How to Use the Ratio Test and Radius of Convergence

For a power series, the ratio test and radius of convergence work together as a two-step routine: apply the ratio test to the coefficients, solve the resulting inequality for where the series converges, and the boundary of that interval is the radius R. Formally, for Σ aₙ(x − c)ⁿ, compute L = lim |aₙ₊₁/aₙ|, and then R = 1/L (with R = ∞ when L = 0 and R = 0 when L = ∞). In practice this reduces almost every problem to a single limit.

Why the Ratio Test Is the Standard Tool

The ratio test asks how the terms of a series change from one to the next. For a series Σ bₙ, you form the ratio |bₙ₊₁/bₙ| and take its limit as n → ∞. If that limit is less than 1, the series converges absolutely; if it’s greater than 1, it diverges; if it equals exactly 1, the test is inconclusive and you need a different method.

The reason it dominates power-series problems is that the variable itself survives the ratio. The x-terms usually cancel to a single power of |x − c| multiplied by the coefficient ratio, so the test’s condition “limit < 1” becomes a literal inequality in x. You aren’t just checking convergence at a number; you’re solving for every number at once. Paul’s Online Math Notes develops this logic in full and is worth reading alongside your textbook.

The Radius of Convergence Procedure

Here is the algorithm, and it never changes shape.

1. Write the series as Σ aₙ(x − c)ⁿ, where aₙ contains everything except the power of x. The center c is the number being subtracted inside the parentheses.
2. Form the ratio |aₙ₊₁(x − c)ⁿ⁺¹ / (aₙ(x − c)ⁿ)| and simplify. The (x − c)ⁿ usually cancels, leaving |x − c| times a leftover coefficient ratio.
3. Take the limit of the coefficient part as n → ∞. Call it L.
4. Set L|x − c| < 1 and solve: |x − c| < 1/L. That value, R = 1/L, is the radius of convergence.

The series is guaranteed to converge absolutely on the open interval (c − R, c + R) and guaranteed to diverge outside it. The only genuinely open question is the two endpoints, where |x − c| = R exactly and the limit equals 1, so the ratio test says nothing.

Worked Example

Take Σ (x − 2)ⁿ / n. Here aₙ = 1/n and c = 2. The ratio is |(x − 2)ⁿ⁺¹/(n+1)| · |n/(x − 2)ⁿ|, which simplifies to |x − 2| · n/(n + 1). As n → ∞, n/(n + 1) → 1, so L = 1 and R = 1/1 = 1. Convergence is guaranteed on (1, 3).

Now check the endpoints by hand. At x = 3 the series becomes Σ 1/n, the harmonic series, which diverges. At x = 1 it becomes Σ (−1)ⁿ/n, the alternating harmonic series, which converges. So the interval of convergence is [1, 3) — half-open, because the two ends behave differently. This is why you must never assume the endpoints share a verdict. MIT OpenCourseWare’s calculus sequence includes this exact style of endpoint check in its series lectures.

Shortcuts That Save Time

For coefficients built from factorials and powers, two limits cover nearly everything:

• lim (1 + 1/n)ⁿ = e, useful when the coefficient ratio produces (n/(n + 1))ⁿ-type expressions.
• For aₙ = kⁿ (a constant to the n), the ratio is k, so R = 1/|k|.

Recognizing these lets you skip the algebra and write R directly, which is exactly what Khan Academy’s practice sets train you to do.

What Can Go Wrong

Three traps account for most lost points.

The first is forgetting the center. If the series is Σ (x + 5)ⁿ, the center is c = −5, not 5. The interval is centered at −5.

The second is mishandling factorial ratios. Recall (n + 1)! = (n + 1)·n!, so the ratio (n + 1)!/n! collapses to n + 1, not 1. The same idea applies to (2n + 2)!/(2n)!, which collapses to (2n + 2)(2n + 1). Getting this wrong changes the limit entirely.

The third is writing the whole interval and quitting. The radius gives you (c − R, c + R) for free, but the interval of convergence may include one endpoint, both, or neither. Always test x = c − R and x = c + R as separate ordinary series. Common endpoint tools are the p-series test, the alternating series test, and the divergence test.

That third row is easy to forget: if the coefficients grow so fast that L is infinite, the series collapses to a single point. The exponential series Σ xⁿ/n! sits in the middle row — its ratio gives 1/(n + 1) → 0, so R = ∞ and it converges everywhere, which is one of the facts that makes eˣ = Σ xⁿ/n! so useful.

Comparing Radius and Interval

Students often use “radius” and “interval” interchangeably, but they measure different things. The radius R is a single nonnegative number — or the symbol ∞ — that describes the distance from the center to the boundary. The interval of convergence is the actual set of x-values where the series converges, including whichever endpoints survive testing. A series can share a radius with another series and still have a different interval if their endpoints behave differently, as the harmonic and alternating harmonic examples demonstrate.

Two checks keep the distinction straight. First, the radius tells you nothing about inclusion: (c − R, c + R) is always open, no matter what the endpoints do. Second, the interval is always centered at c, so if your interval is not symmetric about the center, you have either misidentified c or made a solving error. Khan Academy’s series units and the problem sets in Paul’s Online Math Notes both emphasize testing endpoints separately, since that step is where the radius stops being sufficient information.

Key Takeaways

• Compute L = lim |aₙ₊₁/aₙ|, then set R = 1/L; the interval is guaranteed on (c − R, c + R).
• L = 0 gives R = ∞ (converges everywhere); L = ∞ gives R = 0 (converges only at the center c).
• The ratio test is always inconclusive at the endpoints, so test x = c − R and x = c + R as separate series.
• Watch the center sign and factorial cancellations — those two slips cause the most errors.
• The result is written as an interval of convergence, which may be open, closed, or half-open depending on the endpoints.