$\sum_{n=1}^{\infty} (-1)^n \frac{1}{n^2}$

$\sum_{n=1}^{\infty} (-1)^n \frac{\ln n}{n}$

$\sum_{n=1}^{\infty} (-1)^n \frac{\sin(n)}{n^{1.5}}$

$\sum_{n=1}^{\infty} (-1)^n \frac{1}{n^{0.9}}$