Radius of Convergence

The definition, made concrete

A power series centred at a real number a is an infinite sum of the form

f(x) = ∑_n=0^∞ aₙ (x − a)ⁿ = a₀ + a₁(x−a) + a₂(x−a)² + ⋯

The series always converges at x = a — every term except the first is zero. The question is: how far can x drift from a before the sum stops making sense? The answer is a single number, the radius of convergence R, with three exhaustive cases:

• R = 0. The series converges only at x = a. Example: ∑ n! · xⁿ.
• 0 < R < ∞. Converges absolutely for |x−a| < R, diverges for |x−a| > R, undecided at |x−a| = R.
• R = ∞. Converges absolutely for every real x. Example: the exponential series ∑ xⁿ/n!.

The trichotomy is not a heuristic. It is a theorem (Abel, 1826): the convergence set of a power series is always a disk centred at a in the complex plane, intersected with the real line.

The Cauchy-Hadamard formula

The universal formula for R uses the limit superior:

1/R = limsup_n→∞ |aₙ|^1/n

If the limsup is 0, R = ∞. If the limsup is +∞, R = 0. Otherwise R is the reciprocal of the limsup. The reason this works: the series ∑ |aₙ(x−a)ⁿ| has nth term |aₙ|^1/n · |x−a| under the root test, which converges iff that quantity is < 1.

The advantage of Cauchy-Hadamard over the ratio test is that the limsup always exists, even when ordinary limits don't. For series with sparse or oscillating coefficients — say aₙ = 1 when n is a perfect square and 0 otherwise — the ratio bounces around without settling, but the limsup still pins R down exactly.

The ratio formula

When the limit exists, the ratio test gives a faster computation:

1/R = lim_n→∞ |aₙ₊₁/aₙ|

This works for the most common power series in practice — geometric, exponential, trigonometric, binomial. It fails only when many coefficients are zero or when the ratio oscillates.

Cauchy-Hadamard vs ratio formula

	Cauchy-Hadamard	Ratio formula
Formula	1/R = limsup |aₙ|^(1/n)	1/R = lim |aₙ₊₁/aₙ|
Always applicable	Yes (limsup exists for any bounded sequence)	No (ratio limit may not exist)
Handles sparse coefficients	Yes	No (division by zero)
Computational difficulty	Higher (nth root)	Lower (division of consecutive terms)
Best for	Series with many zero or wildly varying aₙ	Series with smooth factorial or polynomial coefficients
Underlying test	Root test	Ratio test
Returns the same R when both apply	Yes, always	Yes, always

Worked examples

Example 1: ∑ xⁿ/n! has R = ∞

Apply the ratio formula with aₙ = 1/n!:

|aₙ₊₁/aₙ| = (1/(n+1)!) ÷ (1/n!) = 1/(n+1) → 0 as n → ∞.

So 1/R = 0, meaning R = ∞. This is the exponential series — converges everywhere. The same calculation works for sin and cos, whose coefficients are bounded by 1/n!.

Example 2: ∑ n·xⁿ has R = 1

Here aₙ = n:

|aₙ₊₁/aₙ| = (n+1)/n → 1.

So R = 1. The series converges for |x| < 1 and diverges for |x| > 1. At x = ±1 the terms n · (±1)ⁿ do not tend to zero, so both endpoints diverge — the interval of convergence is the open (−1, 1).

Example 3: ∑ xⁿ/n has R = 1 with endpoint asymmetry

Same R as Example 2 — but now check endpoints. At x = 1 the series becomes the harmonic ∑ 1/n, which diverges. At x = −1 it becomes the alternating harmonic ∑ (−1)ⁿ/n, which converges by Leibniz. The interval is [−1, 1).

Endpoint behaviour: the four cases

For finite R, the series at x = a + R and x = a − R can each independently converge or diverge. That gives four interval shapes:

• (a−R, a+R) — both endpoints diverge. Example: ∑ xⁿ.
• [a−R, a+R) — left converges, right diverges. Example: ∑ xⁿ/n at a = 0.
• (a−R, a+R] — right converges, left diverges. Mirror image of the above.
• [a−R, a+R] — both endpoints converge. Example: ∑ xⁿ/n² at a = 0.

To decide which case you are in, substitute x = a ± R into the series and apply a non-power-series test: alternating series test, comparison with a p-series, or absolute-value comparison.

Classical applications

Analytic continuation. A function defined by a power series with radius R is analytic inside its disk. To extend it beyond, recentre at another point inside the disk and produce a new power series whose disk pokes out further. Iterating this idea is how Riemann's zeta function escapes its original half-plane and how Weierstrass built complex analysis.

Singularity detection. The radius of convergence equals the distance from the centre to the nearest singularity in the complex plane — even when you only see the function on the real line. The series for 1/(1+x²) centred at 0 has R = 1, not because of any real bad behaviour, but because of the poles at x = ±i.

Truncation error bounds. Inside the disk of convergence, the tail ∑_n≥N aₙ(x−a)ⁿ is bounded by a geometric series, giving practical error estimates for numerical approximations.

Generating functions. In combinatorics, a sequence's generating function ∑ aₙ xⁿ has a radius of convergence determined by the asymptotic growth rate of aₙ. R = 1/φ for Fibonacci coefficients, where φ is the golden ratio.

Common mistakes

• Confusing R with the full interval width. R is the half-width. The total length of the open interval of convergence is 2R, not R.
• Treating endpoints as automatic. The radius formulas say nothing about |x−a| = R. Always test those points separately.
• Forgetting limsup vs lim. Cauchy-Hadamard uses limsup. For sparse coefficients, ordinary lim does not exist; limsup still does.
• Applying the formula to series that aren't centred power series. A series like ∑ sin(nx) is not a power series — these tests don't apply.
• Assuming uniform convergence on the open interval. Convergence is uniform only on closed sub-intervals strictly inside (a−R, a+R) — not on the whole open interval.