$\sum_{n=1}^{\infty} (-1)^n a_n$ or $\sum_{n=1}^{\infty} (-1)^{n+1} a_n$, where $a_n > 0$ for all $n$.

$\lim_{n \to \infty} a_n = 0$

$a_n$ is a decreasing sequence, i.e., $a_n > a_{n+1}$ for all $n$ beyond some index.

If $\lim_{n \to \infty} a_n = 0$ and $a_n$ is decreasing, then the alternating series $\sum (-1)^n a_n$ converges.

It ensures that the terms are getting smaller in magnitude, allowing the partial sums to converge.

The series diverges by the Divergence Test.

No, it only determines conditional convergence. It doesn't tell us if $\sum |a_n|$ converges.

$\cos(n\pi)$ is equivalent to $(-1)^n$, providing the alternating sign for the series.

1. Identify $a_n = \frac{1}{n}$. 2. Check $\lim_{n \to \infty} \frac{1}{n} = 0$. 3. Verify $\frac{1}{n} > \frac{1}{n+1}$. Since both conditions are met, the series converges.

1. Identify $a_n = \frac{n}{n+1}$. 2. Check $\lim_{n \to \infty} \frac{n}{n+1} = 1 \neq 0$. Since the limit is not 0, the series diverges.

1. Identify $a_n = \frac{1}{\ln(n)}$. 2. Check $\lim_{n \to \infty} \frac{1}{\ln(n)} = 0$. 3. Verify $\frac{1}{\ln(n)} > \frac{1}{\ln(n+1)}$. Since both conditions are met, the series converges.

1. Identify $a_n = \frac{n^2}{2^n}$. 2. Check $\lim_{n \to \infty} \frac{n^2}{2^n} = 0$ (using L'Hopital's rule). 3. Verify $\frac{n^2}{2^n} > \frac{(n+1)^2}{2^{n+1}}$ (true for large n). Since both conditions are met, the series converges.