Why Every Power Series Converges at Its Center

Why the Center Is a Special Point

A power series is built from powers of the difference between its input and a fixed number called its center. At the center, that difference is zero. Every positive-degree term therefore disappears. The degree-zero term is different: it is the constant coefficient, and it remains.

This simple observation proves something important: no matter how complicated the coefficients are, a power series converges at its center, and its sum there is its constant coefficient. The proof does not require a convergence test or any restrictions on how the coefficients behave.

Power Series and Their Centers

A power series centered at $c$ has the form below, where $c$ is a real number and the coefficients $a_0,a_1,a_2,\ldots$ are real numbers.

\sum_{n=0}^{\infty} a_n(x-c)^n

The number $c$ is the center. The coefficient $a_0$ is the constant coefficient, because its term has degree zero. The coefficients $a_1,a_2,\ldots$ multiply terms of positive degree.

Definition: The partial sum through degree $N$ of a power series centered at $c$ is $$ S_N(x)=\sum_{n=0}^{N}a_n(x-c)^n. $$ The infinite series converges at an input $x$ if the sequence of partial sums $S_N(x)$, as $N$ increases, approaches a finite limit. That limit is the sum of the series at $x$.

At $x=c$, the difference $x-c$ equals zero. For each positive integer $n$, its $n$-th power is zero. But the degree-zero term is $a_0$, not zero: it is the constant term in the series. We can express the evaluation without assigning a value to $0^0$: the $n=0$ term is $a_0$, and every term with $n\geq 1$ vanishes.

The Main Result

Theorem: Every power series $$ \sum_{n=0}^{\infty}a_n(x-c)^n $$ converges at its center $x=c$, and its sum there is $a_0$. This holds for every choice of finite coefficients $a_n$; no additional condition on the coefficients is required.

The key is to use the definition of convergence of a series through its partial sums. At the center, every partial sum, not merely some approximation to the sum, is exactly the same value.

Evaluate the center difference.
When $x=c$, we have $x-c=0$.

Separate the constant term.
The $n=0$ term is $a_0$; each term with $1\leq n\leq N$ contains a positive power of zero.

Find every partial sum.
For every $N\geq 0$, the partial sum at the center equals $a_0$.

Take the limit.
A sequence whose every entry is $a_0$ converges to $a_0$. Thus the series converges at $c$ to $a_0$.

Why the Proof Works

For $N\geq 0$, write the partial sum at the center and separate its first term:

S_N(c) = \sum_{n=0}^{N}a_n(c-c)^n = a_0+\sum_{n=1}^{N}a_n0^n.

Every exponent in the remaining finite sum is positive. Thus $0^n=0$ for each $n=1,\ldots,N$, and each product $a_n0^n$ is zero. Therefore

$$ S_N(c)=a_0 $$

for every $N\geq 0$. The sequence of partial sums is consequently the constant sequence $a_0,a_0,a_0,\ldots$, whose limit is $a_0$. By the definition of convergence of a series, the original power series converges at $c$ to $a_0$. This proves the result.

Notice the role of the partial sums: a series is not evaluated by informally manipulating an infinite list of terms. We show that the finite partial sums have a limit. Here this is especially direct because each partial sum is already equal to the same value.

Worked Evaluations

Worked Example: A Series Centered at Four

Evaluate the power series $\sum_{n=0}^{\infty}(n+2)(x-4)^n$ at its center.

The center is $c=4$. The coefficient at degree zero is found by setting $n=0$ in the coefficient formula: $a_0=0+2=2$. To verify directly, substitute $x=4$ into the first terms. The term for $n=0$ is $2(4-4)^0$, understood as the constant term $2$. The term for $n=1$ is $3(4-4)=3\cdot0=0$, and the term for $n=2$ is $4(4-4)^2=4\cdot0^2=0$. All subsequent terms also have positive powers of zero.

S_N(4)=2+0+0+\cdots+0=2 \qquad\text{for every }N\geq 0.

Thus the partial sums converge to $2$. The series converges at its center to its constant coefficient, $2$.

