The Central Idea
A numerical series has fixed terms such as
\[ \sum_{n=0}^{\infty}\frac1{2^n}. \]A power series allows those terms to depend on a variable:
\[ \sum_{n=0}^{\infty}c_n(x-a)^n. \]The value of \(x\) determines which numerical series we actually obtain.
What Is a Power Series?
A power series centered at \(a\) has the form
\[ \boxed{ \sum_{n=0}^{\infty}c_n(x-a)^n }. \]The numbers \(c_n\) are called the coefficients.
The Center
The number \(a\) is the center of the power series.
At the center,
\[ x=a, \]so every term with \(n\ge1\) vanishes.
The Value at the Center
At \(x=a\),
\[ \sum_{n=0}^{\infty}c_n(x-a)^n = c_0. \]Therefore a power series always converges at its center.
A Power Series Centered at Zero
When \(a=0\), the series becomes
\[ \sum_{n=0}^{\infty}c_nx^n. \]Such a series is often called a power series about \(0\).
Example: A Simple Power Series
Consider
\[ \sum_{n=0}^{\infty}x^n = 1+x+x^2+x^3+\cdots. \]This is a power series centered at \(0\) with coefficients
\[ c_n=1. \]Power Series Become Numerical Series
Once a particular value of \(x\) is chosen, a power series becomes an ordinary numerical series.
For example, at \(x=1/2\),
\[ \sum_{n=0}^{\infty}x^n = \sum_{n=0}^{\infty}\left(\frac12\right)^n. \]Convergence Depends on \(x\)
The geometric power series
\[ \sum_{n=0}^{\infty}x^n \]converges when
\[ |x|<1 \]and diverges when
\[ |x|\ge1. \]The Geometric Series Is the Fundamental Model
Recall:
\[ \sum_{n=0}^{\infty}r^n = \frac1{1-r}, \qquad |r|<1. \]Replacing \(r\) with \(x\) gives
\[ \boxed{ \frac1{1-x} = \sum_{n=0}^{\infty}x^n, \qquad |x|<1. } \]A Function Represented by an Infinite Series
The identity
\[ \frac1{1-x} = 1+x+x^2+x^3+\cdots \]means that within the interval \(|x|<1\), the infinite series behaves exactly like the function \(1/(1-x)\).
Power Series Are Functions
Wherever a power series converges, it defines a function:
\[ f(x)=\sum_{n=0}^{\infty}c_n(x-a)^n. \]Its domain is the set of \(x\)-values for which the series converges.
The Basic Convergence Pattern
Every power series has one of three types of convergence behavior:
- it converges only at its center;
- it converges for all real \(x\);
- it converges on an interval centered at \(a\).
Radius of Convergence
The number \(R\) describing the distance from the center to the convergence boundary is called the radius of convergence.
\[ \boxed{|x-a|Possible Radius Values
The radius may be:
- \(R=0\), meaning convergence only at \(x=a\);
- a finite positive number;
- \(R=\infty\), meaning convergence for all real \(x\).
Interval of Convergence
The interval of convergence is the complete set of real \(x\)-values for which the power series converges.
For finite \(R\), the interior is
\[ a-REndpoints Require Separate Testing
The radius tells us what happens strictly inside and outside the interval.
It does not determine convergence at
\[ x=a-R \]or
\[ x=a+R. \]Each endpoint must be tested separately.
