AP Calculus BC

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

The sum of negative 1 to the n over 3 to the n converges absolutely. It is geometric with common ratio negative one third, and the condition is that the size of r is under 1, not that r is under 1. The first term over 1 minus r gives the value, exactly negative one quarter.

∑n=1∞(−1)n3n\sum_{n=1}^{\infty}\frac{(-1)^{n}}{3^{n}}

Converges

sum=−14\text{sum} = -\frac{1}{4}

Settled by the geometric series test.

Fold the sign into the ratio

Both pieces carry the exponent nn, so they collapse into one power and the alternation stops being a separate feature to worry about.

(−1)n3n=(−13)n\frac{(-1)^{n}}{3^{n}} = \left(-\frac{1}{3}\right)^{n}

The common ratio is r=−13r = -\frac{1}{3}, so ∣r∣=13<1|r| = \frac{1}{3} < 1 and the series converges. Taking absolute values leaves a geometric series with ratio 13\frac{1}{3}, which converges, so the convergence is absolute rather than conditional.

The formula handles the signs itself

The first term, at n=1n = 1, is −13-\frac{1}{3}. Feed that and rr into a1−r\frac{a}{1-r} with no special adjustment for the alternating signs.

a1−r=−131−(−13)=−1343=−14\frac{a}{1-r} = \frac{-\frac{1}{3}}{1-\left(-\frac{1}{3}\right)} = \frac{-\frac{1}{3}}{\frac{4}{3}} = -\frac{1}{4}

The condition is on the size of r

Write the condition as the absolute value of r being less than 1. A ratio of negative 2 satisfies r less than 1 and the series diverges anyway, because the terms grow without bound while flipping sign.

The mistakes students make

Every common slip on this page involves the minus sign.

  • Stating the condition as r<1r < 1. It is ∣r∣<1|r| < 1, so r=−2r = -2 fails despite −2<1-2 < 1.
  • Using a=1a = 1. The sum starts at n=1n = 1, so the first term is −13-\frac{1}{3} and the answer is negative.
  • Calling this conditional convergence because the signs alternate. Take absolute values and you get a geometric series with ratio 13\frac{1}{3}, which converges, so the convergence is absolute.

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/3^n converge?

Yes, absolutely, to −14-\frac{1}{4}.

Can a geometric series have a negative ratio?

Yes. Only the size of rr matters, so any rr with ∣r∣<1|r| < 1 works and a1−r\frac{a}{1-r} still returns the sum.

Is this conditional convergence like the alternating harmonic series?

No. Stripping the signs here leaves a convergent geometric series, so the convergence is absolute.