AP Calculus BC

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

The sum of 1 over 2 to the n, starting at n equals 1, converges to exactly 1. It is geometric with first term one half and common ratio one half, and a geometric series converges whenever the absolute value of the ratio is less than 1.

∑n=1∞12n\sum_{n=1}^{\infty}\frac{1}{2^{n}}

Converges

sum=1\text{sum} = 1

Settled by the geometric series test.

Geometric, and the sum formula applies

Each term is exactly half the one before, so the ratio is constant, which is what makes a series geometric.

∑n=1∞arn−1=a1−r,∣r∣<1\sum_{n=1}^{\infty} ar^{n-1} = \frac{a}{1-r}, \qquad |r| < 1

Here the first term is a=12a = \frac{1}{2} and the ratio is r=12r = \frac{1}{2}.

∑n=1∞12n=121−12=1\sum_{n=1}^{\infty}\frac{1}{2^{n}} = \frac{\frac{1}{2}}{1 - \frac{1}{2}} = 1

The starting index changes the sum, not the verdict

Starting at n = 0 instead adds the term 1 and gives a sum of 2. Convergence depends only on r; the value depends on where you start, which is why the formula uses the FIRST term actually present.

The mistakes students make

  • Using a=1a = 1 when the sum starts at n=1n = 1. The first term present is 12\frac{1}{2}, not 11.
  • Writing a1+r\frac{a}{1+r}. The denominator is 1−r1 - r, and a negative ratio takes care of itself through that subtraction.
  • Applying the formula when ∣r∣≥1|r| \ge 1. There is no sum in that case; the series diverges.

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 1/2^n converge?

Yes, to exactly 11 when the sum starts at n=1n = 1.

What if it starts at n = 0?

Then the extra term 11 is included and the sum is 22. The verdict is unchanged.

Why does a geometric series converge only for |r| < 1?

Because rn→0r^{n} \to 0 exactly when ∣r∣<1|r| < 1; otherwise the terms fail to vanish and the nth term test already forces divergence.