Power Series and Radius of Convergence

A power series ∑anxn\sum a_n x^n converges absolutely inside its radius RR and diverges outside — watch three archetypal series pin down their radii, and see why the boundary ∣x∣=R\lvert x\rvert = R must be decided case by case.

1R=lim sup⁡n→∞∣an∣1/n\frac{1}{R}=\limsup_{n\to\infty}\lvert a_n\rvert^{1/n}
∑n=0∞xnn! has R=∞,∑n=0∞n! xn has R=0,∑n=0∞xn has R=1\sum_{n=0}^{\infty}\frac{x^n}{n!}\ \text{has}\ R=\infty,\quad \sum_{n=0}^{\infty}n!\,x^n\ \text{has}\ R=0,\quad \sum_{n=0}^{\infty}x^n\ \text{has}\ R=1

Examples & Exercises

Prove by the ratio test that ∑n=0∞xn/n!\sum_{n=0}^{\infty} x^n/n! has radius of convergence ∞\infty.

∣an+1an∣=1n+1→0\left\lvert\frac{a_{n+1}}{a_n}\right\rvert = \frac{1}{n+1} \to 0
x=2x = 2, N=20N = 20 — the terms shrink fast and S20(2)≈e2S_{20}(2) \approx e^2.
Show proof
  1. Put tn=xn/n!t_n = x^n/n!. For x≠0x \ne 0 the ratio of consecutive terms is ∣tn+1tn∣=∣x∣n+1\left\lvert \frac{t_{n+1}}{t_n} \right\rvert = \frac{\lvert x\rvert}{n+1}.
  2. Since ∣x∣n+1→0<1\frac{\lvert x\rvert}{n+1} \to 0 < 1, the ratio test gives convergence of ∑∣tn∣\sum \lvert t_n\rvert for every x≠0x \ne 0: the series converges absolutely.
    ∣tn+1tn∣=∣x∣n+1⟶0<1\left\lvert\frac{t_{n+1}}{t_n}\right\rvert = \frac{\lvert x\rvert}{n+1} \longrightarrow 0 < 1
  3. At x=0x = 0 the series is 1+0+0+⋯=11 + 0 + 0 + \cdots = 1, which converges.
  4. Hence the series converges absolutely for every real xx, so the radius of convergence is R=∞R = \infty. (Cauchy–Hadamard: 1R=lim sup⁡∣1/n!∣1/n=0\frac{1}{R} = \limsup \lvert 1/n!\rvert^{1/n} = 0.)

Prove ∑n=0∞n! xn\sum_{n=0}^{\infty} n!\,x^n has radius of convergence 00.

∣(n+1)! xn+1n! xn∣=(n+1)∣x∣→∞\left\lvert\frac{(n+1)!\,x^{n+1}}{n!\,x^n}\right\rvert = (n+1)\lvert x\rvert \to \infty
x=0.5x = 0.5, N=10N = 10 — the terms only grow, and the partial sums blow up.
Show proof
  1. Put tn=n! xnt_n = n!\,x^n. For x≠0x \ne 0 we have ∣tn+1tn∣=(n+1)∣x∣→∞\left\lvert \frac{t_{n+1}}{t_n} \right\rvert = (n+1)\lvert x\rvert \to \infty.
  2. For large nn we get ∣tn+1∣≥2∣tn∣\lvert t_{n+1}\rvert \ge 2\lvert t_n\rvert, so ∣tn∣\lvert t_n\rvert cannot tend to 00 and in fact grows without bound. Hence the series diverges by the divergence test.
  3. At x=0x = 0 the series is 1+0+0+⋯=11 + 0 + 0 + \cdots = 1, which converges.
  4. The series converges only at x=0x = 0, so the radius of convergence is R=0R = 0.

Prove ∑n=0∞xn\sum_{n=0}^{\infty} x^n has radius 11 and diverges at both endpoints x=±1x=\pm 1.

R=1,1n↛0,(−1)n↛0R = 1,\quad 1^n \not\to 0,\quad (-1)^n \not\to 0
x=1x = 1, N=20N = 20 — every term is 11, so the partial sums grow linearly to 2121.
Show proof
  1. The ratio test on ∣x∣n\lvert x\rvert^n gives ∣x∣n+1∣x∣n=∣x∣\frac{\lvert x\rvert^{n+1}}{\lvert x\rvert^n} = \lvert x\rvert. It converges for ∣x∣<1\lvert x\rvert < 1 and diverges for ∣x∣>1\lvert x\rvert > 1 (terms grow without bound), so the radius is R=1R = 1.
  2. At the endpoint x=1x = 1 the terms are 1n=1↛01^n = 1 \not\to 0, so the series diverges by the divergence test.
  3. At the endpoint x=−1x = -1 the terms are (−1)n↛0(-1)^n \not\to 0, so the series diverges by the divergence test as well.
  4. Thus the radius is 11, and both endpoints diverge — the boundary must be decided case by case.

8. Power Series