Source: Integral Calculus, University of Texas at Austin
Tags: power series, radius of convergence, interval of convergence, ratio test for power series, root test for power series, endpoint convergence
Difficulty: Intermediate to Advanced | Prerequisites: Series convergence tests (Ratio Test, Root Test, Comparison Test, Alternating Series Test), geometric series, p-series.
Power series are infinite series with a variable in them, and they are the bridge between the "does this series converge?" questions from the first half of the unit and the Taylor/Maclaurin series that come next. The central questions here are: for which values of x does a given power series converge, and how do you find the radius and full interval of convergence? This material builds directly on the convergence tests you already know, applying them with x treated as a parameter.
Every power series Σcₙ(x − a)ⁿ has a radius of convergence R. It converges absolutely for |x − a| < R and diverges for |x − a| > R. The endpoints x = a ± R must be checked individually, usually with the Alternating Series Test, p-series comparison, or the Divergence Test. The Ratio Test or Root Test applied to the general term (with x left as a variable) is the standard method for finding R.
Power series
An infinite series of the form Σcₙ(x − a)ⁿ, where cₙ are the coefficients, a is the centre, and x is the variable. In simple terms, it is a polynomial with infinitely many terms, but it only "works" (converges) for certain values of x.
Centre (of a power series)
The value a in Σcₙ(x − a)ⁿ. The series always converges at x = a (every term is zero except the constant). The interval of convergence is symmetric about this point.
Radius of convergence (R)
The distance from the centre within which the series converges absolutely. If R = 3 and the centre is a = 4, the series converges for |x − 4| < 3, that is, for 1 < x < 7. R can be 0 (converges only at the centre), a positive number, or ∞ (converges everywhere).
Interval of convergence
The set of all x values for which the power series converges. It is always an interval centred at a, with radius R, but the endpoints may or may not be included. You must test each endpoint separately.
Endpoint convergence
At x = a + R and x = a − R, the series might converge or diverge. Substituting each endpoint into the series gives a numerical series, which you test with the tools from the convergence-tests unit (p-series, alternating series, divergence test).
The standard procedure is to apply the Ratio Test to the general term, treating x as a parameter.
Step 1. Write out |aₙ₊₁/aₙ| and simplify, keeping |x − a| (or |x|) as a factor.
Step 2. Take the limit as n → ∞. You will get L = |x − a| · (some constant or expression). For convergence, set L < 1 and solve for |x − a| < R.
Step 3. R is the reciprocal of whatever multiplies |x − a| in the limit.
Worked examples from HW12:
Q14: Σ n⁴(x − 3)ⁿ. Ratio: |aₙ₊₁/aₙ| = ((n+1)/n)⁴ · |x − 3| → |x − 3|. Converges when |x − 3| < 1, so R = 1.
Q16: Σ xⁿ/(n+2)! Ratio: |aₙ₊₁/aₙ| = |x|/(n+3) → 0 for any x. The series converges for all x, so R = ∞.
Q17: Σ xⁿ/(6ⁿn⁵). Ratio: |aₙ₊₁/aₙ| = |x|/6 · (n/(n+1))⁵ → |x|/6. Converges when |x| < 6, so R = 6.
Q20: Σ (−3)ⁿ/(³√n) · (x − 4)ⁿ. Ratio gives |aₙ₊₁/aₙ| → 3|x − 4|. Converges when |x − 4| < 1/3, so R = 1/3.
When to use the Root Test instead. If the general term has the form [f(n)]ⁿ or [g(x)]ⁿ, the Root Test may be cleaner. For Σ (−5)ⁿxⁿ / ⁴√(n+5), the nth root gives |−5x| · 1/(n+5)^(1/(4n)) → 5|x|. So R = 1/5.
Once you have R, the interior of the interval is determined: (a − R, a + R). The real work is checking whether each endpoint is included.
At each endpoint, substitute x into the series. You get a numerical series. Test it for convergence.
