Radius and Interval of Convergence of Power Series

3. Visual Connection

5. Concept Development

BC-only topic. A power series is an infinite polynomial whose convergence can change with \(x\). Its convergent inputs form an interval centered at the power-series center, with each finite endpoint requiring its own decision.

Standard form and center

The series is centered at \(a\). At \(x=a\), every positive-power term vanishes, so the series always converges to its constant term \(c_0\).

Three possible radii

Every real power series has exactly one of these convergence patterns:

• \(R=0\): it converges only at \(x=a\).
• \(0<R<\infty\): it converges absolutely for \(|x-a|<R\), diverges for \(|x-a|>R\), and needs endpoint tests when \(|x-a|=R\).
• \(R=\infty\): it converges for every real \(x\).

Radius versus interval

The radius of convergence is the distance \(R\) from the center to either boundary. The interval of convergence is the full set of real inputs where the series converges, including any accepted endpoints.

is only the open interior. Endpoint brackets are added after separate tests.

Why the interior is symmetric

Convergence inside the radius depends on \(|x-a|\), so all points at a smaller distance from the center converge absolutely. Endpoint behavior can be asymmetric because substituting \(a-R\) and \(a+R\) may produce different signs.

Ratio Test workflow

For \(u_n=c_n(x-a)^n\), compute

1. Simplify until the condition \(L<1\) becomes an inequality in \(|x-a|\).
2. Solve that inequality to find the open interval and radius.
3. Substitute the left endpoint into the original series.
4. Use an ordinary numerical-series test and record convergence or divergence.
5. Repeat independently at the right endpoint.
6. Write both \(R\) and the final interval with correct brackets.

Why the Ratio Test stops at endpoints

At \(|x-a|=R\), the ratio limit is usually \(1\), which is inconclusive. This does not mean the endpoint diverges. It means another test must decide.

Endpoint tests to recognize

• Terms failing to approach zero imply divergence.
• A geometric series is classified by its common ratio.
• A positive \(p\)-series converges exactly when \(p>1\).
• An alternating \(p\)-series may converge conditionally.
• Comparison, limit comparison, the Integral Test, or the Alternating Series Test may be required.

Four finite-endpoint patterns

A finite-radius interval can be \((a-R,a+R)\), \([a-R,a+R)\), \((a-R,a+R]\), or \([a-R,a+R]\). The radius is \(R\) in all four cases.

Absolute and conditional behavior

Every point strictly inside the radius gives absolute convergence. A retained endpoint may be absolutely or conditionally convergent. State that classification when it helps justify the bracket.

Linear expressions must be normalized

If the power is \((mx+b)^n\), rewrite

to reveal the center \(-b/m\), or solve the resulting absolute-value inequality directly. The radius is measured in \(x\)-units, so scaling by \(m\) changes it.

Known convergence information

If a power series centered at \(a\) converges at an input \(x_1\), then it converges absolutely at every point closer to \(a\): \(|x-a|<|x_1-a|\). If it diverges at \(x_2\), it must diverge at every point farther from the center.

Differentiation and integration

Term-by-term differentiation or integration preserves the radius of convergence. Endpoint behavior can change, so the endpoints must be retested for the transformed series.

AP-style checklist

1. Identify the center correctly.
2. Use the Ratio Test or another valid method to find \(R\).
3. Convert \(|x-a|<R\) into an open interval.
4. Show each endpoint substitution as a numerical series.
5. Name the endpoint test and conclusion.
6. Report both the radius and final interval.

6. Detailed Worked Example and Error Check

Example 1: Left endpoint only

For

the Ratio Test gives \(|x-2|/3<1\), so \(R=3\) and \(-1<x<5\). At \(x=-1\), the series becomes \(\sum(-1)^n/(n+1)\), which converges by the AST. At \(x=5\), it becomes \(\sum1/(n+1)\), which diverges. The interval is \([-1,5)\).

Example 2: Both endpoints included absolutely

For \(\sum_{n=1}^{\infty}(x+1)^n/(n^2 2^n)\), the Ratio Test gives \(|x+1|<2\), so \(R=2\). At \(x=-3\) and \(x=1\), the magnitude series is \(\sum1/n^2\). Both endpoints converge absolutely, and the interval is \([-3,1]\).

Example 3: Conditional left endpoint

For \(\sum_{n=1}^{\infty}(x-4)^n/n\), the open interval is \(3<x<5\). At \(x=3\), the alternating harmonic series converges; at \(x=5\), the harmonic series diverges. Thus \(R=1\) and the interval is \([3,5)\).

Example 4: Conditional right endpoint

For \(\sum_{n=1}^{\infty}(-1)^n(x-1)^n/n\), the interior condition is \(|x-1|<1\). At \(x=0\), every term is \(1/n\), so the series diverges. At \(x=2\), it is alternating harmonic and converges. Hence \(R=1\) and the interval is \((0,2]\).

Example 5: Both endpoints included

For \(\sum_{n=1}^{\infty}(x+2)^n/n^3\), the radius is \(1\). At \(x=-3\), the series is \(\sum(-1)^n/n^3\); at \(x=-1\), it is \(\sum1/n^3\). Both converge absolutely, so the interval is \([-3,-1]\).

Example 6: Neither endpoint included

The geometric power series \(\sum_{n=0}^{\infty}x^n\) converges for \(|x|<1\). At \(x=1\), its terms are \(1\); at \(x=-1\), its terms alternate between \(1\) and \(-1\). Neither term sequence approaches zero. Thus \(R=1\) and the interval is \((-1,1)\).

Example 7: Infinite radius

For \(\sum x^n/n!\),

for every real \(x\). Therefore \(R=\infty\) and the interval is \((-\infty,\infty)\).

Example 8: Radius zero

For \(\sum n!(x-3)^n\), the ratio is \((n+1)|x-3|\), which tends to infinity whenever \(x\ne3\). At the center \(x=3\), the series converges to its constant term. Thus \(R=0\) and the interval is the single point \(\{3\}\).

Example 9: Endpoint terms fail the zero test

For \(\sum_{n=1}^{\infty}n(x-2)^n/4^n\), the Ratio Test gives \(|x-2|<4\). At \(x=-2\) and \(x=6\), the terms have magnitudes \(n\), so both endpoint series diverge. Therefore \(R=4\) and the interval is \((-2,6)\).

Example 10: Hidden center and scaled radius

For

the interior condition is \(|2x-1|<3\), or \(-1<x<2\). The center is \(1/2\) and the radius in \(x\)-units is \(3/2\). At \(x=-1\), the series is alternating harmonic; at \(x=2\), it is harmonic. The interval is \([-1,2)\).

Common errors

• Reporting only the radius or only the interval.
• Including endpoints automatically after solving the Ratio Test inequality.
• Discarding endpoints because the ratio limit equals \(1\).
• Testing only one endpoint and copying its conclusion to the other.
• Substituting endpoints into the simplified ratio instead of the original series.
• Forgetting that the center always converges.
• Confusing the center \(a\) with the radius \(R\).
• Reading the radius directly from \((mx+b)^n\) without accounting for \(m\).
• Calling a conditionally convergent endpoint absolutely convergent.
• Writing infinity with a square bracket.
• Assuming the interval itself must be symmetric including brackets; only the open interior is symmetric.
• Forgetting to retest endpoints after differentiating or integrating a power series.