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🎯⭐ INTERACTIVE LESSON

Power Series

Learn step-by-step with interactive practice!

Power Series - Complete Interactive Lesson

Part 1: Core Concepts

Power Series — Definition & Convergence

Part 1 of 7 — Introduction to Power Series

What Is a Power Series?

A power series centered at x=cx = c is:

∑n=0∞an(x−c)n=a0+a1(x−c)+a2(x−c)2+⋯\boxed{\sum_{n=0}^\infty a_n (x - c)^n = a_0 + a_1(x-c) + a_2(x-c)^2 + \cdots}

When c=0c = 0, this is a Maclaurin-type power series: ∑anxn\sum a_n x^n.

Key Terminology

TermMeaning
Center ccThe point about which the series is expanded
Coefficients ana_nThe constants multiplying each power
Radius of convergence RRSeries converges for $
Interval of convergenceFull interval including endpoint analysis

Three Convergence Possibilities

For any power series, exactly ONE is true:

  1. Converges only at x=cx = c (radius R=0R = 0)
  2. Converges for all xx (radius R=∞R = \infty)
  3. Converges for ∣x−c∣<R|x - c| < R and diverges for ∣x−c∣>R|x - c| > R (finite R>0R > 0)

AP Tip: The Ratio Test is the primary tool for finding the radius of convergence.

Finding the Radius with the Ratio Test

For ∑an(x−c)n\sum a_n (x-c)^n, apply the Ratio Test:

L=lim⁡n→∞∣an+1an∣⋅∣x−c∣L = \lim_{n\to\infty} \left|\frac{a_{n+1}}{a_n}\right| \cdot |x - c|

Converges when L<1L < 1, i.e., ∣x−c∣<1lim⁡∣an+1/an∣|x - c| < \frac{1}{\lim |a_{n+1}/a_n|}.

R=1lim⁡n→∞∣an+1/an∣\boxed{R = \frac{1}{\lim_{n\to\infty} |a_{n+1}/a_n|}}

Example: ∑n=0∞xnn!\sum_{n=0}^\infty \frac{x^n}{n!}

∣an+1an∣=1n+1→0\left|\frac{a_{n+1}}{a_n}\right| = \frac{1}{n+1} \to 0

So L=0⋅∣x∣=0<1L = 0 \cdot |x| = 0 < 1 for all xx. Radius R=∞R = \infty. (This is the series for exe^x.)

Example: ∑n=0∞n!xn\sum_{n=0}^\infty n! x^n

∣an+1an∣=n+1→∞\left|\frac{a_{n+1}}{a_n}\right| = n+1 \to \infty

So L=∞L = \infty for any x≠0x \neq 0. Radius R=0R = 0. Converges only at x=0x = 0.

Finding Radius of Convergence

Convergence Analysis

Radius Computation

Summary

  • Power series: ∑an(x−c)n\sum a_n(x-c)^n — an "infinite polynomial"
  • Radius found via Ratio Test: R=1/lim⁡∣an+1/an∣R = 1/\lim|a_{n+1}/a_n|
  • Three possibilities: R=0R = 0, R=∞R = \infty, or finite RR
  • Endpoints must ALWAYS be tested separately

Next: Part 2 — Interval of Convergence (Endpoint Testing).

Part 2: Worked Examples

Power Series — Interval of Convergence

Part 2 of 7 — Endpoint Testing

From Radius to Interval

After finding RR, the open interval (c−R, c+R)( c - R,\ c + R ) is guaranteed. But endpoints need individual testing.

Endpoint Testing Procedure

  1. Find RR using Ratio/Root Test
  2. Substitute x=c−Rx = c - R into ∑an(x−c)n\sum a_n(x-c)^n → get a numerical series
  3. Substitute x=c+Rx = c + R → get another numerical series
  4. Test each for convergence (often pp-series, alternating, geometric, etc.)

Complete Example: ∑n=1∞xnn\sum_{n=1}^\infty \frac{x^n}{n}

Step 1: ∣an+1/an∣=n/(n+1)→1|a_{n+1}/a_n| = n/(n+1) \to 1. So R=1R = 1.

Step 2: At x=1x = 1: ∑1/n\sum 1/n — diverges (harmonic)

Step 3: At x=−1x = -1: ∑(−1)n/n\sum (-1)^n/n — converges (alternating harmonic)

Interval of convergence: [−1,1)\boxed{\text{Interval of convergence: } [-1, 1)}

AP Tip: The interval notation matters! Use brackets [[ for included endpoints, parentheses (( for excluded.