Common endpoint outcomes:
The series becomes an alternating series that satisfies the Leibniz conditions → the endpoint is included.
The series becomes a p-series with p > 1 → converges, endpoint included.
The series becomes the harmonic series or a p-series with p ≤ 1 → diverges, endpoint excluded.
The terms do not go to zero → diverges by the Divergence Test, endpoint excluded.
Worked examples from HW12:
Q12: Σ (−5)ⁿxⁿ / ⁴√(n+5). R = 1/5, so check x = 1/5 and x = −1/5. At x = 1/5: Σ (−1)ⁿ / ⁴√(n+5), an alternating series with decreasing terms going to zero → converges. At x = −1/5: Σ 1/⁴√(n+5), which behaves like Σ 1/n^(1/4), a p-series with p = 1/4 < 1 → diverges. Interval: [−1/5, 1/5).
Q13: Σ xⁿ/√(n+4). R = 1. At x = 1: Σ 1/√(n+4), p-series p = 1/2 → diverges. At x = −1: Σ (−1)ⁿ/√(n+4), alternating with decreasing terms → converges. Interval: [−1, 1).
Q15: Σ n³(x − 4)ⁿ. R = 1. At both endpoints x = 3 and x = 5, the terms n³(±1)ⁿ do not approach zero → diverges at both endpoints. Interval: (3, 5).
Q18: Σ (−1)ⁿ xⁿ/(2n+5). R = 1. At x = 1: Σ (−1)ⁿ/(2n+5), alternating, decreasing, limit zero → converges. At x = −1: Σ 1/(2n+5), behaves like Σ 1/n → diverges. Interval: (−1, 1].
Q19: Σ xⁿ/(7ⁿn³). R = 7. At x = 7: Σ 1/n³, p-series p = 3 → converges. At x = −7: Σ (−1)ⁿ/n³, absolutely convergent. Interval: [−7, 7].
Q21: Σ xⁿ/(n+6)! R = ∞. The series converges for all x. Interval: (−∞, ∞).
When the series is Σcₙ(x − a)ⁿ with a ≠ 0, the only thing that changes is where the interval sits on the number line. The radius R is determined entirely by the coefficients cₙ, not by the centre a.
The Ratio or Root Test gives you R from the coefficients.
The interval is then (a − R, a + R), centred at a.
Endpoint checks are the same process: substitute x = a − R and x = a + R.
Example: Q20, Σ (−3)ⁿ/(³√n) · (x − 4)ⁿ. The centre is a = 4. The ratio of coefficients gives R = 1/3. The interval before endpoint checks is (4 − 1/3, 4 + 1/3) = (11/3, 13/3).
Q22 tests a key idea: if Σcₙxⁿ has radius of convergence R, what is the radius of convergence of Σcₙx²ⁿ?
The substitution is u = x². The series Σcₙuⁿ converges for |u| < R, which means |x²| < R, that is, |x| < √R.
So the new radius of convergence is √R.
In Q22, R = 9 for the original series, so the radius for Σcₙx²ⁿ is √9 = 3.
This principle generalises: replacing x with x^k in a power series replaces R with R^(1/k).
Q23 tests a related idea. If Σcₙ7ⁿ converges, the radius of convergence of Σcₙxⁿ is at least 7 (the series converges at x = 7, so R ≥ 7). Then Σcₙ(−6)ⁿ has |−6| = 6 < 7, which is strictly inside the radius, so it converges absolutely. We cannot say anything definitive about Σcₙ(−7)ⁿ because x = −7 is on the boundary: it might converge or diverge depending on the coefficients.
Radius via Ratio Test: R = 1 / lim |cₙ₊₁/cₙ|, provided the limit exists. If the limit is 0, R = ∞. If the limit is ∞, R = 0.
Radius via Root Test: R = 1 / lim |cₙ|^(1/n).
