AP Calculus BC

Does the Sum of (-1)^n/n^2 Converge?

The sum of negative 1 to the n over n squared converges, and converges ABSOLUTELY. Taking absolute values gives the sum of 1 over n squared, a p-series with p equal to 2, which converges. Its value is negative pi squared over 12.

n=1(1)nn2\sum_{n=1}^{\infty}\frac{(-1)^{n}}{n^{2}}

Converges

sum=π212\text{sum} = -\frac{\pi^{2}}{12}

Settled by the alternating series test.

Test the absolute series first

When a series alternates, checking absolute convergence first is usually the faster route, because absolute convergence implies convergence outright.

n=1(1)nn2=n=11n2=π26\sum_{n=1}^{\infty}\left|\frac{(-1)^{n}}{n^{2}}\right| = \sum_{n=1}^{\infty}\frac{1}{n^{2}} = \frac{\pi^{2}}{6}

That is the convergent pp-series with p=2p = 2, so the original converges absolutely. The alternating series test would also work, but it would give the weaker conclusion.

n=1(1)nn2=π212\sum_{n=1}^{\infty}\frac{(-1)^{n}}{n^{2}} = -\frac{\pi^{2}}{12}

Half the magnitude, opposite sign

The alternating sum is exactly half the size of the absolute one and negative, since π212\frac{\pi^{2}}{12} is half of π26\frac{\pi^{2}}{6}. That is not a coincidence: the even-indexed terms of the absolute series account for the difference.

Absolute beats conditional

Reporting absolute convergence when it holds is strictly more informative, and it is what lets you rearrange terms freely. Only fall back on the alternating series test when the absolute series diverges.

The mistakes students make

  • Calling it conditional because it alternates. Alternating says nothing about which kind of convergence you have.
  • Stopping at the alternating series test. It proves convergence but misses the stronger, more useful fact.

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^2 converge?

Yes, absolutely, to π212-\frac{\pi^{2}}{12}.

Why absolutely rather than conditionally?

Because the absolute series is 1n2\sum \frac{1}{n^{2}}, a convergent pp-series with p=2p = 2.