AP Calculus BC

Does the Sum of (-1)^n/n^3 Converge? Yes, Absolutely

The sum of negative 1 to the n over n cubed converges ABSOLUTELY. Take absolute values and the series becomes the p-series with p equal to 3, which converges because p is greater than 1. Absolute convergence implies convergence, so the verdict is settled in two lines.

∑n=1∞(−1)nn3\sum_{n=1}^{\infty}\frac{(-1)^{n}}{n^{3}}

Converges

Settled by the alternating series test.

Take absolute values first

The signs disappear the moment you take magnitudes, and what is left is a series you already know.

∑n=1∞∣(−1)nn3∣=∑n=1∞1n3\sum_{n=1}^{\infty}\left|\frac{(-1)^{n}}{n^{3}}\right| = \sum_{n=1}^{\infty}\frac{1}{n^{3}}

A p-series converges when p>1p > 1, and p=3p = 3 here. So ∑∣an∣\sum |a_{n}| converges, and a series whose absolute version converges must itself converge. That is the entire argument.

The alternating test applies too, and tells you less

The magnitudes 1n3\frac{1}{n^{3}} decrease to 00, so the alternating series test is available. Its conclusion is that the series converges, full stop. It cannot distinguish this series from ∑(−1)nn\sum \frac{(-1)^{n}}{n}, which also passes the test but converges only conditionally.

Name the argument you used

If a question asks only whether the series converges, either route earns the mark. If it asks whether the convergence is absolute or conditional, the alternating series test cannot answer, and you have to look at the series of absolute values.

The mistakes students make

All three come from applying a rule outside the conditions it was stated under.

  • Answering conditionally convergent because the series alternates. Conditional requires ∑∣an∣\sum |a_{n}| to diverge, and this one converges.
  • Calling ∑(−1)nn3\sum \frac{(-1)^{n}}{n^{3}} a p-series. The p-series rule is stated for positive terms, so take absolute values before quoting it.
  • Writing the rule as p≥1p \ge 1. The harmonic series has p=1p = 1 and diverges, so the cutoff is strict.

Not sure which test a series wants?

The Convergence Test Chooser walks the decision in order: nth term first, then geometric and p-series pattern matching, then alternating structure, then the ratio test, and finally the comparison family.

Frequently asked questions

Does the sum of (-1)^n/n^3 converge?

Yes, absolutely, because ∑1n3\sum \frac{1}{n^{3}} is a convergent p-series.

Do I need the alternating series test here?

You can use it, and it does give convergence, but it stops there. It cannot tell you the convergence is absolute, so you still have to look at ∑1n3\sum \frac{1}{n^{3}}.

What is the sum?

There is no elementary closed form. The related constant ∑1n3\sum \frac{1}{n^{3}} is Apery's constant, and no AP question will ask for its value.