AP Calculus BC

Does the Sum of (-1)^n/n^4 Converge? Yes, Absolutely

The series converges, and absolutely rather than conditionally, because stripping the signs leaves a p-series with p = 4 that converges on its own. Its exact sum is negative seven pi to the fourth over 720.

∑n=1∞(−1)nn4\sum_{n=1}^{\infty}\frac{(-1)^{n}}{n^{4}}

Converges

sum=−7π4720\text{sum} = -\frac{7\pi^{4}}{720}

Settled by the alternating series test.

Check absolute convergence first

With an alternating series, test the absolute version before reaching for the alternating series test. Here ∑∣(−1)nn4∣=∑1n4\sum \left|\frac{(-1)^{n}}{n^{4}}\right| = \sum \frac{1}{n^{4}} is a p-series with p=4>1p = 4 > 1, so it converges.

That single step gives the stronger conclusion. Absolute convergence implies convergence, so the alternating series test is not needed at all, and the answer carries more information: the series stays convergent even with the signs removed or rearranged.

The exact value

The absolute version sums to ζ(4)=π4/90\zeta(4) = \pi^{4}/90. The alternating version is the eta function at 4, related by η(4)=(1−21−4)ζ(4)=78⋅π490=7π4720\eta(4) = \left(1 - 2^{1-4}\right)\zeta(4) = \frac{7}{8}\cdot\frac{\pi^{4}}{90} = \frac{7\pi^{4}}{720}.

∑n=1∞(−1)nn4=−7π4720≈−0.9470\sum_{n=1}^{\infty}\frac{(-1)^{n}}{n^{4}} = -\frac{7\pi^{4}}{720} \approx -0.9470

The leading minus sign is there because the index starts at n=1n = 1, where (−1)1=−1(-1)^{1} = -1, so the first and largest term is negative.

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

Is this series conditionally convergent?

No, absolutely convergent. Conditional convergence means the absolute version DIVERGES, as with ∑(−1)n/n\sum (-1)^{n}/n. Here the absolute version is a convergent p-series.

Why check absolute convergence first?

Because it is usually one step and gives a stronger result. Only when the absolute version diverges do you need the alternating series test to rescue the original.