Most series do not have a formula for their partial sums, and many have terms that are hard to integrate. But you know a lot of benchmark series: geometric series and the $p$-series from the The Integral Test and p-Series. Comparison tests let you borrow their behavior.
Por qué funciona
Suppose $0 \le a_n \le b_n$. The partial sums of $\sum a_n$ increase (the terms are not negative) and never exceed $\sum b_n$. If that is finite, they are bounded, so they converge (by the monotone convergence theorem). If instead $\sum a_n$ is infinite, the bigger $\sum b_n$ must be infinite too.
For the limit version: if $\frac{a_n}{b_n} \to L$ with $0 < L < \infty$, then eventually $\frac{L}{2}b_n < a_n < \frac{3L}{2}b_n$, so each series is trapped by a multiple of the other, and they behave the same way.
Idea clave
Direct comparison ($0 \le a_n \le b_n$): $\sum b_n$ converges $\Rightarrow$ $\sum a_n$ converges; $\sum a_n$ diverges $\Rightarrow$ $\sum b_n$ diverges. Limit comparison ($a_n, b_n > 0$): if $\frac{a_n}{b_n} \to L$ with $0 < L < \infty$, both converge or both diverge.
| Your series is | than a series that | Conclusion |
|---|---|---|
| smaller | converges | converges |
| bigger | diverges | diverges |
| smaller | diverges | nothing |
| bigger | converges | nothing |
Ejemplo resuelto: Direct comparison
Does $\sum_{n=1}^{\infty}\frac{1}{n^{2} + 5}$ converge?
- $n^{2} + 5 > n^{2}$, so $0 < \frac{1}{n^{2} + 5} < \frac{1}{n^{2}}$.
- $\sum\frac{1}{n^{2}}$ converges ($p = 2$), so the smaller series converges.
Respuesta: It converges.
Ejemplo resuelto: Limit comparison
Does $\sum_{n=1}^{\infty}\frac{2n^{2} + 1}{n^{4} + 3}$ converge?
- For large $n$ the terms behave like $\frac{2n^{2}}{n^{4}} = \frac{2}{n^{2}}$, so compare with $b_n = \frac{1}{n^{2}}$.
- $\frac{a_n}{b_n} = \frac{2n^{4} + n^{2}}{n^{4} + 3} \to 2$, a positive finite number.
- $\sum\frac{1}{n^{2}}$ converges, so this series converges too.
Respuesta: It converges.
Error común
Comparing in the wrong direction. "$\frac{1}{n + 1} < \frac{1}{n}$, and $\sum\frac{1}{n}$ diverges, so $\sum\frac{1}{n + 1}$ converges" is not valid: being smaller than a divergent series proves nothing (and in fact $\sum\frac{1}{n + 1}$ diverges). Before concluding, say the inequality out loud and check it against the table above. When the inequality goes the useless way, try limit comparison instead.
Ejemplo resuelto: Picking the comparison
Does $\sum_{n=1}^{\infty}\frac{1}{3n - 1}$ converge?
- The terms behave like $\frac{1}{3n}$. Limit comparison with $\frac{1}{n}$: $\frac{n}{3n - 1} \to \frac{1}{3}$.
- $\sum\frac{1}{n}$ diverges, so this series diverges.
- (Direct comparison also works here: $\frac{1}{3n - 1} > \frac{1}{3n}$, which is bigger than a divergent series.)
Respuesta: It diverges.