Common Endpoint Patterns

SeriesRRAt x=c+Rx = c+RAt x=c−Rx = c-RIOC
∑xn/n\sum x^n/n1∑1/n\sum 1/n div.∑(−1)n/n\sum(-1)^n/n conv.[−1,1)[-1,1)
∑xn/n2\sum x^n/n^21∑1/n2\sum 1/n^2 conv.∑(−1)n/n2\sum(-1)^n/n^2 conv.[−1,1][-1,1]
∑xn\sum x^n1∑1\sum 1 div.∑(−1)n\sum(-1)^n div.(−1,1)(-1,1)
∑n!xn\sum n! x^n0N/AN/A{0}\{0\}
∑xn/n!\sum x^n/n!∞\inftyN/AN/A(−∞,∞)(-\infty,\infty)

Key Insight

At the positive endpoint (x=c+Rx = c + R): all terms are positive → test with pp-series, comparison, etc.

At the negative endpoint (x=c−Rx = c - R): signs alternate → often use AST.

Common outcome: one endpoint includes (alternating convergence), one excludes (divergence).

Endpoint Testing Practice

Interval Determination

Find the Left Endpoint

Summary

  • After finding RR, always test both endpoints
  • Positive endpoint often yields a positive-term series
  • Negative endpoint often yields an alternating series
  • Four possible IOC shapes: (a,b)(a,b), [a,b)[a,b), (a,b](a,b], [a,b][a,b]
  • AP exam ALWAYS expects endpoint testing — don't skip it!

Next: Part 3 — Operations on Power Series.

Part 3: Problem-Solving Patterns

Power Series — Operations

Part 3 of 7 — Differentiation, Integration, and Manipulation

Term-by-Term Differentiation

If f(x)=∑n=0∞an(x−c)nf(x) = \sum_{n=0}^\infty a_n (x-c)^n with radius R>0R > 0, then:

f′(x)=∑n=1∞nan(x−c)n−1\boxed{f'(x) = \sum_{n=1}^\infty n a_n (x-c)^{n-1}}

The derivative has the same radius RR (but possibly different endpoint behavior).

Term-by-Term Integration

∫f(x) dx=C+∑n=0∞an(x−c)n+1n+1\boxed{\int f(x)\,dx = C + \sum_{n=0}^\infty \frac{a_n (x-c)^{n+1}}{n+1}}

Also has radius RR (but possibly different endpoint behavior).

Example: From Geometric to ln⁡\ln

11−x=∑n=0∞xn,∣x∣<1\frac{1}{1-x} = \sum_{n=0}^\infty x^n,\quad |x| < 1

Integrate both sides:

−ln⁡(1−x)=∑n=0∞xn+1n+1=∑n=1∞xnn-\ln(1-x) = \sum_{n=0}^\infty \frac{x^{n+1}}{n+1} = \sum_{n=1}^\infty \frac{x^n}{n}

So ln⁡(1−x)=−∑n=1∞xnn\ln(1-x) = -\sum_{n=1}^\infty \frac{x^n}{n}, or equivalently ln⁡(1+x)=∑n=1∞(−1)n+1xnn\ln(1+x) = \sum_{n=1}^\infty \frac{(-1)^{n+1} x^n}{n}.

AP Tip: Deriving series by differentiating/integrating known series is a VERY common AP technique.

Substitution

Replace xx with an expression in a known series:

11−x=∑xn  ⟹  11+x2=∑(−x2)n=∑(−1)nx2n\frac{1}{1-x} = \sum x^n \implies \frac{1}{1+x^2} = \sum (-x^2)^n = \sum (-1)^n x^{2n}

Then integrate: arctan⁡x=∑n=0∞(−1)nx2n+12n+1\arctan x = \sum_{n=0}^\infty \frac{(-1)^n x^{2n+1}}{2n+1}

Addition and Multiplication

  • Addition: ∑anxn+∑bnxn=∑(an+bn)xn\sum a_n x^n + \sum b_n x^n = \sum (a_n + b_n) x^n (radius = min of the two)
  • Multiplication by xkx^k: xk∑anxn=∑anxn+kx^k \sum a_n x^n = \sum a_n x^{n+k} (radius unchanged)

Radius Under Operations

OperationNew Radius
DifferentiationSame RR
IntegrationSame RR
Substitution x→g(x)x \to g(x)Solve $
Additionmin⁡(R1,R2)\min(R_1, R_2)
Multiplication by polynomialSame RR

Operations Practice

Manipulation Techniques

Series Derivation

Summary

  • Differentiate and integrate power series term by term
  • Radius stays the same (endpoints may change)
  • Substitution lets you build new series from known ones
  • Key chain: 1/(1−x)→ln⁡(1−x)→arctan⁡x1/(1-x) \to \ln(1-x) \to \arctan x via integration/substitution

Next: Part 4 — Representing Functions as Power Series.

