Why Endpoint Checks Matter
A power series has a center and a radius that determine where it converges absolutely. The radius gives a useful first description of its behavior, but it does not settle what happens at the two boundary points. At each boundary, the terms form a particular numerical series, and that series must be tested on its own.
This distinction matters because endpoint behavior can vary. A series might converge at neither endpoint, at exactly one, or at both. To report the interval of convergence correctly, you need the radius and a separate conclusion for each endpoint.
Power Series and the Ratio Test
A power series centered at \(a\) has the form \(\sum_{n=0}^{\infty} c_n(x-a)^n\), where \(a\) and the coefficients \(c_n\) are constants. For a fixed value of \(x\), this is a numerical series. The value \(a\) is the center; the radius of convergence \(R\) describes the distance from the center within which the series converges absolutely.
A common way to find \(R\) is to apply the ratio test to the absolute values of the terms. When the relevant coefficient ratios are defined and \(\lim_{n\to\infty}|c_{n+1}/c_n|=L\) for a finite positive number \(L\), then for each \(x\ne a\) the ratio of consecutive term magnitudes has limit \(L|x-a|\). (At \(x=a\), every term with \(n\ge1\) is zero, so the series converges trivially.) Thus the series converges absolutely when \(L|x-a|<1\), and diverges when \(L|x-a|>1\). The radius is \(R=1/L\). When \(L=0\) or the limit does not exist, other forms of the ratio or root test may be needed; the examples here use an existing positive finite limit.
The ratio test is inconclusive when its limit is \(1\). At either candidate endpoint, \(|x-a|=R\), so the ratio of consecutive term magnitudes has limit \(LR=1\). This is why you cannot use the radius alone to decide whether to include an endpoint.
Why the Radius Formula Works
The ratio test’s conclusions follow from comparison with a geometric series. Suppose the absolute-value terms of a numerical series are \(v_n\), and \(\lim_{n\to\infty}v_{n+1}/v_n=q\), with the terms defined for all sufficiently large \(n\).
Choose a number \(r\) with \(q<r<1\). From some index onward, \(v_{n+1}/v_n<r\). Repeatedly applying this inequality bounds the tail terms by a constant times successive powers of \(r\). The geometric series with ratio \(r\) converges, so the original series converges absolutely.
Choose \(r\) with \(1<r<q\). From some index onward, \(v_{n+1}/v_n>r\). The positive terms then grow by a factor greater than \(r\) at every step, so they do not approach zero. The original series diverges by the nth-term test.
Neither comparison applies: a ratio tending to \(1\) does not establish convergence or divergence. That is the situation at a power series’s boundary, so direct endpoint tests are necessary.
For a power series with coefficient ratio limit \(L\), the term ratio at a fixed \(x\ne a\) has limit \(L|x-a|\). Requiring that limit to be less than \(1\) gives \(|x-a|<1/L\), which explains the radius formula. At equality, the ratio test has reached its inconclusive case, not a verdict about convergence.
A Reliable Endpoint Procedure
Once the radius is known, the remaining work is finite: substitute the left endpoint into the original series, test the resulting numerical series, then repeat independently at the right endpoint. Do not carry over a conclusion from one endpoint to the other.
- Find the center \(a\) and radius \(R\).
- Write the two candidates \(a-R\) and \(a+R\).
- Substitute each candidate into the original series and simplify its terms.
- Use an appropriate test, such as the nth-term test, a p-series test, or the alternating-series test.
- State whether each endpoint is included, then write the full interval.
Worked Examples
Worked Example: A Conditionally Convergent Endpoint
Find the interval of convergence of \(\sum_{n=1}^{\infty}(x-1)^n/n\).
For \(x\ne1\), let \(u_n=(x-1)^n/n\). The ratio of consecutive term magnitudes is
Since \(n/(n+1)\to1\), the ratio limit is \(|x-1|\). The ratio test gives absolute convergence when \(|x-1|<1\), so the center is \(1\), the radius is \(1\), and the endpoint candidates are \(0\) and \(2\).
