A power series centered at a = 2 is found to have radius of convergence R = 3. A student concludes the interval of convergence is (−1, 5). What is wrong with this conclusion?
ANothing is wrong — (−1, 5) is the correct interval of convergence
BThe center and radius were applied incorrectly; the interval should be (−1, 5) is wrong and it should be (2−3, 2+3)
CThe endpoints x = −1 and x = 5 require separate testing before including or excluding them
DA radius of 3 means the interval has length 3, not 6, so the answer should be (2, 5)
The ratio or root test is inconclusive exactly at the boundary points |x − a| = R. At x = −1 and x = 5, the test gives limit = 1, which tells you nothing. Each endpoint produces a fixed numerical series that must be tested independently using other tests (alternating series, p-series, etc.). One endpoint might converge, one might diverge, or both might do either. The interval (−1, 5) is the open interior guaranteed by the ratio test, but the full interval of convergence could be (−1, 5), [−1, 5), (−1, 5], or [−1, 5].
Question 2 Multiple Choice
Applying the ratio test to the power series Σ xⁿ / n gives lim |aₙ₊₁/aₙ| = |x|. Which of the following correctly describes the interval of convergence?
A[−1, 1], because the series converges for all |x| ≤ 1
B(−1, 1), because the ratio test shows convergence for |x| < 1 and the endpoints are not worth checking
C[−1, 1), because x = −1 gives a convergent alternating series and x = 1 gives the divergent harmonic series
D(−1, 1], because x = 1 gives a convergent p-series and x = −1 gives a divergent series
The ratio test gives R = 1, guaranteeing convergence for |x| < 1 and divergence for |x| > 1. At x = 1, the series becomes Σ 1/n (harmonic series), which diverges. At x = −1, the series becomes Σ (−1)ⁿ/n (alternating harmonic series), which converges by the alternating series test. So the interval is [−1, 1) — left endpoint included, right excluded. This illustrates that each endpoint is an independent question.
Question 3 True / False
If the radius of convergence of a power series is R = 5, then the series converges for most x in the closed interval [a − 5, a + 5].
TTrue
FFalse
Answer: False
The radius of convergence guarantees convergence only in the open interval (a − 5, a + 5). The endpoints a − 5 and a + 5 lie exactly at the boundary where the ratio/root test is inconclusive (limit = 1). Each endpoint must be tested separately and may converge or diverge independently. The interval of convergence could be open, half-open, or closed at either end.
Question 4 True / False
The radius of convergence R and the interval of convergence are two distinct concepts: R is a non-negative number (or ∞), while the interval of convergence is a set of real numbers.
TTrue
FFalse
Answer: True
This is a crucial distinction students frequently blur. R = 5 tells you the half-length of the convergence interval, but it does not specify the interval — you still need to determine whether each endpoint is included. For the same R, the interval could be (a−5, a+5), [a−5, a+5), (a−5, a+5], or [a−5, a+5]. R is a scalar; the interval is a subset of ℝ.
Question 5 Short Answer
Why must the endpoints of the interval of convergence be tested separately, rather than being determined by the radius of convergence alone?
Think about your answer, then reveal below.
Model answer: At the endpoints x = a ± R, the ratio (or root) test yields a limit exactly equal to 1, which is the test's inconclusive case. The test only guarantees convergence when the limit is strictly less than 1 and divergence when strictly greater than 1. At the boundary, anything is possible: the series may converge or diverge at each endpoint independently. Each endpoint produces a fixed numerical series (with x replaced by a + R or a − R), which must be analyzed using other tests such as the alternating series test or p-series comparison.
The ratio test is derived from comparison with a geometric series: if the ratio of successive terms is eventually less than some r < 1, the series converges like a geometric series. At the boundary, the ratio approaches exactly 1, which means the comparison with a geometric series breaks down — successive terms are not shrinking fast enough to guarantee convergence, but they aren't growing either. The resulting boundary series may be anything from a p-series (which converges for p > 1, diverges for p ≤ 1) to an alternating series or something else entirely.