Determine whether the series from 0 to infinity of cos(nπ)/(n + 3) converges conditionally or absolutely.
A. The series diverges.
B. The series converges conditionally but not absolutely.
C. The series converges absolutely but not conditionally.
D. The series converges absolutely.
I think the answer is B. (Also, when is C actually possible?)
I agree
To determine whether the series from 0 to infinity of cos(nπ)/(n + 3) converges conditionally or absolutely, we can use the alternating series test and the comparison test.
First, let's apply the alternating series test. For an alternating series to converge, two conditions must be met:
1. The terms must alternate in sign.
2. The absolute value of terms must decrease as n increases.
In this series, the terms alternate in sign since we have the cosine function. However, let's examine the absolute value of the terms to see if they decrease:
|cos(nπ)/(n + 3)| = |(-1)^(n)(1/(n + 3))|
The term 1/(n + 3) is always positive for n ≥ 0. Therefore, the absolute value of the terms in the series is always 1/(n + 3).
Unfortunately, the absolute value of the terms does not decrease as n increases. So, the alternating series test does not provide conclusive evidence about the convergence.
Next, let's apply the comparison test. We can compare the series of cos(nπ)/(n + 3) with the series of 1/(n + 3) since |cos(nπ)/(n + 3)| ≤ 1/(n + 3) for all n.
The series of 1/(n + 3) is a well-known p-series with p = 1. The p-series converges if p > 1 and diverges if p ≤ 1. In this case, p = 1, so the series of 1/(n + 3) diverges.
Since the series of cos(nπ)/(n + 3) is not absolutely convergent, we can conclude that it converges conditionally but not absolutely.
So, your answer, B, is correct.
Regarding option C, a series can only converge conditionally if it does not converge absolutely. Convergence without absolute convergence is possible when the series has alternating terms or oscillating terms that cancel each other out when taking the absolute value.