The ratio calculation above is for \(x\ne1\), because it divides by \(u_n\). At \(x=1\), every term \((1-1)^n/n\) is zero, so the series is identically zero and converges absolutely. This separate check covers the center.
At the left endpoint \(x=0\), substitution gives
The magnitudes \(1/n\) decrease to \(0\), so the alternating-series test proves convergence. The absolute-value series is \(\sum 1/n\), the divergent harmonic series. Therefore this endpoint is conditionally convergent.
At the right endpoint \(x=2\), substitution gives \(\sum 1/n\), which diverges. Thus the complete interval of convergence is \([0,2)\). Checking the endpoints against the original series confirms that \(0\) is included and \(2\) is not.
Worked Example: Both Endpoints Converge
Find the interval of convergence of \(\sum_{n=1}^{\infty}(x+2)^n/n^2\).
For \(x\ne-2\), the ratio of consecutive term magnitudes is
As \(n\to\infty\), this approaches \(|x+2|\). The series converges absolutely if \(|x+2|<1\), giving radius \(1\) and endpoint candidates \(-3\) and \(-1\).
At \(x=-3\), the series becomes \(\sum (-1)^n/n^2\). Its absolute-value series is \(\sum 1/n^2\), which converges by the p-series test because \(p=2>1\). So the endpoint converges absolutely.
At \(x=-1\), the series becomes \(\sum 1/n^2\), which also converges by the p-series test. Both endpoints are included, and the complete interval is \([-3,-1]\). Substituting the endpoints verifies that each resulting series has the stated form.
Worked Example: Neither Endpoint Converges
Find the interval of convergence of \(\sum_{n=1}^{\infty}\frac{n}{5^n}(x-4)^n\).
Let \(u_n=\frac{n}{5^n}(x-4)^n\). When the quotient is defined, the ratio of consecutive term magnitudes is
The limit is \(|x-4|/5\), so the radius is \(5\), with endpoint candidates \(-1\) and \(9\). At the center \(x=4\), every term is zero because \(n\ge1\), so the series converges there as expected.
At \(x=-1\), \((x-4)/5=-1\), and substitution gives the terms \(n(-1)^n\). Their magnitudes are \(n\), which do not approach zero. The series diverges by the nth-term test.
At \(x=9\), \((x-4)/5=1\), and the series becomes \(\sum n\). Its terms also fail to approach zero, so it diverges by the nth-term test. Neither endpoint is included, and the complete interval is \((-1,9)\). The endpoint substitutions confirm the two divergent term sequences.
A Common Endpoint Mistake
A frequent error is to calculate the radius correctly and then assume that both endpoints are included, or that both are excluded. For example, the ratio test in the first worked example gives center \(1\) and radius \(1\). The incorrect conclusion “the interval is \([0,2]\)” treats the radius as an endpoint convergence test. But at \(x=2\), substitution gives \(\sum 1/n\), which diverges. At \(x=0\), the resulting alternating harmonic series converges. The corrected interval is \([0,2)\).
Reporting the Complete Interval
After testing both endpoints, use parentheses for an excluded endpoint and brackets for an included endpoint. The interval notation should agree with the convergence results at both boundaries. If an endpoint is conditionally convergent, it is still included: interval membership depends on convergence, not on absolute convergence.
It is also useful to distinguish what the radius statement guarantees from what it does not. For every point strictly inside the radius, the series converges absolutely. For every point strictly outside, it diverges. At the two boundary points, neither conclusion follows from the radius alone.
Check Your Understanding
For each question, explain your conclusion and show the endpoint substitutions where needed.
- A power series has center \(3\) and radius \(2\). What are its two endpoint candidates?
- Why does a ratio-test limit of \(1\) at an endpoint fail to decide convergence?
- If the left endpoint gives an alternating series whose term magnitudes decrease to zero, what does the alternating-series test establish? What else must be checked to decide whether convergence is absolute?
- At an endpoint, the substituted terms are \(n(-1)^n\). Which test proves divergence, and why?
- A power series converges absolutely for \(|x-a|<R\), diverges for \(|x-a|>R\), and converges at \(a-R\) but diverges at \(a+R\). Write its interval of convergence.
