Power Series

Letting the terms of a series carry a variable turns the series into a function. A power series is a series of the form

where the constants $c_n$ are the coefficients. For each fixed $x$ this is a series of numbers that may converge or diverge; the sum, where it exists, is a function of $x$. A power series looks like a polynomial with infinitely many terms, and much of its behavior follows that analogy.

More generally, a power series centered at $a$ is

adopting the convention $(x-a{)}^0=1$ even at $x=a$. Setting $x=a$ makes every term past the first vanish, so a power series centered at $a$ always converges at $x=a$, to $c_0$. The question is what other $x$ it converges for.

The radius and interval of convergence

The convergence set of a power series is always an interval, centered at $a$. The Ratio Test shows why. Applied to $\sum c_n(x-a{)}^n$, the ratio of successive terms carries a factor $|x-a|$, so the test compares $|x-a|$ to a threshold.

For $\sum n!x^n$ the ratio is $|(n+1)!x^{n+1}/(n!x^n)|=(n+1)|x|\to \infty$ for any $x\ne 0$, so the series converges only at $x=0$ ($R=0$). For the geometric series $\sum x^n$ the ratio is $|x|$, giving convergence on $(-1,1)$ ($R=1$). Between these extremes lies the third case, a finite positive radius with endpoints to settle.

The Ratio (or Root) Test finds $R$ but says nothing at the endpoints $a\pm R$, where it always returns the inconclusive $L=1$. Each endpoint must be checked by substituting it and applying a different test. All four outcomes occur: $(a-R,a+R)$, $[a-R,a+R)$, $(a-R,a+R]$, or $[a-R,a+R]$.

Functions as power series

Power series represent functions in a form that can be differentiated and integrated term by term. The starting point is the geometric series, read from right to left as a formula for a function:

New representations come from three operations: substitution, multiplication by a power of $x$, and factoring a constant out of the denominator.

• Substitution. Replacing $x$ by $-x^2$, $$\frac{1}{1+x^2}=\sum _{n=0}^{\infty }(-1{)}^n{x}^{2n}=1-x^2+x^4-x^6+\cdots ,|x|<1.$$
• Factoring the denominator. To expand $1/(2+x)$, write it as $\frac{1}{2}\cdot \frac{1}{1-(-x/2)}$ and use the geometric formula with ratio $-x/2$: $$\frac{1}{2+x}=\sum _{n=0}^{\infty }\frac{(-1{)}^n}{2^{n+1}}{x}^n,|x|<2.$$
• Multiplying by $x^k$. Then $x^3/(2+x)=\sum (-1{)}^n{x}^{n+3}/2^{n+1}$ on the same interval, since $x^3$ can be taken inside the summation sign.

Differentiating and integrating term by term

Inside its interval of convergence a power series behaves like an infinite polynomial: it can be differentiated and integrated one term at a time, and the radius of convergence is unchanged.

The radius is preserved, but an endpoint can change status: a series may converge at $a+R$ while its derivative diverges there. Two applications produce series that algebraic manipulation alone does not.

Differentiating. Differentiating $1/(1-x)=\sum x^n$ term by term gives

Integrating. Since $\frac{\mathbf{d}}{\mathbf{d}x}\ln (1+x)=\frac{1}{1+x}=1-x+x^2-\cdots$, integrating term by term (and fixing $C=0$ from $x=0$) gives

The same route, starting from $1/(1+x^2)$, produces Gregory's series for the inverse tangent:

At $x=1$ this collapses to the Leibniz formula $\frac{\pi }{4}=1-\frac{1}{3}+\frac{1}{5}-\cdots$.

Beyond generating identities, term-by-term integration evaluates integrals with no elementary antiderivative. Since $1/(1+x^7)=\sum (-1{)}^n{x}^{7n}$,

and this series, being alternating for $0\le x<1$, gives a definite integral to any accuracy with a controlled error.

Series representations obtained

Each representation above came from manipulating the geometric series, not from computing coefficients directly. Tying the coefficients $c_n$ to the derivatives of $f$ at the center is the route of the Taylor and Maclaurin series.