Power Series and Radius of Convergence

Where does a power series converge, and where does it fall apart?

Inside the radius lies a safe zone. There the series converges even with absolute values. Outside, the terms themselves fail to shrink to zero. Points on the boundary each have their own fate.

For example, think of Wi-Fi in a 🏠 house. It works well inside and drops as you walk away. Right at the door it may or may not connect, so you check case by case.

1R=lim sup⁑nβ†’βˆžβˆ£an∣1/n\frac{1}{R}=\limsup_{n\to\infty}\lvert a_n\rvert^{1/n}
βˆ‘n=0∞xnn!Β hasΒ R=∞,βˆ‘n=0∞n! xnΒ hasΒ R=0,βˆ‘n=0∞xnΒ hasΒ 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 βˆ‘n=0∞xn/n!\sum_{n=0}^{\infty} x^n/n! has radius of convergence ∞\infty.

∣an+1an∣=1n+1β†’0\left\lvert\frac{a_{n+1}}{a_n}\right\rvert = \frac{1}{n+1} \to 0
x=2x = 2, N=20N = 20 β€” the terms shrink fast and S20(2)β‰ˆe2S_{20}(2) \approx e^2.
Show proof

  1. Put tn=xn/n!t_n = x^n/n!. For xβ‰ 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+1β†’0<1\frac{\lvert x\rvert}{n+1} \to 0 < 1, the ratio test gives convergence of βˆ‘βˆ£tn∣\sum \lvert t_n\rvert for every xβ‰ 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+β‹―=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 βˆ‘n=0∞n! xn\sum_{n=0}^{\infty} n!\,x^n has radius of convergence 00.

∣(n+1)! xn+1n! xn∣=(n+1)∣xβˆ£β†’βˆž\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 β€” the terms only grow, and the partial sums blow up.
Show proof

  1. Put tn=n! xnt_n = n!\,x^n. For xβ‰ 0x \ne 0 we have ∣tn+1tn∣=(n+1)∣xβˆ£β†’βˆž\left\lvert \frac{t_{n+1}}{t_n} \right\rvert = (n+1)\lvert x\rvert \to \infty.

  2. In particular, for some NN and all nβ‰₯Nn \ge N, ∣tn+1∣β‰₯2∣tn∣\lvert t_{n+1}\rvert \ge 2\lvert t_n\rvert.

    Since xβ‰ 0x \ne 0, tNβ‰ 0t_N \ne 0, and induction gives ∣tN+k∣β‰₯2k∣tN∣\lvert t_{N+k}\rvert \ge 2^k\lvert t_N\rvert for kβ‰₯0k \ge 0, so ∣tnβˆ£β†’βˆž\lvert t_n\rvert \to \infty.

    Hence the terms do not tend to 00, and the series diverges by the divergence test.

  3. At x=0x = 0 the series is 1+0+0+β‹―=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 βˆ‘n=0∞xn\sum_{n=0}^{\infty} x^n has radius 11 and diverges at both endpoints x=Β±1x=\pm 1.

R=1,1nβ†’ΜΈ0,(βˆ’1)nβ†’ΜΈ0R = 1,\quad 1^n \not\to 0,\quad (-1)^n \not\to 0
x=1x = 1, N=20N = 20 β€” every term is 11, so the partial sums grow linearly to 2121.
Show proof

  1. At x=0x = 0 the series is 1+0+0+β‹―=11 + 0 + 0 + \cdots = 1, converging trivially (every term except the constant term is 00).

    Now assume xβ‰ 0x \ne 0 and apply the ratio test to ∣x∣n\lvert x\rvert^n: ∣x∣n+1∣x∣n=∣x∣\frac{\lvert x\rvert^{n+1}}{\lvert x\rvert^n} = \lvert x\rvert (at x=0x = 0 this ratio is 0/00/0, which is why that case was separated).

    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=1β†’ΜΈ01^n = 1 \not\to 0, so the series diverges by the divergence test.

  3. At the endpoint x=βˆ’1x = -1 the terms are (βˆ’1)nβ†’ΜΈ0(-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 β€” the boundary must be decided case by case.

8. Power Series