Find the infinite sum accurate to three decimal places.
-$$\sum _ { n = 1 } ^ { \infty } ( - 1 ) ^ { n } \frac { 1 } { ( 4 n + 1 ) ^ { 3 } }$$
A) 0.007
B) -0.010
C) -0.007
D) 0.010

Question

Quizplus · Accepted Answer

Using the alternating series test, we can see that the given series is convergent. Let's use the formula for the sum of an infinite geometric series:
$$\sum _ { n = 1 } ^ { \infty } ( - 1 ) ^ { n } \frac { 1 } { ( 4 n + 1 ) ^ { 3 } } = \frac{1}{125} - \frac{1}{729} + \frac{1}{4913} - \cdots$$
We can use the formula for the sum of an alternating series to approximate the sum to three decimal places:
$$|R_{n+1}| = \frac{1}{(4n+5)^3} < \frac{1}{4^3(n+1)^3}$$
$$|S-S_n| \leq |R_{n+1}| < \frac{1}{4^3(n+1)^3}$$
We want:
$$\frac{1}{4^3(n+1)^3} < 0.0005$$
$$n > 10$$
So, we can use the sum of the first 11 terms to approximate the sum to three decimal places:
$$\sum _ { n = 1 } ^ { \infty } ( - 1 ) ^ { n } \frac { 1 } { ( 4 n + 1 ) ^ { 3 } } \approx \frac{1}{125} - \frac{1}{729} + \frac{1}{4913} - \cdots - \frac{1}{23298085122481} \approx -0.007$$
Therefore, the best choice is $\boxed{	extbf{(C)}}.$

Find the Infinite Sum Accurate to Three Decimal Places ∑n=1∞(−1)n1(4n+1)3\sum _ { n = 1 } ^ { \infty } ( - 1 ) ^ { n } \frac { 1 } { ( 4 n + 1 ) ^ { 3 } }n=1∑∞​(−1)n(4n+1)31​

Find the Infinite Sum Accurate to Three Decimal Places $\sum _ { n = 1 } ^ { \infty } ( - 1 ) ^ { n } \frac { 1 } { ( 4 n + 1 ) ^ { 3 } }$