What Is a Power Series?

A Series Whose Terms Depend on an Input

In earlier tutorials, a series was a fixed sequence of terms, such as $\sum a_n$. A power series introduces an input: for each real number $x$, it produces a numerical series whose terms depend on $x$. This lets one study a whole family of series together. The center determines where the powers are measured from, and the coefficients determine how those powers are weighted.

The expression is useful only after an input is chosen. At some inputs, the resulting numerical series converges; at others, it may diverge. We will first make this distinction precise, then prove a basic transfer principle: convergence at one point guarantees absolute convergence at every point strictly closer to the center. The behavior at the boundary is a separate question.

Definition: A power series centered at $a$ is an expression of the form $$ \sum_{n=0}^{\infty} c_n(x-a)^n, $$ where $a$ and each coefficient $c_n$ are real numbers. For a fixed input $x$, the associated numerical series is obtained by substituting that value of $x$ into every term. The power series converges at $x$ if this numerical series converges, and it converges absolutely at $x$ if the series of absolute values of its terms converges.

The number $a$ is the center, and $(c_n)_{n=0}^{\infty}$ is the coefficient sequence. The term with index $n=0$ is $c_0$, since $(x-a)^0=1$. When we evaluate at a fixed input, it can be helpful to write the result as

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

For each nonnegative integer $N$, the finite partial sum $S_N(x)=\sum_{n=0}^{N}c_n(x-a)^n$ is a polynomial in $x$. The infinite power series is not automatically a function on all of $\mathbb{R}$: its value is defined by the limit of these partial sums only at inputs where the associated numerical series converges. Thus, the same symbolic expression can be convergent at one input and divergent at another.

What Happens at the Center?

At the center, all positive powers vanish. This gives an immediate value, regardless of how complicated the coefficient sequence is.

Proposition (Value at the Center): Every power series $\sum_{n=0}^{\infty}c_n(x-a)^n$ converges at $x=a$, and its sum there is $c_0$.

Proof. Substituting $x=a$, the term with $n=0$ is $c_0$, while for every $n\geq1$, $(a-a)^n=0^n=0$. Therefore the numerical series at the center is

c_0+\sum_{n=1}^{\infty}c_n(a-a)^n = c_0+\sum_{n=1}^{\infty}0 = c_0.

Its partial sums are all equal to $c_0$, so they converge to $c_0$. $\square$

This result is a useful check when reading a power series: the coefficient $c_0$ is its value at the center. It does not imply anything about convergence at other inputs. For that, we need to compare the sizes of the powers $(x-a)^n$.

Convergence at One Point Controls Closer Points

Suppose the series converges at an input $x_0$ different from its center. Its terms $c_n(x_0-a)^n$ must tend to zero by the Necessary Condition for Series Convergence. In particular, those terms are bounded. At a closer input $y$, each positive-index term is the corresponding term at $x_0$, multiplied in absolute value by a geometric factor. The Comparison Principle for Positive-Term Series then gives the result.

Theorem (Convergence Transfers Inward): Suppose the power series $\sum_{n=0}^{\infty}c_n(x-a)^n$ converges at some $x_0\neq a$. Then it converges absolutely at every $y$ satisfying $|y-a|<|x_0-a|$.

Proof. Since the series converges at $x_0$, its terms tend to zero by the Necessary Condition for Series Convergence. Therefore the sequence $u_n=c_n(x_0-a)^n$, for $n\geq1$, is bounded. Choose $M\geq0$ such that $|u_n|\leq M$ for every $n\geq1$.

Fix $y$ with $|y-a|<|x_0-a|$. If $y=a$, every term of positive index is zero, so the series at $y$ converges absolutely. Otherwise, define $r=|y-a|/|x_0-a|$; then $0<r<1$. For every $n\geq1$,

|c_n(y-a)^n| = |c_n(x_0-a)^n| \left(\frac{|y-a|}{|x_0-a|}\right)^n = |u_n|r^n \leq Mr^n.

The geometric series $\sum_{n=1}^{\infty}Mr^n$ converges because $0<r<1$, by the Convergence Criterion for a Geometric Series. The Comparison Principle for Positive-Term Series now shows that $\sum_{n=1}^{\infty}|c_n(y-a)^n|$ converges. The index-zero term $|c_0|$ is a single finite term, so including it preserves convergence. Thus the power series converges absolutely at $y$. $\square$

The strict inequality in the hypothesis matters. When $|y-a|=|x_0-a|$, the factor $r$ in the proof equals $1$; the geometric comparison no longer provides a convergent bound. The theorem therefore makes no claim about inputs exactly as far from the center as $x_0$.

