Two Extreme Convergence Patterns
A power series is centered at a particular number \(c\), and its terms involve powers of the distance \(x-c\). The coefficients influence how those powers combine as \(n\) increases. In many examples, the radius of convergence is a positive finite number. There are also two extremes: a series may converge only at its center, or it may converge for every real input.
These extremes are useful for understanding what a radius of convergence describes. It measures how far convergence extends from the center; it is not a value of the series and it does not, by itself, describe the sum.
The endpoint distinction in the finite-radius case does not arise in the same way at the extremes. When \(R=0\), every point other than the center is outside the radius. When \(R=\infty\), every real input is at a finite distance from the center and therefore lies within the radius.
Rapidly Growing Coefficients: Radius Zero
Suppose the coefficients grow so quickly that, for any fixed nonzero distance from the center, the terms of the series fail to approach zero. Then the series cannot converge away from its center. The term test says that if the terms of a series do not tend to zero, the series diverges. It does not say that terms tending to zero guarantee convergence, so we will use it only to prove divergence in these examples.
Worked Example: Coefficients \(n^n\)
Consider the power series \(\sum_{n=1}^{\infty} n^n x^n\), centered at \(0\). At \(x=0\), every term is \(n^n 0^n=0\), so the series converges and its sum is \(0\).
Now fix any \(x\ne 0\). The absolute value of the \(n\)th term is
Because \(|x|>0\), choose an integer \(N\) large enough that \(n|x|\ge 2\) for every \(n\ge N\). Then for every such \(n\),
In particular, the terms do not tend to zero. The term test proves divergence for every \(x\ne 0\). At the center the series converges, and everywhere else it diverges, so its radius is \(0\). Check: the center value was evaluated directly, and divergence at each nonzero input follows from the necessary term condition for convergence.
The argument works for negative as well as positive \(x\), since the term test uses absolute values. Signs may change, but they cannot make terms whose absolute values grow at least as fast as \(2^n\) tend to zero.
Worked Example: Coefficients \(2^{n^2}\) at a Shifted Center
Consider \(\sum_{n=1}^{\infty}2^{n^2}(x-1)^n\), centered at \(c=1\). At \(x=1\), every term is zero, so the series converges with sum \(0\). For any \(x\ne 1\), let \(d=|x-1|>0\). The absolute value of the \(n\)th term is
Choose \(N\) so large that \(2^n d\ge 2\) whenever \(n\ge N\). Such an \(N\) exists because \(2^n d\) grows without bound for fixed \(d>0\). Thus, for \(n\ge N\),
The terms do not tend to zero, so the series diverges for every \(x\ne 1\). It converges at its center and nowhere else; its radius is \(0\). Check: substituting \(x=1\) makes every term zero, while at any other input the displayed lower bound rules out convergence.
Why the Center Still Works
A radius of zero does not mean a power series fails to converge everywhere. A power series centered at \(c\) has the form \(\sum_{n=0}^{\infty}a_n(x-c)^n\). At the center, \(x=c\), every term with \(n\ge1\) is zero. The series there reduces to its constant term \(a_0\), so it converges. This is why radius zero means convergence only at the center, not convergence nowhere.
Rapidly Decaying Coefficients: Infinite Radius
For the opposite extreme, consider coefficients that decay rapidly enough to counteract any fixed real value of \(x-c\). To prove convergence at every input, it is not enough to say that the coefficients become small: the powers \((x-c)^n\) also change with \(n\). A useful approach is to fix an arbitrary real input, bound the terms from some index onward by a convergent geometric series, and then note that adding finitely many earlier terms does not affect convergence.
Worked Example: Coefficients \(1/n^n\)
Consider the power series
which is centered at \(c=-2\). Fix any real \(x\), and set \(d=|x+2|\). If \(d=0\), the series is just \(1\), so it converges. Suppose instead \(d>0\), and write \(u_n=d^n/n^n\) for the absolute value of the \(n\)th term. Consecutive terms satisfy
The inequality follows from \(n^n\le(n+1)^n\), which gives \(n^n/(n+1)^{n+1}\le 1/(n+1)\). Choose \(N\) so that \(N+1\ge 2d\). For every \(n\ge N\), \(d/(n+1)\le1/2\), and therefore \(u_{n+1}\le u_n/2\). Repeatedly applying this inequality gives \(u_{N+k}\le u_N(1/2)^k\) for every integer \(k\ge0\). The tail is bounded by
This geometric bound is finite, and the finitely many terms before \(N\) are finite too. The series of absolute values converges, so the original series converges absolutely at the chosen \(x\). Since \(x\) was arbitrary, it converges for every real \(x\), and its radius is \(\infty\). Check: the argument covers \(x=-2\) directly and every other real input through a finite geometric bound.
The General Reason the Two Conclusions Follow
The radius description connects these examples to the full convergence region of a power series. Its essential logic can be stated briefly. Once a radius has been identified, points strictly closer to the center than that radius are convergence points, while points strictly farther away are divergence points. The center always converges. Applying these facts at the two extreme values gives the following conclusions.
The center \(c\) converges. Every \(x\ne c\) has \(|x-c|>0=R\), so it lies outside the radius and diverges. Thus the center is the sole convergence point.
Every real \(x\) has a finite distance \(|x-c|\) from the center, which is less than \(\infty\). Therefore every real \(x\) lies within the convergence region.
This reasoning also explains why no endpoint test is needed for either extreme. There is no real point at a positive distance from the center that is exactly distance zero, and no real input is infinitely far from a real center. For a finite positive radius, by contrast, the points at distance exactly \(R\) must be checked separately.
A Common Mistake
A common error is to interpret “radius zero” as “the series does not converge.” For the series \(\sum_{n=1}^{\infty} n^n x^n\), that would incorrectly classify the center. Substituting \(x=0\) gives
The correction is to distinguish the center from all other inputs: this series converges at \(0\) and diverges at every \(x\ne0\). Its radius is zero precisely because the convergence set contains only its center. A related mistake is to use a finite number of successful substitutions to claim that a series converges for every real number. The infinite-radius example required an argument for an arbitrary fixed \(x\), not a check of a few sample inputs.
How to Recognize the Extreme Cases
The examples suggest two practical strategies. Rapidly growing coefficients may make the terms fail the term test at every noncentral input. Rapidly decaying coefficients may allow a geometric comparison for each fixed input. These are proof strategies, not automatic rules based only on the words “growing” or “decaying”: the actual terms must be examined, and the convergence argument must cover every required input.
| Radius | What to establish | Typical argument |
|---|---|---|
| \(R=0\) | Convergence at the center and divergence at every other input | Show the terms fail to tend to zero away from the center |
| \(R=\infty\) | Convergence for an arbitrary real input | Bound the tail by a convergent geometric series |
In both cases, state clearly which input is fixed and why the argument applies to every input in the claimed set. That step is what turns a suggestive pattern in the coefficients into a conclusion about the entire convergence region.
Check Your Understanding
Use the center and the term or comparison tests where appropriate.
- For \(\sum_{n=1}^{\infty}n^n(x-3)^n\), what is the radius, and at which real input does the series converge?
- Why does showing that the terms fail to tend to zero at every noncentral input prove divergence there?
- For a power series centered at \(c\) with radius \(0\), explain why its center is still a convergence point.
- In the series \(1+\sum_{n=1}^{\infty}(x+2)^n/n^n\), why is it necessary to fix an arbitrary \(x\) before choosing the geometric bound?
- What does an infinite radius imply about convergence at the real inputs \(x=0\) and \(x=100\), regardless of the center?