Some series have signs that are neither all positive nor neatly alternating, like $\sum\frac{\sin n}{n^{2}}$. The way in is to look at the sizes $|a_n|$, where the comparison tests apply, and to remember the alternating series test for the cases where sizes alone are not enough.
Por qué funciona
If $\sum|a_n|$ converges, then so does $\sum a_n$. Here is why: $0 \le a_n + |a_n| \le 2|a_n|$, so $\sum(a_n + |a_n|)$ converges by comparison, and $a_n = (a_n + |a_n|) - |a_n|$ is the difference of two convergent series. Cancellation can only help a series converge, never hurt it.
Idea clave
$\sum a_n$ is absolutely convergent if $\sum|a_n|$ converges (then $\sum a_n$ converges too). It is conditionally convergent if $\sum a_n$ converges but $\sum|a_n|$ diverges. Otherwise it diverges.
- Step 1: test $\sum|a_n|$ (comparison, $p$-series, ratio test, …). If it converges, you are done: absolute convergence.
- Step 2: if $\sum|a_n|$ diverges, test $\sum a_n$ itself, usually with the alternating series test. If it converges: conditional.
- Step 3: if the terms do not go to 0, it diverges.
Ejemplo resuelto: Absolutely convergent
Classify $\sum_{n=1}^{\infty}\frac{(-1)^{n}}{n^{2}}$.
- $\sum\left|\frac{(-1)^{n}}{n^{2}}\right| = \sum\frac{1}{n^{2}}$ converges ($p = 2$).
- So the series converges absolutely.
Respuesta: Absolutely convergent
Ejemplo resuelto: Conditionally convergent
Classify $\sum_{n=1}^{\infty}\frac{(-1)^{n}}{\sqrt{n}}$.
- $\sum\frac{1}{\sqrt{n}}$ diverges ($p = \frac{1}{2}$), so it is not absolutely convergent.
- But $\frac{1}{\sqrt{n}}$ decreases to 0, so the alternating series converges.
- So it converges conditionally.
Respuesta: Conditionally convergent
Ejemplo resuelto: Irregular signs
Classify $\sum_{n=1}^{\infty}\frac{\sin n}{n^{2}}$.
- The signs follow no simple pattern, so the alternating series test does not apply.
- $\left|\frac{\sin n}{n^{2}}\right| \le \frac{1}{n^{2}}$, and $\sum\frac{1}{n^{2}}$ converges, so $\sum\left|\frac{\sin n}{n^{2}}\right|$ converges by comparison.
- So the series converges absolutely.
Respuesta: Absolutely convergent
Error común
Calling a series absolutely convergent because it passes the alternating series test. The alternating series test only shows convergence; $\sum\frac{(-1)^{n}}{n}$ passes it, yet $\sum\frac{1}{n}$ diverges, so that series is conditionally convergent. Absolute convergence is a statement about $\sum|a_n|$, and you have to test that series separately.
The difference matters: an absolutely convergent series has the same sum however you reorder its terms. A conditionally convergent one can be rearranged to add up to any number you like (Riemann's rearrangement theorem), so its sum depends on the order.