Interval structure: Always (a − R, a + R) before endpoint checks. Endpoints tested individually.
Substitution rule: If Σcₙxⁿ has radius R, then Σcₙx^(kn) has radius R^(1/k).
Factorial coefficient rule: If cₙ = 1/(n+k)! for any fixed k, then R = ∞. Factorials in the denominator always beat any polynomial or exponential in the numerator.
Inside the radius: If Σcₙaⁿ converges and |b| < |a|, then Σcₙbⁿ converges absolutely.
Students often assume the interval of convergence is always open, forgetting to check the endpoints. The endpoints can go either way, and the interval can be open, closed, half-open on the left, or half-open on the right.
Students sometimes confuse the radius with the interval. R = 5 does not mean the interval is (−5, 5). If the centre is a = 3, the interval is (−2, 8) before endpoint checks.
Students think that if the series has (n+2)! or (n+6)! in the denominator, R must equal 2 or 6. It does not. The factorial dominates everything, and R = ∞ regardless of the shift inside the factorial.
Students forget that convergence at one endpoint tells you nothing about the other. Each endpoint is its own separate series with its own convergence behaviour.
⚠️ Endpoint checking is where most points are lost. Always check both endpoints, even if it seems obvious.
⚠️ Know the three possible R values: 0 (converges only at centre), a finite positive number, and ∞ (converges everywhere). A factorial in the denominator almost always means R = ∞.
⚠️ The substitution question (changing xⁿ to x²ⁿ, as in Q22) is a common exam question. Remember: R becomes √R.
⚠️ Q23-style reasoning (if convergence at x = c is known, what can you conclude about x = d?) tests your understanding of radius of convergence as a threshold. If |d| < |c| and the series converges at c, it converges absolutely at d.
True or false: The radius of convergence depends on the centre a. (False. R depends only on the coefficients.)
Fill in the blank: If Σcₙxⁿ has R = 9, then Σcₙx²ⁿ has R = ___. (3.)
True or false: If R = ∞, the series converges for every real number x. (True.)
True or false: If a power series converges at x = 5, it must converge at x = −4. (True, because |−4| < |5|, so −4 is inside the radius.)
Fill in the blank: The interval of convergence of Σxⁿ/(n+6)! is ___. ((−∞, ∞).)
Q: Find the radius of convergence of Σ n⁴(x − 3)ⁿ.
A: R = 1. The Ratio Test gives L = |x − 3|, which is less than 1 when |x − 3| < 1.
Q: Find the interval of convergence of Σ xⁿ/√(n+4).
A: [−1, 1). R = 1. At x = 1, Σ1/√(n+4) diverges (p = 1/2). At x = −1, the alternating series converges.
Q: What is the radius of convergence of Σ xⁿ/(n+2)!?
A: R = ∞. The Ratio Test gives L = 0 for any x, because dividing by (n+3) drives the ratio to zero.
Q: Find the interval of convergence of Σ xⁿ/(7ⁿn³).
A: [−7, 7]. R = 7. At x = 7: Σ1/n³ converges (p = 3). At x = −7: Σ(−1)ⁿ/n³ converges absolutely.
Q: If Σcₙxⁿ has R = 9, what is R for Σcₙx²ⁿ?
A: R = 3. Replace x with x², so |x²| < 9 means |x| < 3.
Q: If Σcₙ7ⁿ converges, must Σcₙ(−6)ⁿ converge?
A: Yes. Since the series converges at x = 7, R ≥ 7. |−6| = 6 < 7, so x = −6 is inside the interval of convergence and the series converges absolutely there.
Power series are the foundation for Taylor and Maclaurin series, which come next in the course. Once you can find the interval of convergence, you can represent functions as power series and differentiate or integrate them term by term within that interval. This connects to differential equations, where power series solutions are a standard technique. The convergence tests from the first set of notes (Ratio, Root, Comparison, Alternating Series) are the tools you use at every step here.
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