Part 4: Graphs and Interpretation

Power Series — Function Representation

Part 4 of 7 — Building Series from Known Functions

The Essential Known Series

Memorize these — they're the building blocks:

FunctionSeriesIOC
11−x\frac{1}{1-x}∑n=0∞xn\sum_{n=0}^\infty x^n(−1,1)(-1,1)
exe^x∑n=0∞xnn!\sum_{n=0}^\infty \frac{x^n}{n!}(−∞,∞)(-\infty, \infty)
sin⁡x\sin x∑n=0∞(−1)nx2n+1(2n+1)!\sum_{n=0}^\infty \frac{(-1)^n x^{2n+1}}{(2n+1)!}(−∞,∞)(-\infty, \infty)
cos⁡x\cos x∑n=0∞(−1)nx2n(2n)!\sum_{n=0}^\infty \frac{(-1)^n x^{2n}}{(2n)!}(−∞,∞)(-\infty, \infty)
ln⁡(1+x)\ln(1+x)∑n=1∞(−1)n+1xnn\sum_{n=1}^\infty \frac{(-1)^{n+1} x^n}{n}(−1,1](-1, 1]
arctan⁡x\arctan x∑n=0∞(−1)nx2n+12n+1\sum_{n=0}^\infty \frac{(-1)^n x^{2n+1}}{2n+1}[−1,1][-1, 1]

Most AP power series problems reduce to manipulating these six.\boxed{\text{Most AP power series problems reduce to manipulating these six.}}

AP Tip: You'll often need to find a series by relating the function to one of these through substitution, differentiation, or integration.

Technique: Partial Fractions + Geometric

Find the series for f(x)=32−xf(x) = \frac{3}{2-x}:

32−x=32⋅11−x/2=32∑n=0∞(x2)n=∑n=0∞3xn2n+1\frac{3}{2-x} = \frac{3}{2} \cdot \frac{1}{1 - x/2} = \frac{3}{2} \sum_{n=0}^\infty \left(\frac{x}{2}\right)^n = \sum_{n=0}^\infty \frac{3 x^n}{2^{n+1}}

IOC: ∣x/2∣<1  ⟹  ∣x∣<2|x/2| < 1 \implies |x| < 2

Technique: Composition

Find the series for ex2e^{x^2}:

eu=∑un/n!e^u = \sum u^n/n!. Set u=x2u = x^2:

ex2=∑n=0∞x2nn!=1+x2+x42+x66+⋯e^{x^2} = \sum_{n=0}^\infty \frac{x^{2n}}{n!} = 1 + x^2 + \frac{x^4}{2} + \frac{x^6}{6} + \cdots

Technique: Integration of Known Series

Find the series for ∫0xe−t2 dt\int_0^x e^{-t^2}\,dt (no elementary form!):

e−t2=∑(−1)nt2n/n!e^{-t^2} = \sum (-1)^n t^{2n}/n!. Integrate: ∫0x=∑n=0∞(−1)nx2n+1(2n+1)n!\int_0^x = \sum_{n=0}^\infty \frac{(-1)^n x^{2n+1}}{(2n+1) n!}

This is related to the error function erf(x)\text{erf}(x) — series representation gives exact computation!

Function Representation

Series Building

Coefficient Finding

Summary

  • Six essential series to memorize (geometric, exe^x, sin⁡\sin, cos⁡\cos, ln⁡\ln, arctan⁡\arctan)
  • Build new series via substitution, multiplication, differentiation, integration
  • Partial fractions reduce rational functions to geometric-type series
  • Series let you "compute" functions with no elementary antiderivative

Next: Part 5 — Power Series and Differential Equations.

Part 5: Applications

Power Series — Differential Equations & AP Strategies

Part 5 of 7 — Series Solutions & Exam Techniques

Power Series Solutions to DEs

Some AP problems ask you to find coefficients of a power series solution to a DE.

