AP Calculus BC

Does the Sum of 1/3^n Converge? Yes, to 1/2

The sum of 1 over 3 to the n, starting at n equals 1, converges to exactly one half. It is geometric with first term one third and common ratio one third, and since the ratio is less than 1 in absolute value the sum formula applies.

∑n=1∞13n\sum_{n=1}^{\infty}\frac{1}{3^{n}}

Converges

sum=12\text{sum} = \frac{1}{2}

Settled by the geometric series test.

Applying the sum formula

First term a=13a = \frac{1}{3}, ratio r=13r = \frac{1}{3}, and ∣r∣<1|r| < 1.

∑n=1∞13n=131−13=1323=12\sum_{n=1}^{\infty}\frac{1}{3^{n}} = \frac{\frac{1}{3}}{1-\frac{1}{3}} = \frac{\frac{1}{3}}{\frac{2}{3}} = \frac{1}{2}

A smaller ratio means faster decay and a smaller total. Compare ∑12n=1\sum \frac{1}{2^{n}} = 1: halving the ratio from 12\frac{1}{2} to 13\frac{1}{3} halves the sum.

The mistakes students make

  • Answering 13\frac{1}{3}, which is the first term rather than the sum.
  • Forgetting to simplify the compound fraction and leaving 1/32/3\frac{1/3}{2/3} unresolved.
  • Treating 13n\frac{1}{3^{n}} as a pp-series. The variable is in the EXPONENT, which makes it geometric, not a power.

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 1/3^n converge?

Yes, to 12\frac{1}{2} starting from n=1n = 1.

Is 1/3^n a p-series?

No. A pp-series has nn in the base; here nn is the exponent, which makes it geometric.