You now have a full toolbox: the The Divergence Test, geometric and telescoping sums, the The Integral Test and p-Series and $p$-series, comparison tests, the alternating series test, and the ratio and root tests. Choosing well saves a lot of work.
Why it works
Each test is built to detect one kind of structure. Terms that do not shrink: divergence test. Terms that are a fixed ratio apart: geometric. Terms like a power of $n$: compare with a $p$-series. Factorials and exponentials: the ratio test, because the ratio cancels most of them. Whole expressions raised to the $n$th power: the root test undoes the power. Alternating signs: the alternating series test. So look at the shape of the term first, and the test usually picks itself.
Key idea
Ask in order: (1) Do the terms go to 0? (2) Is it geometric, a $p$-series or telescoping? (3) Is it alternating? (4) Factorials or $c^{n}$: ratio test. $(\ldots)^{n}$: root test. (5) A rational or root expression: compare (limit comparison) with a $p$-series. (6) An easy integral: integral test.
| The term looks like | Try |
|---|---|
| $\frac{n^{2}}{3n^{2} + 1}$ (limit not 0) | divergence test |
| $\frac{3^{n}}{4^{n}}$, $ar^{n}$ | geometric series |
| $\frac{n + 1}{n^{3} + 2}$, $\frac{\sqrt{n}}{n^{2} + 1}$ | limit comparison with $\frac{1}{n^{p}}$ |
| $\frac{(-1)^{n}}{\ln n}$ | alternating series test |
| $\frac{n^{5}}{2^{n}}$, $\frac{n!}{10^{n}}$ | ratio test |
| $\left(\frac{n}{2n + 1}\right)^{n}$ | root test |
| $\frac{1}{n\ln n}$, $ne^{-n^{2}}$ | integral test |
Worked example: Quick decisions
Decide whether each converges: (a) $\sum\frac{n}{n^{2} + 1}$, (b) $\sum\frac{n^{2}}{2^{n}}$, (c) $\sum\frac{(-1)^{n}}{\ln n}$ ($n \ge 2$), (d) $\sum\left(1 + \frac{1}{n}\right)^{n}$.
- (a) Like $\frac{1}{n}$: limit comparison gives $\frac{n^{2}}{n^{2} + 1} \to 1$, so it diverges.
- (b) Ratio test: $\frac{(n + 1)^{2}}{2n^{2}} \to \frac{1}{2} < 1$, so it converges.
- (c) $\frac{1}{\ln n}$ decreases to 0, so it converges by the alternating series test (conditionally, since $\frac{1}{\ln n} > \frac{1}{n}$).
- (d) The terms tend to $e \ne 0$, so it diverges.
Answer: (a) diverges, (b) converges, (c) converges, (d) diverges
Common mistake
Reaching for the ratio test on everything. On $\sum\frac{n}{n^{2} + 1}$ it returns $L = 1$ and tells you nothing, after some algebra. The ratio test pays off when there are factorials or exponentials; for rational expressions, compare with a $p$-series right away. And always run the divergence test first: it takes seconds.
Worked example: Combining tests
Does $\sum_{n=1}^{\infty}\frac{(-1)^{n}n}{n^{2} + 4}$ converge absolutely, conditionally, or not at all?
- Sizes: $\frac{n}{n^{2} + 4}$ behaves like $\frac{1}{n}$ (limit comparison gives 1), so $\sum|a_n|$ diverges.
- The sizes go to 0 and decrease once $n \ge 2$ (the derivative of $\frac{x}{x^{2} + 4}$ is $\frac{4 - x^{2}}{(x^{2} + 4)^{2}} \le 0$ there).
- So it converges by the alternating series test: conditionally.
Answer: Conditionally