Convergence of a Power Series vs Convergence at One x
A worked explanation of convergence of a power series vs convergence at one x, including the key limitation and a checkable example.
Convergence of a Power Series vs Convergence at One x
The same power series converges for some x-values and diverges for others because the terms cn(x - a)n grow or shrink depending on x. Apply the Ratio Test: lim |cn+1 / cn| · |x - a| = L|x - a|. Convergence requires L|x - a| < 1, which is a condition on x. Change x and you change the test's outcome. This is why the convergence of a power series vs convergence at one x is a variable-dependent property: the series is an exact object whose behavior shifts with the input.
Why the Same Series Behave Differently for Different Inputs
A power series centered at a is Σ cn (x - a)n. The coefficients cn are fixed, but the factor (x - a)n changes with x. For small |x - a|, the terms shrink fast enough for the infinite sum to settle to a finite limit. For large |x - a|, the terms grow and the sum blows up. The boundary is the radius of convergence R. Inside |x - a| < R the series converges absolutely; outside |x - a| > R it diverges. At the endpoints |x - a| = R, separate testing is required.
A Worked Example
Work through Σ xn / n. Apply the Ratio Test: limit of |an+1 / an| = |x| · n/(n+1), which approaches |x|. So the series converges for |x| < 1 and diverges for |x| > 1. At x = 0.5, the series becomes Σ (0.5)n / n. The Ratio Test gives 0.5 < 1, so it converges absolutely. At x = 2, the series becomes Σ 2n / n. The terms do not approach zero, they blow up, so the series diverges. The same power series yields two completely different outcomes based purely on the input x.
Calculation Steps
To calculate convergence for a new x, use a power series expansion calculator. Enter the series and the x-value, and the tool applies the Ratio or Root Test to the numeric series. For Σ xn / n at x = 0.5, the calculator confirms absolute convergence because the limit is 0.5. At x = -0.5, the series becomes Σ (-0.5)n / n, an alternating series. The Alternating Series Test applies: terms decrease to zero, so it converges conditionally. The same calculation at x = 1 gives Σ 1/n, the divergent harmonic series. At x = -1, the series is Σ (-1)n / n, which converges conditionally by the Alternating Series Test. Endpoint testing is essential because the Ratio Test is inconclusive at L = 1.
Why the Method Works
The ratio L|x - a| from the Ratio Test is the key. The coefficients cn determine L. The variable x determines |x - a|. Multiply them and you get a number that either falls below 1 (convergence) or exceeds 1 (divergence). That is why the exact same power series can converge for some inputs and diverge for others: the test's outcome is a product of two factors, one fixed and one variable. Changing x changes the product.
Failure Case: When the Ratio Test Is Inconclusive
For Σ xn / n2, the Ratio Test gives |x|, so R = 1. At x = 1, the series is Σ 1/n2, which converges (p-series, p = 2). At x = -1, the series is Σ (-1)n / n2, which converges absolutely. The Ratio Test correctly gave R = 1 but did not decide the endpoints. A common failure: students assume convergence at x = 1 because the Ratio Test limit was 1, but the test is inconclusive at that point. You must test endpoints separately using the p-series test or the Alternating Series Test.
| X-Value | Series Form | Test Applied | Result |
|---|---|---|---|
| 0.5 | Σ (0.5)<sup>n</sup> / n | Ratio Test | Converges absolutely |
| -0.5 | Σ (-0.5)<sup>n</sup> / n | Alternating Series Test | Converges conditionally |
| 1 | Σ 1/n | p-series test | Diverges (harmonic series) |
| -1 | Σ (-1)<sup>n</sup> / n | Alternating Series Test | Converges conditionally |
| 2 | Σ 2<sup>n</sup> / n | Divergence Test | Diverges (terms do not approach zero) |
What Most Often Goes Wrong
Newcomers apply the Ratio Test to a power series, get a limit L, compute R = 1/L, and stop. They never test the endpoints. That is the failure case. For Σ xn / n, the interval of convergence is [-1, 1), not [-1, 1]. At x = 1, the series is the divergent harmonic series. At x = -1, it converges conditionally. The Ratio Test gives the radius, not the interval. Always test x = a ± R separately using a p-series test, the Alternating Series Test, or a comparison test. If you skip endpoint testing, your interval is wrong.
Common Questions
Why does the same power series converge for one x and diverge for another?
The Ratio Test gives a condition L|x - a| < 1 for convergence. L is fixed from the coefficients; |x - a| changes with x. Different x-values push the product above or below the threshold.
Can a power series converge at every real x?
Yes. If the limit in the Ratio or Root Test is zero, the radius R is infinite. The series converges for all x. Example: Σ x<sup>n</sup> / n! has R = ∞.
What happens if the Ratio Test gives limit = 1?
The test is inconclusive. You must test the endpoints x = a ± R separately using other tests like the p-series test or the Alternating Series Test.
Does the Divergence Test ever prove convergence of a power series?
No. The Divergence Test only proves divergence when the limit of the terms is not zero. A zero limit does not guarantee convergence; the series could still diverge.