Why Endpoints Are Special
Ratio and root tests often reduce the interior condition to
\[ |x-a|Using the Ratio Test
For
\[ \sum c_n(x-a)^n, \]apply the ratio test to the absolute values:
\[ L= \lim_{n\to\infty} \left| \frac{c_{n+1}(x-a)^{n+1}} {c_n(x-a)^n} \right|. \]Simplifying the Ratio
The variable factor usually separates:
\[ L= |x-a| \lim_{n\to\infty} \left|\frac{c_{n+1}}{c_n}\right|. \]Convergence occurs when
\[ L<1. \]Example: Finding a Radius
Consider
\[ \sum_{n=1}^{\infty}\frac{x^n}{3^n}. \]This is geometric:
\[ \sum_{n=1}^{\infty}\left(\frac{x}{3}\right)^n. \]Therefore it converges when
\[ \left|\frac{x}{3}\right|<1. \]Hence
\[ |x|<3, \qquad \boxed{R=3}. \]Example: Shifted Center
Consider
\[ \sum_{n=0}^{\infty}\frac{(x-2)^n}{5^n}. \]The series converges when
\[ \left|\frac{x-2}{5}\right|<1. \]Therefore
\[ |x-2|<5, \qquad \boxed{R=5}. \]Convert Radius Form to Interval Form
From
\[ |x-2|<5, \]we obtain
\[ -5Interior Convergence Is Absolute
A power series converges absolutely at every point strictly inside its radius of convergence.
\[ |x-a|Outside the Radius
If
\[ |x-a|>R, \]the power series diverges.
What Happens at the Boundary?
At
\[ |x-a|=R, \]several outcomes are possible:
- both endpoints converge;
- both endpoints diverge;
- one endpoint converges and the other diverges.
Example with Divergent Endpoints
Consider
\[ \sum_{n=1}^{\infty}\frac{x^n}{n}. \]The radius is \(R=1\).
At \(x=1\),
\[ \sum\frac1n \]diverges.
At \(x=-1\),
\[ \sum\frac{(-1)^n}{n} \]converges conditionally.
Endpoint Behavior Can Differ
The previous example has interval of convergence
\[ \boxed{[-1,1)}. \]One endpoint is included and the other is not.
Example with Both Endpoints Convergent
Consider
\[ \sum_{n=1}^{\infty}\frac{x^n}{n^2}. \]The radius is \(R=1\).
At both \(x=1\) and \(x=-1\), the resulting series converges absolutely.
Therefore the interval is
\[ \boxed{[-1,1]}. \]Example with Both Endpoints Divergent
Consider
\[ \sum_{n=0}^{\infty}x^n. \]The radius is \(R=1\).
At \(x=1\), the series becomes
\[ 1+1+1+\cdots, \]and at \(x=-1\),
\[ 1-1+1-1+\cdots. \]Both diverge, so the interval is
\[ \boxed{(-1,1)}. \]Using the Root Test
The root test can also determine a power series radius:
\[ L= \limsup_{n\to\infty} \sqrt[n]{|c_n(x-a)^n|}. \]This separates as
\[ L= |x-a| \limsup_{n\to\infty}\sqrt[n]{|c_n|}. \]Cauchy-Hadamard Radius Formula
In advanced form, the radius is given by
\[ \boxed{ \frac1R= \limsup_{n\to\infty}\sqrt[n]{|c_n|} }. \]This formula is useful when coefficient ratios do not settle cleanly.
Ratio Formula for the Radius
When the coefficient ratio limit exists and is positive,
\[ \lim_{n\to\infty} \left|\frac{c_n}{c_{n+1}}\right| =R. \]This is a convenient shortcut derived from the ratio test.
Example: Factorial Coefficients
Consider
\[ \sum_{n=0}^{\infty}\frac{x^n}{n!}. \]The ratio test gives
\[ \left| \frac{x^{n+1}/(n+1)!}{x^n/n!} \right| = \frac{|x|}{n+1}\to0. \]Therefore the series converges for every real \(x\), so
\[ \boxed{R=\infty}. \]Example: Coefficients Growing Too Fast
Consider
\[ \sum_{n=0}^{\infty}n!x^n. \]The ratio of consecutive absolute terms is
\[ (n+1)|x|. \]For any \(x\ne0\), this eventually exceeds \(1\).
The series converges only at \(x=0\), so
\[ \boxed{R=0}. \]A Power Series Can Have Any Radius
The coefficients determine the radius.
By changing \(c_n\), we can obtain:
- radius \(0\);
- any chosen finite positive radius;
- infinite radius.
Geometric-Series Substitution
Starting from
\[ \frac1{1-r} = \sum_{n=0}^{\infty}r^n, \]we can replace \(r\) with an expression involving \(x\).
