How to determine whether an infinite series converges or diverges?

Here is a systematic decision tree for series convergence tests. **Step 1: Check $\lim_{n \ o \infty} a_n$.** If the limit is not zero (or does not exist), the series diverges (Divergence Test). **Step 2: Look at the form of $a_n$.** **Is it a $p$-series $\sum 1/n^p$?** - Converges if $p > 1$, diverges if $p \leq 1$ **Is it a geometric series $\sum ar^{n-1}$?** - Converges if $|r| < 1$, diverges if $|r| \geq 1$ **Does it contain $n!$ or $c^n$?** → Use the **Ratio Test**: $\lim |a_{n+1}/a_n| < 1$ converges, $> 1$ diverges **Does it contain $n^n$ or $n!$?** → Use the **Root Test**: $\lim \sqrt[n]{|a_n|} < 1$ converges, $> 1$ diverges **Does it look like $\int f(n)$?** → Use the **Integral Test** if $a_n = f(n)$ where $f$ is positive, continuous, decreasing. **Is it an alternating series?** → Use the **Alternating Series Test**: if $|a_n|$ decreases to $0$, it converges (conditionally). **Can you bound it by a simpler series?** → Use **Comparison Test** or **Limit Comparison Test**. **Applications:** 1. $\sum \frac{n^2}{2^n}$: Ratio Test $$\lim \frac{(n+1)^2/2^{n+1}}{n^2/2^n} = \lim \frac{(n+1)^2}{2n^2} = \frac12 < 1 \implies \ ext{converges}$$ 2. $\sum \frac{1}{n \ln n}$: Integral Test $$\int_2^{\infty} \frac{dx}{x \ln x} = \lim_{b \ o \infty} [\ln(\ln x)]_2^b = \infty \implies \ ext{diverges}$$ 3. $\sum \frac{(-1)^n}{n}$: Alternating Series Test Terms $1/n$ decrease to $0 \implies$ converges conditionally (harmonic series $\sum 1/n$ diverges).