Which of the following is true for the infinite series $\sum _ { n = 1 } ^ { \infty } \frac { \cos ( n \pi ) } { n } ?$
A)Its sequence of partial sums is not bounded from above.
B)Its sequence of partial sums is decreasing.
C)It is absolutely convergent.
D)It is conditionally convergent.
E)It is divergent.

Question

Which of the following is true for the infinite series $\sum _ { n = 1 } ^ { \infty } \frac { \cos ( n \pi ) } { n } ?$ 
A)Its sequence of partial sums is not bounded from above.
B)Its sequence of partial sums is decreasing.
C)It is absolutely convergent.
D)It is conditionally convergent.
E)It is divergent.

Quizplus · Accepted Answer

The given series $\sum _ { n = 1 } ^ { \infty } \frac { \cos ( n \pi ) } { n }$ alternates between positive and negative terms because $\cos(n\pi)$ equals 1 for even $n$ and -1 for odd $n$, effectively making the series $\sum _ { n = 1 } ^ { \infty } \frac { (-1)^{n+1} } { n }$, which is the alternating harmonic series. This series is known to be conditionally convergent, meaning it converges, but not absolutely. Absolute convergence would require that the series of absolute values, $\sum _ { n = 1 } ^ { \infty } \frac { | \cos ( n \pi ) | } { n } = \sum _ { n = 1 } ^ { \infty } \frac { 1 } { n }$, also converges, which it does not (it's the harmonic series, which diverges).

Which of the Following Is True for the Infinite Series ∑n=1∞cos⁡(nπ)n?\sum _ { n = 1 } ^ { \infty } \frac { \cos ( n \pi ) } { n } ?∑n=1∞​ncos(nπ)​?

Which of the Following Is True for the Infinite Series $\sum _ { n = 1 } ^ { \infty } \frac { \cos ( n \pi ) } { n } ?$