Example: Replace \(r\) with \(2x\)
\[ \frac1{1-2x} = \sum_{n=0}^{\infty}(2x)^n = \sum_{n=0}^{\infty}2^nx^n. \]The geometric condition is
\[ |2x|<1, \]so
\[ |x|<\frac12. \]Example: Shifted Geometric Representation
Write
\[ \frac1{3-x} \]in geometric form:
\[ \frac1{3-x} = \frac13\frac1{1-x/3}. \]Therefore
\[ \boxed{ \frac1{3-x} = \sum_{n=0}^{\infty}\frac{x^n}{3^{n+1}}, \qquad |x|<3. } \]Example: Centering at a Nonzero Point
Consider
\[ \frac1{4-(x-1)}. \]Then
\[ \frac1{4-(x-1)} = \frac14 \frac1{1-(x-1)/4}. \]Hence
\[ \frac1{4-(x-1)} = \sum_{n=0}^{\infty} \frac{(x-1)^n}{4^{n+1}}, \qquad |x-1|<4. \]Power Series Can Be Added
Suppose
\[ \sum a_n(x-a)^n \]and
\[ \sum b_n(x-a)^n \]both converge in a common interval.
Then
\[ \sum (a_n+b_n)(x-a)^n \]represents the sum of the two functions there.
Power Series Can Be Scaled
Multiplying a power series by a constant \(k\) gives
\[ k\sum c_n(x-a)^n = \sum kc_n(x-a)^n. \]Power Series Can Be Multiplied
The product of two power series uses coefficient convolution.
If
\[ A(x)=\sum_{n=0}^{\infty}a_nx^n \]and
\[ B(x)=\sum_{n=0}^{\infty}b_nx^n, \]then formally
\[ A(x)B(x) = \sum_{n=0}^{\infty} \left( \sum_{k=0}^{n}a_kb_{n-k} \right)x^n. \]The Cauchy Product
The coefficient
\[ c_n= \sum_{k=0}^{n}a_kb_{n-k} \]is called the Cauchy-product coefficient.
Term-by-Term Differentiation
Inside the radius of convergence,
\[ \sum_{n=0}^{\infty}c_n(x-a)^n \]may be differentiated term by term:
\[ \boxed{ \frac{d}{dx} \sum_{n=0}^{\infty}c_n(x-a)^n = \sum_{n=1}^{\infty} nc_n(x-a)^{n-1} }. \]Differentiation Preserves the Radius
The differentiated power series has the same radius of convergence as the original power series.
Endpoint behavior may change and must again be checked separately.
Term-by-Term Integration
Inside the radius of convergence,
\[ \boxed{ \int \sum_{n=0}^{\infty}c_n(x-a)^n\,dx = C+ \sum_{n=0}^{\infty} \frac{c_n}{n+1}(x-a)^{n+1} }. \]Integration Also Preserves the Radius
The integrated power series has the same radius of convergence as the original.
Again, endpoints require separate analysis.
Generating New Series from \(1/(1-x)\)
Start with
\[ \frac1{1-x} = \sum_{n=0}^{\infty}x^n. \]Differentiate:
\[ \frac1{(1-x)^2} = \sum_{n=1}^{\infty}nx^{n-1}. \]A Reindexed Derivative Series
Replacing \(n-1\) by \(n\) gives
\[ \frac1{(1-x)^2} = \sum_{n=0}^{\infty}(n+1)x^n, \qquad |x|<1. \]Generating a Logarithmic Series
Integrate
\[ \frac1{1+x} = 1-x+x^2-x^3+\cdots \]from \(0\) to \(x\):
\[ \ln(1+x) = x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\cdots. \]Power Series and Approximation
Truncating a power series after finitely many terms produces a polynomial.
This polynomial can approximate the represented function near the center.
Partial Sums of a Power Series
The \(N\)-th partial sum is
\[ P_N(x)=\sum_{n=0}^{N}c_n(x-a)^n. \]This is an ordinary polynomial.
