AP Calculus BC

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

The sum of negative 1 to the n over 2 to the n, starting at n equals 1, converges to negative one third. It is geometric with ratio negative one half, and since the absolute value of that ratio is less than 1 the series converges absolutely.

∑n=1∞(−1)n2n\sum_{n=1}^{\infty}\frac{(-1)^{n}}{2^{n}}

Converges

sum=−13\text{sum} = -\frac{1}{3}

Settled by the geometric series test.

A negative ratio is still geometric

Combining the two powers gives a single base, which happens to be negative. Nothing about the geometric test changes.

(−1)n2n=(−12)n\frac{(-1)^{n}}{2^{n}} = \left(-\frac{1}{2}\right)^{n}

The test asks only about ∣r∣|r|, and ∣−12∣=12<1\left|-\frac{1}{2}\right| = \frac{1}{2} < 1. The sum formula then handles the sign automatically.

∑n=1∞(−12)n=−121−(−12)=−1232=−13\sum_{n=1}^{\infty}\left(-\frac{1}{2}\right)^{n} = \frac{-\frac{1}{2}}{1-\left(-\frac{1}{2}\right)} = \frac{-\frac{1}{2}}{\frac{3}{2}} = -\frac{1}{3}

Absolute, not merely conditional

Taking absolute values gives ∑12n\sum \frac{1}{2^{n}}, which converges to 11. Since the absolute series converges, this one converges ABSOLUTELY, a stronger statement than the alternating series test would give.

Alternating is not the same as conditional

Plenty of alternating series converge absolutely. Conditional convergence requires the absolute version to DIVERGE, which is what separates the alternating harmonic series from this one.

The mistakes students make

  • Requiring r>0r > 0. The test is on ∣r∣|r|, so a negative ratio is fine.
  • Writing 1−r1 - r as 1−121 - \frac{1}{2} and losing the double negative. With r=−12r = -\frac{1}{2} the denominator is 32\frac{3}{2}, not 12\frac{1}{2}.
  • Calling it conditionally convergent just because the signs alternate.

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/2^n converge?

Yes, absolutely, to −13-\frac{1}{3} from n=1n = 1.

Can the common ratio be negative?

Yes. Only ∣r∣<1|r| < 1 matters, and the sum formula handles the sign through the 1−r1 - r denominator.