Why geometric endpoints deserve a separate check
A power series has a radius of convergence, but that radius alone does not determine what happens at its endpoints. Inside the radius, a geometric power series converges and has a familiar closed-form sum. At either endpoint, its ratio has absolute value \(1\), and the behavior changes: the terms no longer approach zero.
This makes geometric power series a clear case in which to distinguish convergence strictly inside the radius from behavior exactly on the boundary. In the geometric case, both endpoints diverge, although the partial sums behave differently at the two ends.
Geometric power series and their radius
The convergence condition for a geometric series is \(|r|<1\). Here this becomes \(|\lambda(x-c)|<1\), or \(|x-c|<1/|\lambda|\). Thus the radius is \(R=1/|\lambda|\), and the candidate endpoints are \(c-R\) and \(c+R\). Substituting them into \(r\) gives ratios with values \(1\) and \(-1\), in an order that depends on the sign of \(\lambda\).
For \(|r|<1\), the sum is \(A/(1-r)\). The endpoint ratios do not meet this convergence condition. To see exactly why they diverge, it helps to examine the finite partial sums.
Why the interior sum formula works
Let \(S_N=A+Ar+\cdots+Ar^N\) be the partial sum through the term with index \(N\). Multiplying by \(r\) and subtracting gives a finite identity:
When \(r\ne 1\), division gives \(S_N=A(1-r^{N+1})/(1-r)\). If \(|r|<1\), then \(r^{N+1}\to 0\) as \(N\to\infty\). Therefore the partial sums converge to \(A/(1-r)\).
For the endpoint claim, the two possible ratios are \(r=1\) and \(r=-1\). At \(r=1\), each term is \(A\), so the terms do not approach zero; the partial sums are \(A(N+1)\). At \(r=-1\), the terms alternate between \(A\) and \(-A\). The partial sums through index \(N\) are \(A\) when \(N\) is even and \(0\) when \(N\) is odd, so they have no limit. In either case the series diverges.
These two endpoint calculations also agree with the term test: if a series converges, its terms must approach zero. At either geometric boundary point the term has constant nonzero magnitude \(|A|\), so that necessary condition fails.
Worked examples at both endpoints
Worked Example: A geometric series centered at 2
Find the interval of convergence of \(\sum_{n=0}^{\infty} ((x-2)/3)^n\), and give its sum where it converges.
The common ratio is \(r=(x-2)/3\). Convergence requires \(|(x-2)/3|<1\), so \(|x-2|<3\). The radius is \(3\), and the candidate endpoints are \(2-3=-1\) and \(2+3=5\).
At the left endpoint, \(x=-1\), the ratio is \(r=(-1-2)/3=-1\). The series becomes \(\sum_{n=0}^{\infty}(-1)^n\). Its partial sums through index \(N\) are \((1-(-1)^{N+1})/2\): they equal \(1\) for even \(N\) and \(0\) for odd \(N\). There is no limit, so this endpoint diverges.
At the right endpoint, \(x=5\), the ratio is \(r=(5-2)/3=1\). The series becomes \(\sum_{n=0}^{\infty}1\), whose partial sum through index \(N\) is \(N+1\). These partial sums grow without bound, so this endpoint diverges too.
For \(-1<x<5\), the geometric sum formula gives \(1/(1-(x-2)/3)=3/(5-x)\). As a check, at \(x=2\) the series is \(1+0+0+\cdots=1\), and the formula gives \(3/(5-2)=1\). The interval of convergence is therefore \((-1,5)\), with sum \(3/(5-x)\) on that interval.
Worked Example: A negative scale factor
Analyze \(\sum_{n=0}^{\infty}4[-2(x+1)]^n\), including both endpoints.
Here \(c=-1\), \(A=4\), and \(\lambda=-2\), so the ratio is \(r=-2(x+1)\). The condition \(|r|<1\) is \(|x+1|<1/2\). Thus the radius is \(1/2\), and the endpoints are \(-1-1/2=-3/2\) and \(-1+1/2=-1/2\).