Check: Every partial sum shown is exactly $2$, so its limit is $2=a_0$, as claimed.

Worked Example: Alternating Coefficients Do Not Change the Center Value

Evaluate $\sum_{n=0}^{\infty}(2n+3)(-1)^n(x+1)^n$ at its center.

Since $x+1=x-(-1)$, the center is $c=-1$. The degree-zero coefficient is $(2\cdot0+3)(-1)^0=3$. At $x=-1$, the next terms are $(2\cdot1+3)(-1)^1(-1+1)=5(-1)\cdot0=0$ and $(2\cdot2+3)(-1)^2(-1+1)^2=7\cdot1\cdot0^2=0$. The same reasoning applies to every term of degree at least one.

S_N(-1)=3+0+0+\cdots+0=3 \qquad\text{for every }N\geq 0.

Therefore the series converges at $x=-1$ to $3$. The alternating signs in the coefficients do not affect this conclusion: every coefficient beyond the constant one is multiplied by zero at the center.

Check: The partial sums are all $3$, and $3$ is precisely the degree-zero coefficient.

Worked Example: Fractional Coefficients at a Shifted Center

Evaluate $\sum_{n=0}^{\infty}\frac{n^2+1}{3^n}(x-2)^n$ at its center.

The center is $c=2$. Evaluate the coefficient formula at $n=0$: $\frac{0^2+1}{3^0}=\frac{1}{1}=1$. The term for $n=1$, when $x=2$, is $\frac{1^2+1}{3^1}(2-2)=\frac{2}{3}\cdot0=0$. The term for $n=2$ is $\frac{2^2+1}{3^2}(2-2)^2=\frac{5}{9}\cdot0^2=0$. Each later term likewise contains a positive power of $2-2=0$.

S_N(2)=1+0+0+\cdots+0=1 \qquad\text{for every }N\geq 0.

Hence the series converges at $x=2$ to $1$, its constant coefficient.

Check: The value $1$ is the limit of the partial sums, since every partial sum at $2$ is $1$.

A Common Mistake: Losing the Constant Term

A tempting but incorrect calculation is to substitute $x=c$ and conclude that the whole series is zero because $x-c=0$. For example, one might write

\sum_{n=0}^{\infty}a_n(c-c)^n = \sum_{n=0}^{\infty}a_n0^n = 0.

The error is treating the degree-zero term as though it contained a positive power of zero. It does not: the term for $n=0$ is $a_0$. Only the terms for $n\geq1$ vanish. For instance, for the series $\sum_{n=0}^{\infty}(n+2)(x-4)^n$, the incorrect calculation would give $0$, while the correct partial sum at $x=4$ is $2+0+\cdots+0=2$.

Takeaway: At a power series’ center, retain the degree-zero term $a_0$. Every positive-degree term vanishes, so the series converges to $a_0$.

Why This Result Matters

The conclusion depends only on the form of a power series and the meaning of its center. The coefficients may grow rapidly, alternate in sign, or follow any other pattern; the proof still works because at the center each positive-degree term is exactly zero. The convergence conclusion comes from the partial sums being constant, not from comparing the coefficients with some other series.

This gives a reliable first evaluation whenever a power series is presented: identify its center, identify the coefficient of the degree-zero term, and substitute the center. The answer is that coefficient. Keep the distinction between the constant term and the positive-degree terms explicit, and the evaluation is both short and rigorous.

Check Your Understanding

Use partial sums to justify each answer, and distinguish the constant term from the terms of positive degree.

For $\sum_{n=0}^{\infty}(5-n)(x-3)^n$, identify the center and find the sum at the center.
For $\sum_{n=0}^{\infty}\frac{(-2)^n}{n+1}(x+4)^n$, what is the center value, and which term supplies it?
Explain why every partial sum at $x=c$ equals $a_0$ for $\sum_{n=0}^{\infty}a_n(x-c)^n$.
A student claims that any power series evaluates to zero at its center. Identify the error and correct the claim.
Does the theorem require the coefficients $a_n$ to approach zero? Explain using the partial-sum argument.