Chapter 5 — Power Series, Radius & Interval of Convergence

By the end of this chapter you will:

1. Write a power series in standard form
2. Use the ratio test to find the radius of convergence $R$
3. Test endpoints to determine the full interval of convergence $I$
4. Handle every standard AP power-series question

5.1 What is a Power Series?

A series whose terms are powers of a variable $x$:

$$\sum_{n=0}^{\infty} c_n (x - a)^n = c_0 + c_1(x-a) + c_2(x-a)^2 + \cdots$$

• $a$ — the center
• $c_n$ — the coefficients
• $x$ — the variable
• When $a = 0$ we say "power series about 0", written $\sum c_n x^n$

🎵 Audio 1 — Definition — audio/ch5_01_intro.mp3

The fundamental question

A power series is a family of series — one for each $x$. The same series can converge for some $x$ and diverge for others. So the central question is:

"For which $x$ does the series converge?"

The answer is always an interval $I$, centered at $a$, with width determined by the radius of convergence $R$.

5.2 The Radius of Convergence Theorem

Theorem: every power series $\sum c_n(x-a)^n$ falls into exactly one of three cases:

1. Converges only at $x=a$ ($R = 0$)
2. Converges for all $x$ ($R = \infty$)
3. There is some $R > 0$: absolute convergence for $\lvert x-a\rvert < R$, divergence for $\lvert x-a\rvert > R$, and the endpoints $x = a \pm R$ must be tested individually.

🔑 Insight: the interior is determined by $R$. The endpoints take separate work — and that's where most AP exam points are won or lost.

🎵 Audio 2 — Three-case theorem — audio/ch5_02_radius.mp3

5.3 Finding $R$ via the Ratio Test

For $\sum c_n(x-a)^n$:

$$L = \lim_{n\to\infty} \left\lvert \frac{c_{n+1}(x-a)^{n+1}}{c_n(x-a)^n} \right\rvert = \lvert x - a \rvert \cdot \lim \left\lvert \frac{c_{n+1}}{c_n} \right\rvert$$

Let $K = \lim \lvert c_{n+1}/c_n \rvert$ (when it exists).

Convergence requires $L < 1$:
$$\lvert x - a \rvert \cdot K < 1 \Rightarrow \lvert x - a \rvert < \frac{1}{K}$$

So $R = \dfrac{1}{K}$ ($K = 0 \Rightarrow R = \infty$; $K = \infty \Rightarrow R = 0$).

🎵 Audio 3 — Ratio test for $R$ — audio/ch5_03_ratio_method.mp3

Example 1

$\displaystyle\sum_{n=0}^{\infty} \frac{(x-3)^n}{n+1}$.

$$L = \lvert x - 3\rvert \cdot \lim \frac{n+1}{n+2} = \lvert x - 3\rvert$$

$L < 1 \Rightarrow \lvert x-3\rvert < 1 \Rightarrow R = 1$. Center $a = 3$, interior $(2, 4)$.

Example 2 ($R = \infty$)

$\displaystyle\sum_{n=0}^{\infty} \frac{x^n}{n!}$.

$$L = \lvert x\rvert \cdot \lim \frac{1}{n+1} = 0$$

Holds for every $x$, so $R = \infty$. (This is the Maclaurin series of $e^x$.)

Example 3 ($R = 0$)

$\sum n! \cdot x^n$.

$$L = \lvert x\rvert \cdot \lim (n+1) = \infty \text{ unless } x = 0$$

So $R = 0$, the series converges only at $x = 0$.

5.4 Endpoint Testing — Where AP Points Are Won

After you find $R$, the interior is $(a - R, a + R)$. The endpoints $x = a - R$ and $x = a + R$ must be plugged in and analyzed individually.

There are 4 possible outcomes:

🎵 Audio 4 — Why endpoints matter — audio/ch5_04_endpoints.mp3

Full worked example

Find the interval of convergence of $\displaystyle\sum_{n=1}^{\infty} \frac{(x-2)^n}{n \cdot 3^n}$.