At \(x=-3/2\), \(x+1=-1/2\), so \(r=-2(-1/2)=1\). Every term is \(4\), and the partial sum through index \(N\) is \(4(N+1)\). It does not converge.
At \(x=-1/2\), \(x+1=1/2\), so \(r=-2(1/2)=-1\). The series is \(4-4+4-4+\cdots\). Its partial sums are \(4\) for even \(N\) and \(0\) for odd \(N\), so this endpoint also diverges.
For \(-3/2<x<-1/2\), its sum is \(4/(1+2(x+1))=4/(2x+3)\). At the center \(x=-1\), the original series is \(4+0+0+\cdots=4\), and the formula gives \(4/(2(-1)+3)=4\), as required. The interval is open because neither endpoint converges.
Worked Example: The signs of the endpoint ratios can reverse
Determine the interval of convergence and sum of \(\sum_{n=0}^{\infty}(-1)^n((x-4)/2)^n\).
Combining the factors in the \(n\)th term gives the common ratio \(r=-(x-4)/2=(4-x)/2\). The convergence condition is \(|(x-4)/2|<1\), which is equivalent to \(|x-4|<2\). The radius is \(2\), and the endpoints are \(4-2=2\) and \(4+2=6\).
At \(x=2\), \(r=(4-2)/2=1\). Every term equals \(1\), so the partial sum through index \(N\) is \(N+1\), and the series diverges.
At \(x=6\), \(r=(4-6)/2=-1\). The partial sums are \(1\) for even \(N\) and \(0\) for odd \(N\), so they do not converge. Hence both endpoints are excluded.
Inside \(2<x<6\), the sum is \(1/(1-r)=1/(1-(4-x)/2)=2/(x-2)\). At \(x=4\), the original series is \(1+0+0+\cdots=1\); the formula gives \(2/(4-2)=1\). Thus the interval of convergence is \((2,6)\), and the stated sum is valid throughout it.
What differs at the two boundary points
At one endpoint the ratio is \(1\); at the other it is \(-1\). Both series diverge, but it is useful not to conflate their partial sums. When \(r=1\), the partial sums grow linearly. When \(r=-1\), they alternate between two values. Neither pattern has a finite limit.
| Ratio at the point | Terms | Partial sums through index \(N\) | Conclusion |
|---|---|---|---|
| \(r=1\) | \(A,A,A,\ldots\) | \(A(N+1)\) | Diverges |
| \(r=-1\) | \(A,-A,A,-A,\ldots\) | \(A\) for even \(N\); \(0\) for odd \(N\) | Diverges |
The endpoint behavior is not a contradiction of the sum formula. That formula was derived as a limit under the condition \(|r|<1\). At a boundary point, \(r\) is \(1\) or \(-1\), so the condition required for the limit no longer holds. Substituting a boundary value into a closed-form expression cannot establish convergence there.
A common endpoint mistake
The correction is to check the actual partial sums, or first apply the necessary term test. Here the terms alternate but do not approach zero, so the series cannot converge. A different common error is to include both endpoints merely because the radius gives a closed-looking candidate interval. The radius identifies where to test; each endpoint still requires its own analysis.
Why this boundary case matters
A geometric power series makes the role of the radius especially transparent. Strictly inside the radius, the ratio has magnitude less than \(1\), the powers \(r^n\) approach zero, and the finite-sum formula has a limit. At the endpoints, the magnitude of each term remains \(|A|\), and convergence fails. This gives a useful benchmark when analyzing other power series: the radius describes the interior region, but boundary behavior must be determined by substituting and testing the endpoint series.
Check Your Understanding
For each question, use the common ratio and, at endpoints, the behavior of the partial sums.
- For \(\sum_{n=0}^{\infty}[(x+1)/4]^n\), find the radius and both endpoints.
- At the left and right endpoints in Question 1, what are the respective ratios?
- What are the partial sums through index \(N\) when the ratio is \(1\)?
- Why does the series \(\sum_{n=0}^{\infty}(-1)^n\) diverge, even though its terms alternate?
- State the sum of \(\sum_{n=0}^{\infty}A[\lambda(x-c)]^n\) strictly inside its radius, including the condition on \(x\).