Setup: Assume y=∑anxn=a0+a1x+a2x2+a3x3+⋯y = \sum a_n x^n = a_0 + a_1 x + a_2 x^2 + a_3 x^3 + \cdots

Then y′=∑n=1∞nanxn−1=a1+2a2x+3a3x2+⋯y' = \sum_{n=1}^\infty n a_n x^{n-1} = a_1 + 2a_2 x + 3a_3 x^2 + \cdots

Example: y′=yy' = y, y(0)=1y(0) = 1

Substituting: a1+2a2x+3a3x2+⋯=a0+a1x+a2x2+⋯a_1 + 2a_2 x + 3a_3 x^2 + \cdots = a_0 + a_1 x + a_2 x^2 + \cdots

Matching coefficients: a1=a0=1a_1 = a_0 = 1, 2a2=a1⇒a2=1/22a_2 = a_1 \Rightarrow a_2 = 1/2, 3a3=a2⇒a3=1/63a_3 = a_2 \Rightarrow a_3 = 1/6

Pattern: an=1/n!a_n = 1/n! → solution is y=exy = e^x ✓

AP Tip: These problems typically ask for the first 3 or 4 nonzero terms, not the general pattern.

Common AP FRQ Formats

Type 1: "Write the first four nonzero terms..."

  • Use known series + operations
  • Example: First 4 terms of exsin⁡xe^x \sin x → multiply truncated series

Type 2: "Find the coefficient of xnx^n..."

  • Use Taylor formula: an=f(n)(0)/n!a_n = f^{(n)}(0)/n!
  • Or manipulate known series

Type 3: "Use the series to approximate..."

  • Evaluate at specific xx, bound error
  • Use alternating series error bound when applicable

Type 4: "Find the interval of convergence"

  • Ratio test for RR, then test endpoints

Quick AP Checks

f(x)=∑n=0∞f(n)(c)n!(x−c)n  ⟹  an=f(n)(c)n!\boxed{f(x) = \sum_{n=0}^\infty \frac{f^{(n)}(c)}{n!}(x-c)^n \implies a_n = \frac{f^{(n)}(c)}{n!}}

So if you know the series, you know the derivatives at the center: f(n)(c)=n!⋅anf^{(n)}(c) = n! \cdot a_n

AP-Style Practice

Series & Derivatives

DE Series Solution

Summary

  • Power series can solve DEs by matching coefficients
  • f(n)(c)=n!⋅anf^{(n)}(c) = n! \cdot a_n connects series coefficients to derivatives
  • AP FRQ: "first four nonzero terms" is the most common format
  • Build series from known ones rather than computing derivatives

Next: Part 6 — Problem-Solving Workshop.

Part 6: Exam Strategy

Power Series — Problem-Solving Workshop

Part 6 of 7 — Mixed Practice

Workshop Focus Areas

SkillWhat to Practice
Finding RRRatio Test on coefficients
Endpoint testingSubstitute x=c±Rx = c \pm R, test convergence
Series manipulationSubstitution, differentiation, integration
Coefficient extractionan=f(n)(c)/n!a_n = f^{(n)}(c)/n!
Series buildingFrom known series to new functions

Workshop Problems

IOC Workshop

Series Evaluation

Workshop Takeaways

  • Ratio Test is the go-to for finding RR
  • Endpoint testing is mandatory on the AP exam
  • Building series from known ones is faster than computing derivatives
  • f(n)(c)=n!⋅anf^{(n)}(c) = n! \cdot a_n is a powerful shortcut

Next: Part 7 — Comprehensive Review.

Part 7: Mixed Review

Power Series — Comprehensive Review

Part 7 of 7 — Complete Topic Review

Power Series Checklist

SkillKey Points
Definition∑an(x−c)n\sum a_n(x-c)^n; converges in interval around cc
RadiusRatio Test: $R = 1/\lim
EndpointsTest separately; four possible IOC shapes
OperationsDifferentiate/integrate term-by-term; same RR
Known series1/(1−x)1/(1-x), exe^x, sin⁡x\sin x, cos⁡x\cos x, ln⁡(1+x)\ln(1+x), arctan⁡x\arctan x
Coefficientsan=f(n)(c)/n!a_n = f^{(n)}(c)/n!
DE solutionsMatch coefficients after substituting series

Power series = the bridge between algebra and analysis on the AP exam.\boxed{\text{Power series = the bridge between algebra and analysis on the AP exam.}}

Comprehensive MC Review

Final Review Drill

Final Challenge

Power Series — Complete Summary

You've mastered:

  • Definition & radius — Ratio Test for RR, three convergence scenarios
  • Endpoint testing — individual analysis, four IOC shapes
  • Operations — differentiate, integrate, substitute term-by-term
  • Known series — six essential Maclaurin series
  • Function representation — building new series from old
  • DE connections — coefficient matching for series solutions

Key Fact: Power series questions appear in 5+ MC questions and at least 1 FRQ on every BC exam. This is arguably the most important BC-specific topic.

Up Next: Taylor & Maclaurin Series — the general construction formula.