The Infinite Series as a Limit of Polynomials
Where the power series converges,
\[ f(x)=\lim_{N\to\infty}P_N(x). \]In this sense, the function is built from an infinite sequence of polynomial approximations.
Why the Center Matters for Approximation
Powers of
\[ x-a \]become small when \(x\) is close to \(a\).
This often makes lower-degree terms dominate near the center.
Radius as a Domain of Reliable Series Representation
The radius of convergence describes how far the power-series representation extends from its center before the infinite sum ceases to converge.
Radius Is Not an Error Bound
The statement
\[ |x-a|It does not by itself tell us how many terms are needed for a desired numerical accuracy.
Convergence and Approximation Are Separate Questions
First determine whether the power series converges at the chosen \(x\).
Then determine how rapidly its partial sums approach the limiting function.
Later lessons develop systematic error estimates.
Common Error: Forgetting the Center
In
\[ \sum c_n(x-4)^n, \]the center is \(4\), not \(0\).
Common Error: Reporting Only the Radius
If a problem asks for the interval of convergence, \(R\) alone is incomplete.
The endpoints must be tested separately.
Common Error: Automatically Including Endpoints
The condition
\[ |x-a|Endpoint inclusion depends on separate numerical-series tests.
Common Error: Automatically Excluding Endpoints
Endpoints do not always diverge.
Some power series converge at one or both boundary points.
Common Error: Using the Ratio Test at \(L=1\)
When the ratio-test limit is \(1\), the test is inconclusive.
This is exactly why endpoint tests often require comparison, \(p\)-series, alternating-series, or other methods.
Common Error: Confusing Radius with Interval Length
If the radius is \(R\), the interior convergence interval has total length
\[ 2R, \]centered at \(a\).
Common Error: Losing an Index Shift
When differentiating, integrating, or reindexing, carefully track the exponent, coefficient, and starting index.
A Reliable Radius-and-Interval Procedure
- Identify the center \(a\).
- Apply the ratio or root test to the general term.
- Solve the resulting inequality for \(|x-a|\).
- Read off the radius \(R\).
- Convert the inequality into an open interval.
- Test the left endpoint separately.
- Test the right endpoint separately.
- Write the final interval with correct bracket or parenthesis notation.
A Reliable Representation Procedure
- Begin with a known power series, usually geometric.
- Rewrite the target function to match that structure.
- Substitute the appropriate expression for the geometric ratio.
- Multiply by any outside constants.
- State the convergence condition.
- Differentiate, integrate, or reindex if needed.
Worked Example: Radius and Interval
Find the radius and interval of convergence of
\[ \sum_{n=1}^{\infty}\frac{(x-3)^n}{n2^n}. \]The ratio test gives the interior condition
\[ \left|\frac{x-3}{2}\right|<1. \]Thus
\[ |x-3|<2, \]so \(R=2\) and the candidate interval is
\[ 1Worked Example: Left Endpoint
At \(x=1\),
\[ \sum_{n=1}^{\infty} \frac{(1-3)^n}{n2^n} = \sum_{n=1}^{\infty}\frac{(-1)^n}{n}. \]This converges by the Alternating Series Test.
Worked Example: Right Endpoint
At \(x=5\),
\[ \sum_{n=1}^{\infty} \frac{(5-3)^n}{n2^n} = \sum_{n=1}^{\infty}\frac1n. \]This diverges.
Therefore the interval of convergence is
\[ \boxed{[1,5)}. \]Worked Example: Infinite Radius
Find the radius of
\[ \sum_{n=0}^{\infty}\frac{(x+2)^n}{n!}. \]The ratio test gives
\[ \frac{|x+2|}{n+1}\to0. \]This is less than \(1\) for every real \(x\), so
\[ \boxed{R=\infty}. \]Worked Example: Radius Zero
Find the radius of
\[ \sum_{n=0}^{\infty}n!(x-1)^n. \]The ratio of magnitudes is
\[ (n+1)|x-1|. \]For every \(x\ne1\), this eventually exceeds \(1\).
Therefore
\[ \boxed{R=0}. \]Worked Example: Geometric Representation
Represent
\[ \frac1{2+x} \]as a power series centered at \(0\).
