } }$