Key Concepts of Power Series

Why This Matters

Power series let you transform complicated functions into infinite polynomials that you can differentiate, integrate, and manipulate with ease. In Calculus II, you're expected to determine where a series converges, how to represent functions as series, and why term-by-term operations work. These concepts connect directly to approximation theory, differential equations, and the fundamental relationship between discrete sums and continuous functions.

When you see a power series problem, immediately think about convergence behavior, the relationship between a function and its series representation, and how operations like differentiation and integration affect the series.

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Foundational Structure

Power series have a specific anatomy that determines everything else: their convergence, their usefulness, and how you manipulate them. You need to understand the basic form before tackling applications.

Definition of a Power Series

A power series is an infinite sum of the form:

$\sum_{n=0}^{\infty} a_n (x - c)^n = a_0 + a_1(x-c) + a_2(x-c)^2 + \cdots$

Here $a_n$ are the coefficients and $c$ is the center. Each term is a polynomial-like piece built on powers of $(x - c)$, and the whole series expresses a function as this infinite sum.

The series will converge for some values of $x$ and diverge for others. Figuring out that domain is a huge part of working with power series.

Geometric Series

The geometric series is the simplest power series:

$\sum_{n=0}^{\infty} ar^n = a + ar + ar^2 + \cdots$

• Convergence condition: converges only when $|r| < 1$, diverges when $|r| \geq 1$
• Closed-form sum: $\frac{a}{1 - r}$

This formula shows up constantly. For example, $\frac{1}{1-x} = \sum_{n=0}^{\infty} x^n$ for $|x| < 1$. Many series manipulation problems start by recognizing a geometric series hiding inside a more complex expression.

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Convergence Analysis

The central question for any power series is where does it converge? This determines the domain where your series actually represents a meaningful function.

Radius of Convergence

The radius of convergence $R$ tells you how far from the center $c$ the series converges. In practice, you'll almost always find $R$ using the Ratio or Root Test (see below) rather than the formal definition.

• The series converges absolutely when $|x - c| < R$
• The series diverges when $|x - c| > R$
• Special cases: $R = 0$ means convergence only at $x = c$; $R = \infty$ means convergence for all $x$

Interval of Convergence

The interval of convergence includes every $x$ value where the series converges. Typically this looks like $(c - R,\; c + R)$, but the endpoints need separate attention.

• The Ratio/Root Test is inconclusive at $|x - c| = R$, so you must plug in $x = c - R$ and $x = c + R$ and test each one individually.
• Use parentheses for excluded endpoints and brackets for included ones. For instance, $[c - R,\; c + R)$ means the left endpoint converges but the right one diverges.

Convergence Tests for Power Series

Here's how to find $R$ and handle endpoints:

1. Apply the Ratio Test. Compute $L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|$. The series converges absolutely when $L \cdot |x - c| < 1$, which gives $R = \frac{1}{L}$. This test works especially well when coefficients involve factorials (e.g., $\frac{n!}{3^n}$).

2. Or apply the Root Test. Compute $\lim_{n \to \infty} \sqrt[n]{|a_n|}$. This handles expressions like $n^n$ or $a^n$ more cleanly than the Ratio Test.

3. Test the endpoints. Once you have $R$, substitute each endpoint into the series and use whichever test fits: the alternating series test, the $p$-series test, direct comparison, or limit comparison.

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Function Representation

Power series aren't just abstract objects. They represent real functions. Taylor and Maclaurin series formalize how to build a power series from any sufficiently smooth function.

Taylor Series

The Taylor series of $f(x)$ centered at $c$ is:

$f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(c)}{n!} (x - c)^n$

Each coefficient $a_n = \frac{f^{(n)}(c)}{n!}$ is determined by the $n$th derivative of $f$ evaluated at the center. To build a Taylor series:

1. Compute successive derivatives $f(c),\; f'(c),\; f''(c),\; f'''(c), \ldots$
2. Divide each by $n!$
3. Multiply by $(x - c)^n$ and sum

Truncating after a finite number of terms gives a Taylor polynomial, which approximates $f(x)$ near $c$. More terms means better accuracy in that neighborhood.

Maclaurin Series

A Maclaurin series is just a Taylor series centered at $c = 0$:

$f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} x^n$

You should memorize these standard Maclaurin series:

• $e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}$ (converges for all $x$)
• $\sin x = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!}$ (converges for all $x$)
• $\cos x = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!}$ (converges for all $x$)
• $\frac{1}{1-x} = \sum_{n=0}^{\infty} x^n$ (converges for $|x| < 1$)
• $\ln(1+x) = \sum_{n=1}^{\infty} \frac{(-1)^{n+1} x^n}{n}$ (converges for $-1 < x \leq 1$)

These are building blocks. Many exam problems ask you to modify one of these (by substitution, differentiation, or integration) rather than build a series from scratch.

Representation of Functions as Power Series

Exponential, trigonometric, logarithmic, and rational functions can all be expressed as power series. The practical payoff: once a function is in series form, you can apply polynomial techniques to integrate or differentiate it. This is especially useful for functions like $e^{-x^2}$ that have no elementary antiderivative. Power series solutions also come up when standard methods for differential equations fail.

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Operations on Power Series

One of the most useful features of power series is that you can differentiate and integrate them term by term, as long as you stay within the interval of convergence.

Differentiation of Power Series

Taking the derivative term by term:

$\frac{d}{dx}\sum_{n=0}^{\infty} a_n (x - c)^n = \sum_{n=1}^{\infty} n\, a_n (x - c)^{n-1}$

Notice the index shifts to $n = 1$ (the constant term drops out). The radius of convergence stays the same, though endpoint behavior may change. For example, a series that converges at an endpoint might diverge there after differentiation.

This is how you'd derive the series for $\frac{1}{(1-x)^2}$: differentiate the geometric series $\frac{1}{1-x} = \sum x^n$ to get $\frac{1}{(1-x)^2} = \sum_{n=1}^{\infty} n\,x^{n-1}$.

Integration of Power Series

Integrating term by term:

$\int \sum_{n=0}^{\infty} a_n (x - c)^n\, dx = \sum_{n=0}^{\infty} \frac{a_n}{n+1} (x - c)^{n+1} + C$

Again, the radius of convergence is preserved, though endpoints may now converge where they didn't before. Don't forget the constant of integration $C$.

This is how you'd find the series for $\ln(1-x)$: integrate $\frac{-1}{1-x} = -\sum x^n$ term by term to get $\ln(1-x) = -\sum_{n=1}^{\infty} \frac{x^n}{n} + C$, then use the initial condition $\ln(1) = 0$ to find $C = 0$.

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Quick Reference Table

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Self-Check Questions

1. What is the relationship between the radius of convergence and the interval of convergence, and why must you test endpoints separately?

2. Compare Taylor series and Maclaurin series. When would you choose one over the other for approximating a function?

3. If you differentiate a power series, what happens to its radius of convergence? What about its interval of convergence (including endpoints)?

4. Which convergence test would you use for a series with coefficients involving factorials versus one with coefficients involving $n$th powers?

5. Starting from the Maclaurin series for $\frac{1}{1-x}$, how would you derive the series for $\frac{1}{(1-x)^2}$ and for $\ln(1-x)$ using term-by-term operations?