Rewrite:
\[ \frac1{2+x} = \frac12\frac1{1+x/2}. \]Since
\[ \frac1{1-(-x/2)} = \sum_{n=0}^{\infty}\left(-\frac{x}{2}\right)^n, \]we obtain
\[ \boxed{ \frac1{2+x} = \sum_{n=0}^{\infty} \frac{(-1)^n x^n}{2^{n+1}}, \qquad |x|<2. } \]Practice 1: Identify the Center
Identify the center of
\[ \sum_{n=0}^{\infty}\frac{(x-5)^n}{n+1}. \]Practice 2: Geometric Radius
Find the radius and interval of convergence of
\[ \sum_{n=0}^{\infty}\left(\frac{x}{4}\right)^n. \]Practice 3: Shifted Geometric Series
Find the radius and interval of convergence of
\[ \sum_{n=0}^{\infty}\frac{(x+1)^n}{3^n}. \]Practice 4: Endpoint Testing
Find the interval of convergence of
\[ \sum_{n=1}^{\infty}\frac{x^n}{n}. \]Practice 5: Both Endpoints
Find the interval of convergence of
\[ \sum_{n=1}^{\infty}\frac{x^n}{n^2}. \]Practice 6: Factorial Denominator
Find the radius of convergence of
\[ \sum_{n=0}^{\infty}\frac{(x-2)^n}{n!}. \]Practice 7: Factorial Numerator
Find the radius of convergence of
\[ \sum_{n=0}^{\infty}n!(x+3)^n. \]Practice 8: Build a Representation
Express
\[ \frac1{1+2x} \]as a power series centered at \(0\), and state its interval of convergence.
Practice 9: Differentiate a Power Series
Starting from
\[ \frac1{1-x} = \sum_{n=0}^{\infty}x^n, \]differentiate term by term to obtain a series for
\[ \frac1{(1-x)^2}. \]Practice 10: Explain the Endpoint Rule
Explain why the radius of convergence alone does not determine whether the endpoints belong to the interval of convergence.
Practice Answers
Practice 1
The series is written in powers of
\[ x-5, \]so the center is
\[ \boxed{5}. \]Practice 2
This is geometric with ratio \(x/4\). Require
\[ \left|\frac{x}{4}\right|<1. \]Therefore
\[ |x|<4, \qquad \boxed{R=4}, \]and both endpoints diverge, so
\[ \boxed{(-4,4)}. \]Practice 3
Require
\[ \left|\frac{x+1}{3}\right|<1. \]Thus
\[ |x+1|<3, \]so
\[ \boxed{R=3}. \]
The candidate interval is \(-4
The radius is \(1\). At \(x=1\), the harmonic series diverges. At \(x=-1\), the
alternating harmonic series converges. Therefore
The radius is \(1\). Both endpoint series converge absolutely, so
The ratio test gives
for every real \(x\). Therefore
The ratio of consecutive magnitudes is
This diverges for every \(x\ne-3\). Hence
Write
Therefore
for
or
Differentiate:
Reindexing gives
The ratio or root test determines convergence for
Practice 4
Practice 5
Practice 6
Practice 7
Practice 8
Practice 9
Practice 10
What to Remember
A power series has the form
\[ \boxed{ \sum_{n=0}^{\infty}c_n(x-a)^n }. \]- \(a\) is the center.
- \(R\) is the radius of convergence.
- The series converges absolutely when \(|x-a|
- It diverges when \(|x-a|>R\).
- Endpoints must be tested separately.
- Within the radius, power series can be added, scaled, differentiated, and integrated term by term.
The essential shift is conceptual: a power series is not merely an infinite sum of numbers. It is an infinite process that defines a function across a region of \(x\)-values.
Where We Go Next
Power series give a general language for representing functions with infinite polynomial-like expressions. The next lesson develops the most important systematic way to generate their coefficients from derivatives.
The next lesson, Taylor and Maclaurin Series, builds power series directly from a function's derivatives at a chosen center and connects local derivative information to global approximation.