The series for absolute and conditional convergence

Question

$$\sum_{k=1}^{\infty} \frac{\cos(\frac{\pi k}{3})}{2^{\ln(k)}} $$

Answer 1

HINT

We have that

$$\sum_{k=1}^{\infty} \frac{\cos(\frac{\pi k}{3})}{2^{\ln(k)}}=\sum_{k=1}^{\infty} \frac{(-1)^{3k}}{2^{\ln(3k)}}+\frac12\sum_{k=0}^{\infty} \frac{(-1)^{1+3k}}{2^{\ln(1+3k)}}-\frac12\sum_{k=0}^{\infty} \frac{(-1)^{2+3k}}{2^{\ln(2+3k)}}$$

$$\sum_{k=1}^{\infty} \left| \frac{\cos(\frac{\pi k}{3})}{2^{\ln(k)}} \right|\ge \frac12 \sum_{k=1}^{\infty} \frac{1}{2^{\ln(k)}} $$