1. Is $\sum_{m-1}^{\infty} \frac{\cos n \pi}{n}$ absolute convergent, convergent, or divergent? Explain why. [20 points]

Step 1 ：The series in question is \(\sum_{m-1}^{\infty} \frac{\cos n \pi}{n}\). This is a series of complex numbers, where the real part is \(\cos n \pi\) and the imaginary part is 0.

Step 2 ：To determine whether this series is absolutely convergent, convergent, or divergent, we can use the comparison test. The comparison test states that if \(0 \leq a_n \leq b_n\) for all \(n\) and the series \(\sum_{n=1}^{\infty} b_n\) converges, then the series \(\sum_{n=1}^{\infty} a_n\) also converges.

Step 3 ：We can compare the series \(\sum_{m-1}^{\infty} \frac{\cos n \pi}{n}\) with the series \(\sum_{m-1}^{\infty} \frac{1}{n}\), which is a harmonic series and is known to be divergent.

Step 4 ：If the series \(\sum_{m-1}^{\infty} \frac{\cos n \pi}{n}\) is less than or equal to the series \(\sum_{m-1}^{\infty} \frac{1}{n}\), then the series \(\sum_{m-1}^{\infty} \frac{\cos n \pi}{n}\) is also divergent.

Step 5 ：We know that \(-1 \leq \cos n \pi \leq 1\) for all \(n\). Therefore, \(-\frac{1}{n} \leq \frac{\cos n \pi}{n} \leq \frac{1}{n}\) for all \(n\). Taking the absolute value, we get \(0 \leq |\frac{\cos n \pi}{n}| \leq \frac{1}{n}\) for all \(n\).

Step 6 ：Therefore, the series \(\sum_{m-1}^{\infty} \frac{\cos n \pi}{n}\) is less than or equal to the series \(\sum_{m-1}^{\infty} \frac{1}{n}\), and therefore, the series \(\sum_{m-1}^{\infty} \frac{\cos n \pi}{n}\) is divergent.

Step 7 ：Final Answer: The series \(\sum_{m-1}^{\infty} \frac{\cos n \pi}{n}\) is \(\boxed{\text{divergent}}\).