Find an approximation of the sum of the series accurate to two decimal places. $\sum _ { n = 1 } ^ { \infty } \frac { ( - 1 ) ^ { n } } { n ^ { 3 } }$
A) -0.83
B) -0.90
C) -0.96
D) -1.02

Question

Quizplus · Accepted Answer

Using the Alternating Series Test, we can show that the series converges. We can use the alternating series error bound to find an approximation of the sum:
$|S-S_n|<a_{n+1}$ where $S$ is the actual sum, $S_n$ is the sum of the first $n$ terms, and $a_{n+1}$ is the absolute value of the $(n+1)$th term.
Thus, for $n\geq1$, we have
$|S-S_n|<\frac{1}{(n+1)^3}$
To approximate the sum accurate to two decimal places, we want to find the smallest $n$ such that
$\frac{1}{(n+1)^3}<0.005$
Solving for $n$, we get $n>54.76$, so we choose $n=55$. Then,
$S_{55}=-\frac{1}{1^3}+\frac{1}{2^3}-\frac{1}{3^3}+\frac{1}{4^3}-\cdots+\frac{(-1)^{55}}{55^3}$
is an approximation of the sum accurate to two decimal places.
Using a calculator or computer program, we get $S_{55}\approx -0.899862$, which rounded to two decimal places is $-0.90$. Therefore, the best choice is B.

Find an Approximation of the Sum of the Series Accurate ∑n=1∞(−1)nn3\sum _ { n = 1 } ^ { \infty } \frac { ( - 1 ) ^ { n } } { n ^ { 3 } }∑n=1∞​n3(−1)n​

Find an Approximation of the Sum of the Series Accurate $\sum _ { n = 1 } ^ { \infty } \frac { ( - 1 ) ^ { n } } { n ^ { 3 } }$