Worked Examples: Evaluating and Comparing Power Series

Worked Example: A Geometric Power Series

Consider the power series centered at $1$

\sum_{n=0}^{\infty}\frac{(x-1)^n}{4^n}.

Here $a=1$ and $c_n=1/4^n$. At $x=2$, the input is one unit from the center, and substitution gives

\sum_{n=0}^{\infty}\frac{(2-1)^n}{4^n} = \sum_{n=0}^{\infty}\left(\frac14\right)^n = \frac{1}{1-\frac14} = \frac43.

The last equality uses the Sum of a Convergent Geometric Series. Since the series converges at $x_0=2$, Convergence Transfers Inward: it converges absolutely at every $y$ with $|y-1|<1$. This conclusion can also be checked directly: the absolute values of the terms at $y$ are $(|y-1|/4)^n$, a convergent geometric series whenever $|y-1|<1$.

At the center, $x=1$, the sum is $1$, as the Value at the Center proposition predicts. At $x=5$, the terms are all $1$, so they do not tend to zero and the series diverges by the Necessary Condition for Series Convergence. These checks show how changing the input changes the numerical series.

Worked Example: A Power Series That Converges Conditionally at One Point

Consider

\sum_{n=1}^{\infty}\frac{(-1)^{n-1}(x+2)^n}{n},

which is centered at $-2$. At $x=-1$, we have $x+2=1$, so the series becomes

\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}.

The Alternating Series Test applies: the positive terms $1/n$ decrease and tend to zero. Thus the series converges. But it does not converge absolutely, since the series of absolute values is the harmonic series $\sum_{n=1}^{\infty}1/n$, which diverges by the $p$-Series Convergence Criterion with $p=1$.

Because it converges at $x_0=-1$, Convergence Transfers Inward and guarantees absolute convergence whenever $|y+2|<1$. At the other point at the same distance, $y=-3$, substitution gives

\frac{(-1)^{n-1}(-1)^n}{n} = -\frac{1}{n} \qquad\text{for each }n\geq1.

The resulting series is the negative harmonic series and diverges. The theorem does not apply at $y=-3$, because $|y+2|=1=|x_0+2|$. This example illustrates why the boundary cannot be settled just by knowing convergence at a point on it.

Worked Example: Using the Necessary Condition to Detect Divergence

Consider the power series centered at $3$

\sum_{n=1}^{\infty} n(x-3)^n.

At $x=4$, its $n$th term is $n(4-3)^n=n$. Since $n$ does not tend to zero, the Necessary Condition for Series Convergence shows that the series diverges there. At $x=3$, by contrast, every term is zero and the sum is $0$, in agreement with the Value at the Center proposition.

The theorem Convergence Transfers Inward cannot be used to reach a conclusion at $x=4$: its required starting point is a point $x_0\neq3$ where the series is already known to converge, and $x=4$ is not such a point. This example emphasizes that the transfer theorem is a sufficient condition, not a procedure for declaring convergence at every input.

What the Definition Reveals—and What It Does Not

A power series can be viewed as a collection of numerical series indexed by $x$. The center is the one input whose value is known immediately; the coefficient sequence records the weights on successive powers of the displacement from that center. Partial sums provide polynomial approximations at each input, but an infinite sum exists only where those partial sums converge.

The inward-transfer theorem gives a strong structural fact without requiring a new test for every input. Once convergence is established at a point away from the center, all strictly closer points are points of absolute convergence. In particular, conditional convergence at a point cannot occur if there is another known convergence point strictly farther from the center: the transfer theorem would force absolute convergence at the closer point.

There is an important limitation. The theorem does not decide what happens at inputs at the same distance from the center as a known convergence point, and it does not say that the series converges at every farther point. The second worked example showed that two inputs equally far from the center can behave differently. At any proposed input, substitution gives a numerical series, and the convergence tests developed earlier in this course remain essential.

Takeaway: A power series is a numerical series for each chosen input. It always converges at its center, with value $c_0$; if it converges at any other point, it converges absolutely at every point strictly closer to the center. Inputs at the same or greater distance require separate analysis.

Check Your Understanding

Use the definitions and results in this tutorial to answer the following questions.

For a power series centered at $a$, what is its value at $x=a$, and why?
If a power series converges at $x_0\neq a$, what can be concluded at an input $y$ satisfying $|y-a|<|x_0-a|$?
In the proof of Convergence Transfers Inward, why are the terms $c_n(x_0-a)^n$ bounded?
Why does that theorem not determine convergence at a point $y$ with $|y-a|=|x_0-a|$?
For the series $\sum_{n=1}^{\infty}(-1)^{n-1}(x+2)^n/n$, what numerical series results at $x=-3$, and does it converge?

What You'll Learn