Writing The Exact Interval Of Convergence

From Endpoint Verdicts to an Interval

A power series has a center and a radius of convergence. The radius identifies the region where the series converges automatically, but it does not decide what happens at the two boundary values. Each boundary must be checked separately. Once those two endpoint verdicts are known, the remaining task is to write the exact set of real numbers where the series converges.

This final step is more than a notation detail. A round bracket means its endpoint is excluded; a square bracket means its endpoint is included. Reversing or overlooking one of those symbols changes the set being described.

Definition: For a power series centered at \(c\) with finite radius of convergence \(R>0\), the boundary values are \(c-R\) and \(c+R\). Its interval of convergence is the set of real \(x\)-values for which the series converges, including any boundary values where it converges.

The Interior and the Two Boundary Values

For a power series \(\sum_{n=0}^{\infty} a_n(x-c)^n\) with radius \(R\), the radius theorem says the series converges absolutely when \(|x-c|<R\), and diverges when \(|x-c|>R\). Since \(|x-c|<R\) is equivalent to \(c-R<x<c+R\), every point strictly between the two boundary values is included, and every point outside them is excluded.

The radius theorem does not determine convergence at \(x=c-R\) or \(x=c+R\), where \(|x-c|=R\). Substitute each value into the original series and use the resulting endpoint verdict. Include that boundary if the series converges there, whether the convergence is absolute or conditional. Exclude it if the series diverges there.

Call the verdict at \(c-R\) the left-endpoint verdict and the verdict at \(c+R\) the right-endpoint verdict. The notation follows directly from whether each boundary belongs to the convergence set:

Left endpoint \(c-R\)	Right endpoint \(c+R\)	Interval of convergence
Diverges	Diverges	\((c-R,c+R)\)
Converges	Diverges	\([c-R,c+R)\)
Diverges	Converges	\((c-R,c+R]\)
Converges	Converges	\([c-R,c+R]\)

Here, “converges” means the series converges as a series of real numbers. Absolute convergence is one way to converge, but it is not required for endpoint inclusion.

Why These Four Possibilities Are Complete

The radius theorem gives all points strictly between the boundary values and rules out all points strictly outside them. Thus, after applying that theorem, the only membership decisions still to make are whether \(c-R\) and \(c+R\) belong to the convergence set. Each endpoint has two possible verdicts: included if the series converges, excluded if it diverges. The two independent choices give four cases.

Interval notation records those choices one endpoint at a time. A square bracket at the left records inclusion of \(c-R\); a round bracket records exclusion. The same rule applies at the right endpoint. This is why the middle of every interval in the table is unchanged, while the brackets depend on the endpoint tests.

\text{left bracket}= \begin{cases} [ & \text{if the series converges at }c-R,\\ ( & \text{if the series diverges at }c-R, \end{cases} \qquad \text{right bracket}= \begin{cases} ] & \text{if the series converges at }c+R,\\ ) & \text{if the series diverges at }c+R. \end{cases}

Worked Examples

Worked Example: A Half-Open Interval with the Left Endpoint Included

Find the interval of convergence of \(\sum_{n=1}^{\infty}\frac{(x-2)^n}{n}\).

The center is \(c=2\), and the coefficient of \((x-2)^n\) is \(a_n=1/n\). The root test gives the radius from \(\limsup_{n\to\infty}|a_n|^{1/n}\):

\lim_{n\to\infty}\left(\frac{1}{n}\right)^{1/n}=1, \qquad R=\frac{1}{1}=1.

The boundary values are therefore \(2-1=1\) and \(2+1=3\). For \(1<x<3\), the series converges absolutely by the radius theorem; for \(x<1\) or \(x>3\), it diverges.

At the left endpoint, substitute \(x=1\):

\sum_{n=1}^{\infty}\frac{(1-2)^n}{n} = \sum_{n=1}^{\infty}\frac{(-1)^n}{n}.

This alternating harmonic series converges by the alternating-series test, but not absolutely, since the absolute-value series is \(\sum_{n=1}^{\infty}1/n\), which diverges. The left endpoint is included.

At the right endpoint, substitute \(x=3\):

\sum_{n=1}^{\infty}\frac{(3-2)^n}{n} = \sum_{n=1}^{\infty}\frac{1}{n}.

This is the harmonic series, so it diverges. The right endpoint is excluded. The left bracket is square and the right bracket is round, giving \([1,3)\). Checking the endpoint substitutions against the notation confirms that \(1\) is included and \(3\) is not.

