AP Calculus BC

Does the Sum of (-1)^n/5^n Converge? Yes, to -1/6

The series converges absolutely, to exactly -1/6. The alternating signs make no difference to the verdict: this is geometric with ratio -1/5, and a geometric series converges whenever the size of the ratio is less than one.

∑n=1∞(−1)n5n\sum_{n=1}^{\infty}\frac{(-1)^{n}}{5^{n}}

Converges

sum=−16\text{sum} = -\frac{1}{6}

Settled by the geometric series test.

Do not reach for the alternating series test

The signs alternate, so the alternating series test would work. It is the wrong tool anyway. Folding the sign into the ratio gives r=−1/5r = -1/5, and the geometric test then delivers something the alternating test cannot: the exact sum.

It also delivers a stronger verdict. The alternating test proves convergence; the geometric test with ∣r∣<1|r| < 1 proves ABSOLUTE convergence, because ∑∣(−1/5)n∣=∑(1/5)n\sum |(-1/5)^{n}| = \sum (1/5)^{n} converges too.

The sum, and the sign

−1/51−(−1/5)=−1/56/5=−16\frac{-1/5}{1 - (-1/5)} = \frac{-1/5}{6/5} = -\frac{1}{6}

The one minus r in the denominator becomes one PLUS a fifth, which is where the 6 comes from. Dropping that sign is the most common slip on negative-ratio geometric series and turns −1/6-1/6 into −1/4-1/4.

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 or absolutely convergent?

Absolutely. Stripping the signs leaves ∑(1/5)n\sum (1/5)^{n}, which is geometric with ratio 1/51/5 and converges. Conditional convergence only happens when the absolute version diverges.

Why not use the alternating series test here?

It would give the right verdict and nothing else. The geometric test gives the exact sum and the stronger absolute conclusion, for the same effort.