Why the Center Matters
A power series need not be written in powers of \(x\). It may instead be written in powers of \(x-c\), where \(c\) is its center. When \(c\) is not zero, the radius still measures a distance, but the convergence interval is located around \(c\), not around zero.
This distinction is easy to lose when translating a radius into an interval. A radius of \(3\) means that convergence occurs within distance \(3\) of the center. If the center is \(2\), the corresponding open interval is \((-1,5)\), not \((-3,3)\).
For real \(x\), the radius describes how far \(x\) is from \(c\). Thus the open interval inside the radius is \((c-R,c+R)\). If the series converges at one or both boundary points, those points are included in its full interval of convergence; if it diverges there, they are excluded.
The Center Is a Special Point
At the center, \(x=c\), every positive power of \(x-c\) is zero. The series therefore reduces to its constant term \(a_0\). This fact does not require a ratio test.
In particular, every power series converges at its center. When using a ratio test on the terms \(a_n(x-c)^n\), handle \(x=c\) separately: the ratio of consecutive terms at that point can be undefined, since it may take the form \(0/0\).
Finding the Radius from Coefficients
The center changes where the interval is located, while the coefficients determine its radius. Two standard coefficient tests express this relationship. If the coefficients are eventually nonzero and the limit \(L=\lim_{n\to\infty}|a_{n+1}/a_n|\) exists, the ratio test gives \(R=1/L\), with the usual conventions \(1/0=\infty\) and \(1/\infty=0\). If the ratio limit is unavailable, the root-test formula is \(R=1/\limsup_{n\to\infty}|a_n|^{1/n}\), using the same conventions.
For a fixed \(x\ne c\), and for sufficiently large \(n\) such that \(a_n\ne0\) and \(a_{n+1}\ne0\), the ratio of the absolute values of consecutive terms is
When the coefficient ratio tends to \(L\), the limit of this expression is \(L|x-c|\). The ratio test gives absolute convergence if \(L|x-c|<1\), and divergence if \(L|x-c|>1\). These inequalities are equivalent to \(|x-c|<1/L\) and \(|x-c|>1/L\), respectively. At equality, the ratio test is inconclusive.
Why the Radius Is a Distance
The shifted power \((x-c)^n\) is the key. For any fixed \(x\ne c\), its \(n\)th root in absolute value is exactly \(|x-c|\). Consequently, applying the root test to the general term gives the following reasoning.
For \(x\ne c\), \(\bigl|a_n(x-c)^n\bigr|^{1/n}=|a_n|^{1/n}|x-c|\).
Let \(\rho=\limsup_{n\to\infty}|a_n|^{1/n}\). The limsup of the roots of the terms is \(\rho|x-c|\).
If \(\rho|x-c|<1\), the series converges absolutely; if \(\rho|x-c|>1\), it diverges because its terms do not tend to zero.
The dividing distance is \(R=1/\rho\). The point \(x=c\) is handled separately, and equality at the boundary is not decided by this test.
This also explains why translating the center does not make the radius an endpoint coordinate. The coefficients set the critical distance \(|x-c|\); the center \(c\) then places that distance on the real line.
Worked Examples
Worked Example: A Geometric Series Centered at 2
Find the radius and interval of convergence of \(\sum_{n=0}^{\infty}(x-2)^n/3^n\).
Write the terms as \(a_n(x-2)^n\), where \(a_n=1/3^n\). For every \(n\), \(|a_{n+1}/a_n|=(1/3^{n+1})/(1/3^n)=1/3\). Thus the ratio-test radius is \(R=1/(1/3)=3\).
Equivalently, the series is geometric with ratio \(q=(x-2)/3\). It converges when \(|q|<1\), or \(|x-2|<3\). Solving the inequality gives \(-1<x<5\). At \(x=-1\), \(q=-1\), so the terms alternate between \(1\) and \(-1\) and do not tend to zero. At \(x=5\), \(q=1\), so every term is \(1\) and again the terms do not tend to zero. Both endpoints diverge.
Therefore \(R=3\) and the interval of convergence is \((-1,5)\). As a check inside the interval, at \(x=3\) the series is \(\sum_{n=0}^{\infty}(1/3)^n\), whose sum is \(1/(1-1/3)=3/2\). The geometric-series formula applies because \(|1/3|<1\).
Worked Example: Linear Coefficient Growth Centered at Minus 3
Find the radius and interval of convergence of \(\sum_{n=0}^{\infty}(n+1)(x+3)^n/2^n\).