Worked Example: A Closed Interval

Find the interval of convergence of \(\sum_{n=1}^{\infty}\frac{(x+1)^n}{n^2}\).

The center is \(c=-1\), and the coefficients have magnitudes \(1/n^2\). The root test gives:

\lim_{n\to\infty}\left(\frac{1}{n^2}\right)^{1/n}=1, \qquad R=\frac{1}{1}=1.

The boundary values are \(-1-1=-2\) and \(-1+1=0\). Every \(x\) strictly between them is included, while points outside them are excluded.

At \(x=-2\), the substitution gives:

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

This converges absolutely because \(\sum_{n=1}^{\infty}|(-1)^n/n^2|=\sum_{n=1}^{\infty}1/n^2\), a convergent \(p\)-series with \(p=2>1\). Thus, \(-2\) is included.

At \(x=0\), the substitution gives:

\sum_{n=1}^{\infty}\frac{(0+1)^n}{n^2} = \sum_{n=1}^{\infty}\frac{1}{n^2}.

This is also a convergent \(p\)-series with \(p=2>1\), so \(0\) is included. Both endpoints are included, and the interval is \([-2,0]\). The square brackets match the two endpoint convergence verdicts.

Worked Example: A Half-Open Interval with the Right Endpoint Included

Find the interval of convergence of \(\sum_{n=1}^{\infty}\frac{(-1)^n(x-1)^n}{n}\).

The center is \(c=1\), and the coefficient magnitudes are \(1/n\). Therefore:

\lim_{n\to\infty}\left(\frac{1}{n}\right)^{1/n}=1, \qquad R=1.

The boundary values are \(1-1=0\) and \(1+1=2\). The series converges for \(0<x<2\) and diverges for \(x<0\) or \(x>2\).

At \(x=0\), the terms simplify as follows:

\frac{(-1)^n(0-1)^n}{n} = \frac{(-1)^n(-1)^n}{n} = \frac{1}{n}.

The endpoint series is harmonic and diverges, so \(0\) is excluded.

At \(x=2\), the terms simplify to:

\frac{(-1)^n(2-1)^n}{n} = \frac{(-1)^n}{n}.

This alternating harmonic series converges, so \(2\) is included. The interval is \((0,2]\). Substitution confirms the bracket choices: the left endpoint diverges and the right endpoint converges.

A Common Bracket Error

A frequent mistake is to write the open interval \((c-R,c+R)\) as the final answer without checking whether either endpoint converges. For the first example, that shortcut would give \((1,3)\). But at \(x=1\), the series becomes \(\sum_{n=1}^{\infty}(-1)^n/n\), which converges. The correct interval must include \(1\), giving \([1,3)\).

Another possible error is to treat the endpoints as if they had the same substitution. They do not: at \(c-R\), \(x-c=-R\); at \(c+R\), \(x-c=R\). The resulting series may have different signs and different convergence behavior. Substitute each boundary value separately before choosing brackets.

Takeaway: Find the boundary values \(c-R\) and \(c+R\), use the endpoint verdict at each one independently, then choose a square bracket for convergence and a round bracket for divergence.

When the Radius Is Zero or Infinite

The four-case table assumes a finite, positive radius. If \(R=0\), there is no interval of positive length: the radius theorem gives no convergence away from the center, so check the series at \(x=c\) and report \(\{c\}\) if it converges there, or the empty set if it does not. For an ordinary power series, convergence at the center is automatic because all positive-power terms vanish there and only the constant term remains.

If \(R=\infty\), the radius theorem says the series converges absolutely for every real \(x\). Its interval of convergence is \((-\infty,\infty)\). Neither case requires choosing brackets for two finite boundary values.

Check Your Understanding

For each question, focus on how the endpoint verdicts determine the notation.

A series has center \(c=4\), radius \(R=3\), converges at both endpoints, and diverges outside the radius. What is its interval of convergence?
A series has center \(c=-2\), radius \(R=5\), converges at the left endpoint and diverges at the right endpoint. Write its interval of convergence.
Why does convergence at an endpoint include it in the interval even when that convergence is conditional?
A series has center \(c=1\), radius \(R=2\), and diverges at both endpoints. Write its interval of convergence.
For a series centered at \(c\), what two values must be substituted to test the boundaries of a finite-radius interval?