core29 min studyLesson 175 in Mathematics

Power series and their radius of convergence

Use coefficient growth to locate an interval of analytic control while checking its boundary independently.

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What you’ll understand

  • Compute radii using coefficient roots or ratios.
  • Separate interior absolute convergence from endpoint behavior.
  • Use compact interior intervals to justify uniform error bounds and calculus.

A power series looks like an infinite polynomial, but its validity is local. Moving the input away from the expansion center amplifies higher powers, while the coefficients may suppress or reinforce them. The radius of convergence is the balance point between those effects. Inside it, the series has unusually strong analytic behavior; outside it, even its terms fail to support convergence. At the boundary, coefficient details decide the answer and must be examined separately.

Coefficient growth sets one radial threshold

Take the limsup of the nth roots of the coefficient magnitudes. Multiplying a coefficient by the nth power of the distance from the center multiplies its nth-root size by that distance. The root test therefore proves absolute convergence when the resulting product is below one and divergence when it is above one. The reciprocal coefficient-growth rate is the radius, including the conventions that zero growth gives infinite radius and infinite growth gives zero radius.

The limsup handles missing coefficients and irregular growth. A ratio formula is convenient when coefficients are eventually nonzero and the consecutive ratio |cₙ₊₁/cₙ| has a limit, but it is not the definition and may fail to exist. For a series containing only even powers, ratios involving zero coefficients are unhelpful; the root formula still works directly. One should also distinguish coefficient index from the exponent when reindexing a sparse series, since an incorrect index changes the inferred radius.

R=1lim sup⁡n→∞∣cn∣1/n,∑n=0∞cn(x−a)n converges absolutely if ∣x−a∣<RR=\frac{1}{\limsup_{n\to\infty}|c_n|^{1/n}},\qquad \sum_{n=0}^{\infty}c_n(x-a)^n\ \text{converges absolutely if }|x-a|<R
Use 1/0=∞ and 1/∞=0. At |x−a|=R the root test is inconclusive and the endpoint requires a separate argument.↔ Scroll wide equations horizontally.

The boundary remembers what roots forget

Multiplying coefficients by a polynomial power of n does not change their nth-root growth, yet it can completely change convergence at the boundary. Coefficients one, one over n, and one over n squared all give radius one for expansions centered at zero. At x=1 their series are respectively divergent, divergent, and convergent. At x=−1, alternation changes the middle example to conditional convergence. A radius alone never specifies whether either real endpoint is included.

The same distinction matters for representing a named function. Summing a geometric series gives one over one minus x only inside its convergence interval. Algebraically extending that rational formula beyond the interval does not make the original series converge there. The function and a particular expansion are separate objects. Likewise, a smooth function need not equal its Taylor series: smoothness controls finite derivatives, while equality with an infinite Taylor expansion requires an appropriate remainder or analyticity argument.

∑n=0Nxn=1−xN+11−x,∣x∣<1:∑n=0∞xn=11−x\sum_{n=0}^{N}x^n=\frac{1-x^{N+1}}{1-x},\qquad |x|<1:\quad\sum_{n=0}^{\infty}x^n=\frac{1}{1-x}
The finite identity holds for x≠1. Its infinite version additionally requires x^(N+1)→0, which is the missing convergence step outside the unit interval.↔ Scroll wide equations horizontally.

Stay strictly inside for uniform control

Choose a compact interval whose distance from the center is at most r<R. A slightly larger radius still inside R yields geometric domination of the coefficient terms on that compact interval. The Weierstrass M-test then proves uniform convergence there. This is local uniform convergence: a single estimate works throughout every fixed compact subinterval, though it need not work on the whole open convergence interval. Near a boundary, the required number of terms can grow without bound.

Termwise differentiation and integration preserve the radius because multiplying or dividing coefficients by n changes their nth-root growth by a factor tending to one. Derivative identities are justified on compact interior intervals by uniform control, not by treating an infinite expression as an ordinary polynomial. In the spectral lab, changing coefficient decay similarly changes the contribution of high modes, but its sine basis is not a power basis and it does not compute a power-series radius.

One radius, two different endpoint outcomes

Find the interval of convergence of ∑n=1∞(x−2)n/(n3n)\sum_{n=1}^{\infty}(x-2)^n/(n3^n).

  1. The nth root of the coefficient magnitude is 1/[3n1/n]1/[3n^{1/n}], which tends to 1/3. The radius about 2 is therefore 3, giving absolute convergence for −1<x<5.

  2. At x=5 the terms become 1/n, so the harmonic series diverges. This endpoint is excluded even though it lies exactly at the calculated radius.

  3. At x=−1 the series becomes ∑(−1)n/n\sum(-1)^n/n, which converges by the alternating theorem but not absolutely. Thus only the left endpoint is included.

The result

The interval of convergence is [−1,5). Convergence is absolute in its interior and conditional at its left endpoint.

Explore the model

Interactive lab

Interactive lab · spectral

A coefficient-decay analogy with a different basis

Increase p in Σ sin(nx−t)/nᵖ and predict how the shorter-wavelength modes are suppressed. Vary N to expose the retained cutoff. This finite trigonometric family illustrates coefficient control; it is not a power series, and no radius of convergence can be read from its spatial plot.

A coefficient-decay analogy with a different basisA finite sum of sin(nx − t)/nᵖ. Playback changes t; increasing the cutoff includes finer spatial scales. This is not a proof about a continuum limit or renormalization. Current readings: Sum at x = π/2: 0.867; Smallest wavelength: 2π / 5; Highest-mode coefficient: 0.2. All parameters are adjustable with the labeled controls below.01.573.144.716.28−2.56−1.2801.282.56x · radiansFinite sine sum
Finite sumHighest included mode
Sum at x = π/2
0.867
Smallest wavelength
2π / 5
Highest-mode coefficient
0.2
t = 0 s

A finite sum of sin(nx − t)/nᵖ. Playback changes t; increasing the cutoff includes finer spatial scales. This is not a proof about a continuum limit or renormalization. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forCoefficient behavior matters together with the basis and the domain.

Check your understanding

Two power series centered at zero have radius one. What must be checked independently?

Try it yourself

Find the radius of ∑n=0∞xn/n!\sum_{n=0}^{\infty}x^n/n! and explain why that conclusion does not require testing two finite endpoints.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

For fixed x, inspect the ratio of successive term magnitudes.

Compare with the solution

The ratio equals |x|/(n+1), which approaches zero for every fixed real x. Thus the series converges absolutely at every real input, and its radius is infinite. There are no finite boundary points to test. The ratio reasoning concerns each fixed input; uniform convergence on a specified bounded interval follows from a common bound using the interval’s largest |x|.

Take it with you

  • Coefficient nth-root growth determines the radius, including sparse or irregular coefficients.
  • Endpoints require their own series tests.
  • Uniform control and termwise calculus are naturally local inside the radius.

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