Step 1 — find $R$:
$$L = \lvert x-2\rvert \cdot \lim \frac{n \cdot 3^n}{(n+1) \cdot 3^{n+1}} = \frac{\lvert x-2\rvert}{3}$$

$L < 1 \Rightarrow \lvert x-2\rvert < 3 \Rightarrow R = 3$. Interior $(-1, 5)$.

Step 2 — left endpoint $x = -1$:
$$\sum_{n=1}^{\infty} \frac{(-3)^n}{n \cdot 3^n} = \sum \frac{(-1)^n}{n}$$

Alternating harmonic — converges by AST. ✓

Step 3 — right endpoint $x = 5$:
$$\sum_{n=1}^{\infty} \frac{3^n}{n \cdot 3^n} = \sum \frac{1}{n}$$

Harmonic — diverges. ✗

Conclusion: interval of convergence is $\boxed{[-1, 5)}$.

🔑 Strategy: after substitution, endpoints typically reduce to a p-series, geometric series, or alternating series — apply the standard tests from chapters 2–4.

5.5 Convergence-Interval Structure (memorize this picture)

        diverges            converges            diverges
   ←———————|—————————————————————|———————→
        a - R        a            a + R
       (endpoint)  (center)     (endpoint)

• Interior $\lvert x - a\rvert < R$: absolute convergence
• Endpoints $x = a \pm R$: test individually
• Outside $\lvert x - a\rvert > R$: divergence

🎵 Audio 5 — Interval structure — audio/ch5_05_structure.mp3

5.6 Mixed Examples (AP-style)

Find the interval of convergence:

(a) $\displaystyle\sum_{n=0}^{\infty} \frac{(2x)^n}{n^2 + 1}$

(b) $\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^n (x+1)^n}{\sqrt{n}}$

Solution (a):

$L = \lvert 2x\rvert$. $L < 1 \Rightarrow \lvert x\rvert < \frac{1}{2}$, $R = \frac{1}{2}$, interior $(-\frac{1}{2}, \frac{1}{2})$.

Endpoint $x = \frac{1}{2}$: $\sum \frac{1}{n^2+1}$ converges by comparison with $\frac{1}{n^2}$.

Endpoint $x = -\frac{1}{2}$: $\sum \frac{(-1)^n}{n^2+1}$ — absolutely convergent.

Interval: $\boxed{[-\frac{1}{2}, \frac{1}{2}]}$.

Solution (b):

$L = \lvert x+1\rvert \cdot \lim \frac{\sqrt{n}}{\sqrt{n+1}} = \lvert x+1\rvert$. $R = 1$, interior $(-2, 0)$.

Endpoint $x = 0$: $\sum \frac{(-1)^n}{\sqrt{n}}$ — conditionally convergent (AST).

Endpoint $x = -2$: $\sum \frac{(-1)^n(-1)^n}{\sqrt{n}} = \sum \frac{1}{\sqrt{n}}$ — diverges ($p = 1/2$).

Interval: $\boxed{(-2, 0]}$.

🎵 Audio 6 — Worked examples — audio/ch5_06_examples.mp3

5.7 Trap Alerts ⚠️

1. Skipping endpoint checks is the #1 way AP students lose points on power-series free-response.
2. The center isn't always 0. Solve $\lvert x - a\rvert < ?$ correctly.
3. Bracket vs. parenthesis: closed at converging endpoints, open at diverging ones.
4. $R = \infty$: the interval is $(-\infty, \infty)$ — no endpoints.
5. $R = 0$: the interval collapses to the single point $\{a\}$ — write it as a set.

🎵 Audio 7 — Traps + recap — audio/ch5_07_traps_recap.mp3

5.8 Mnemonic

"Ratio for R; test both endpoints; brackets follow convergence."

• Ratio test: $L = \lvert x-a\rvert K < 1 \Rightarrow R = 1/K$
• Test both endpoints by substitution
• Use $[\,]$ for converging endpoints, $(\,)$ for diverging ones

Media Inventory

End of chapter. Next: Taylor and Maclaurin series.