Skip to main content
CalculusBy Educator Today

C46 · Unit C10 Sequences and series

Ratio test and convergence intervals

BC extensionReview editionNot yet human-reviewed
Time
About 45 minutes, untimed
Course framework reference
AP 10.8, 10.13 · BC
Builds on
Part of

Progress on this lesson counts in every pathway listed.

Understand the idea

The ratio test uses L equals lim|a sub n plus 1 /a sub n |: L<1 gives absolute convergence, L>1 gives divergence, and L equals 1 is inconclusive. Applied to a power series it usually determines an open interval |x minus c|<R. Test both endpoints separately with other tests. Radius R describes distance from the center; the interval may include one, both or neither endpoint.

Worked example

For sum sub n ₌ sub 1 ∞(x minus 2) to the power n /n, the ratio limit is |x minus 2|, giving |x minus 2|<1. At x equals 1 the series is alternating harmonic and converges; at x equals 3 it is harmonic and diverges. Thus the interval is [1,3), radius 1.

Explore the mathematics

Ratio test and convergence intervals

Compare consecutive terms and interpret the limiting ratio.

Setting 3 of 5

Scroll sideways to see the full diagram.

What does the ratio of consecutive Taylor terms tell you?xy-2-112eˣ and Taylor degree 3 about 0

Predict before continuing

What does the ratio of consecutive Taylor terms tell you?

Read the complete explanation
  1. What does the ratio of consecutive Taylor terms tell you?
  2. For xᵏ/k!, |aₖ₊₁/aₖ|=|x|/(k+1).
  3. The ratio tends to 0 for each fixed real x, so the exponential series converges for every real x.
  4. For a power series with a finite radius, test each endpoint separately. A ratio-test limit of 1 is inconclusive.
Ratio test and convergence intervals · visual worked method
Ratio test and convergence intervals · visual worked methodFor Σₙ₌₁∞(x−2)ⁿ/n, the ratio limit is |x−2|, giving |x−2|<1. At x=1 the series is alternating harmonic and converges; at x=3 it is harmonic and diverges. Thus the interval is [1,3), radius 1.Worked example · 1/2For Σₙ₌₁∞(x−2)ⁿ/n, the ratio limitis |x−2|, giving |x−2|<1. At x=1the series is alternating harmonicand converges; at x=3Worked example · 2/2it is harmonic and diverges. Thusthe interval is [1,3), radius 1.

For Σₙ₌₁∞(x−2)ⁿ/n, the ratio limit is |x−2|, giving |x−2|<1. At x=1 the series is alternating harmonic and converges; at x=3 it is harmonic and diverges. Thus the interval is [1,3), radius 1.

Discuss: explain what each label represents, then draw a related example with different values.

A common misconception

Ratio limit 1 does not prove divergence. Endpoint behavior is a separate part of a power-series answer.

Investigate and explain

Compare two power series with the same radius but different endpoint behavior. Explain why radius alone is incomplete.
Identify one step in the worked example, name the mathematical fact that permits it, and state its domain conditions.