Here \(c=-3\) and \(a_n=(n+1)/2^n\). The coefficients are nonzero, and their consecutive ratio is
For \(x\ne-3\), the term ratio therefore tends to \(|x+3|/2\). The ratio test gives absolute convergence when \(|x+3|<2\) and divergence when \(|x+3|>2\). Hence \(R=2\), and the open interval is \((-5,-1)\).
Check the endpoints directly. At \(x=-5\), the \(n\)th term is \((n+1)(-1)^n\), whose magnitude \(n+1\) does not tend to zero. At \(x=-1\), the \(n\)th term is \(n+1\), which also does not tend to zero. The endpoints diverge, so the full interval is \((-5,-1)\).
At the center \(x=-3\), the \(n=0\) term is \(1\), while every term with \(n\ge1\) is zero. The series converges there to \(1\), as required. This check also shows why the center should not be tested by dividing consecutive terms.
Worked Example: A Root-Test Calculation Centered at 2
Consider the sparse power series \(\sum_{n=0}^{\infty}(x-2)^{2n}/9^n\). Find its radius and interval of convergence.
To write it in the standard form \(\sum_{k=0}^{\infty}a_k(x-2)^k\), the coefficients are \(a_{2n}=1/9^n\) and \(a_{2n+1}=0\). Along even indices,
Along odd indices, \(|a_{2n+1}|^{1/(2n+1)}=0\). The limsup over all indices is therefore \(1/3\), so \(R=1/(1/3)=3\).
The series itself is geometric with ratio \(q=(x-2)^2/9\). It converges when \(|q|<1\), which is equivalent to \(|x-2|<3\), or \(-1<x<5\). At either endpoint, \(q=1\), so each term of the geometric series is \(1\) and the terms do not tend to zero. Thus the interval is \((-1,5)\).
As a center check, at \(x=2\) the term for \(n=0\) is \(1\) and all terms for \(n\ge1\) are zero. The sum is \(1\). The computed endpoints are also three units from the center: \(2-3=-1\) and \(2+3=5\).
Radius, Open Interval, and Full Interval
Once a radius is known, the open interval inside it follows by solving \(|x-c|<R\). For \(R>0\), this gives \(c-R<x<c+R\), so the interval is \((c-R,c+R)\). The radius is a nonnegative distance; the endpoints are the two real numbers \(c-R\) and \(c+R\).
| Quantity | Meaning | Example when \(c=-3\) and \(R=2\) |
|---|---|---|
| Center \(c\) | The value about which powers are taken | \(-3\) |
| Radius \(R\) | The distance from the center to either boundary | \(2\) |
| Open interval inside the radius | The points satisfying \(|x-c|<R\) | \((-5,-1)\) |
| Full interval of convergence | The open interval, with whichever endpoints converge included | Requires separate endpoint checks |
The radius test settles points strictly inside and strictly outside the radius. At the boundary, where \(|x-c|=R\), it does not determine convergence. The two endpoints can behave differently, so each must be substituted into the original series and tested on its own.
A Common Mistake: Treating the Radius as an Endpoint
For the first example, a mistaken calculation might say, “The radius is \(3\), so the interval is \((-3,3)\).” This uses the radius as though it were the right endpoint and silently assumes the center is zero. The series is centered at \(2\), so its interval is centered at \(2\), not at \(0\): the endpoints are \(2-3=-1\) and \(2+3=5\).
A second related error is to use the term-ratio formula at \(x=c\). At that point every term after the constant term is zero, and a consecutive-term quotient may be \(0/0\). The correction is to evaluate the series at the center directly: it equals \(a_0\). The ratio calculation applies for \(x\ne c\), under its coefficient conditions, not automatically at the center.
Check Your Understanding
Use the center, coefficient behavior, and endpoint checks carefully in each question.
- A power series has center \(2\) and radius \(4\). What is the open interval strictly inside its radius of convergence?
- For a power series centered at \(-3\), what is the value of the series at its center in terms of its coefficients?
- If \(\lim_{n\to\infty}|a_{n+1}/a_n|=1/5\), what is the radius of the series \(\sum a_n(x+3)^n\), assuming the coefficients are eventually nonzero?
- Why must the point \(x=c\) be treated separately when forming a ratio of consecutive terms \(a_n(x-c)^n\)?
- A series has center \(-3\) and radius \(2\). Which two points must be tested separately to decide whether its full interval of convergence includes either endpoint?