$\sum_{n=1}^{\infty} (-1)^n a_n$ 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.

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.

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