Why Distance from the Center Matters
A power series changes in a systematic way as its input moves away from its center. Each term contains a power of the distance from that center, so moving closer or farther multiplies the terms by a geometric factor. That factor explains why convergence happens throughout an interior region, while points farther out cannot converge.
For a power series centered at \(a\), the main question is not initially whether a particular endpoint converges. It is how the terms behave at different distances \(d=|x-a|\). The result is a sharp division: every distance less than the radius gives absolute convergence, and every distance greater than it gives divergence.
Power Series and Radius of Convergence
A power series centered at \(a\) has the form
where the coefficients \(c_n\) are fixed and \(x\) is the input. At distance \(d=|x-a|\), the absolute values of the terms are \(|c_n|d^n\). Thus absolute convergence depends only on the distance from the center, not on which side of the center the input lies.
At the center, \(d=0\), the series consists of \(c_0\) followed by zero terms, so it converges. This description handles the center directly; no convention about \(0^0\) is needed.
The definition uses absolute convergence so that the radius describes a property determined by the term magnitudes. The central theorem says that every distance strictly below \(R\) has absolute convergence and every distance strictly above \(R\) has divergence. It makes no general claim about distance exactly \(R\).
Why Convergence at One Distance Forces Convergence Closer In
First consider a known convergent point at distance \(s>0\). Write its terms as \(c_n(x-a)^n\), where \(|x-a|=s\). If the series converges there, its terms tend to zero and therefore are bounded: for some constant \(M\), \(|c_n|s^n\leq M\) for every \(n\). At a smaller positive distance \(r<s\), let \(q=r/s\). Then \(0<q<1\), and
The geometric series \(\sum Mq^n\) converges: the ratio of consecutive terms is \(q<1\), so the ratio test applies. By comparison, the power series converges absolutely at distance \(r\). At distance \(r=0\), convergence follows separately because only \(c_0\) can be nonzero. This proof shows why even ordinary convergence at a farther point is enough to guarantee absolute convergence at every strictly closer point.
Worked Example: Move Inward from a Known Convergent Point
Consider \(\sum_{n=1}^{\infty}\frac{x^n}{n^2}\). At \(x=1\), the series is \(\sum_{n=1}^{\infty}1/n^2\), which converges by the \(p\)-series test with \(p=2\). Now examine the series at \(x=1/2\). Its absolute terms are \(a_n=(1/2)^n/n^2\). The ratio of consecutive terms is
Since \(1/2<1\), the ratio test proves absolute convergence at \(x=1/2\). This agrees with the distance comparison: the distance \(1/2\) is smaller than the known convergent distance \(1\). Checking the terms at \(x=1/2\) in the original series gives exactly \(a_n=(1/2)^n/n^2\), as required.
Why the Series Diverges Beyond the Radius
Now suppose a power series converges at some point \(x\) at distance \(r>0\). Its terms \(c_n(x-a)^n\) must tend to zero, and hence their absolute values are bounded by some \(M\). At any smaller distance \(t<r\), set \(q=t/r\). For \(t>0\),
Here \(0<q<1\), so the geometric series used as a bound converges by the ratio test. Thus the power series converges absolutely at every distance \(t<r\). If \(t=0\), it converges by the center case.
This observation proves divergence beyond the radius. If the series converged at a distance \(r>R\), it would converge absolutely at every \(t<r\). We could choose \(t\) with \(R<t<r\), which would contradict the definition of \(R\) as the supremum of distances with absolute convergence. Therefore the series cannot converge at any point whose distance from the center is greater than \(R\).
How Ratios Change with Distance
The ratio test offers another view of the same distance effect. At a positive distance \(d\), write the absolute value of the \(n\)th term as \(u_n(d)=|c_n|d^n\). When the relevant coefficients are nonzero, the ratio of consecutive terms is
Consequently, for two positive distances \(r\) and \(s\), the ratio at \(r\) is \(r/s\) times the ratio at \(s\), whenever both ratios are defined. If a limiting ratio exists, moving inward scales that limit down, while moving outward scales it up. The ratio test then gives convergence when the limit is less than \(1\), and divergence when it is greater than \(1\). A limit equal to \(1\) is inconclusive.
This ratio comparison is useful when the coefficient pattern makes consecutive ratios easy to compute. The broader radius theorem does not require such a limit to exist: its proof uses bounded terms and a geometric comparison instead. That distinction matters because some power series have irregular coefficient ratios.
Worked Example: Ratio Test at Two Distances
Consider the power series centered at \(a=-1\)
At distance \(d=|x+1|>0\), the absolute \(n\)th term is \(u_n(d)=2^nd^n/(n+1)\). Its consecutive-term ratio is
At \(x=-0.6\), the distance is \(d=|-0.6+1|=0.4\), so the ratio limit is \(2(0.4)=0.8<1\). The ratio test proves absolute convergence there. Substitution gives the ratio sequence \(0.8(n+1)/(n+2)\), which indeed tends to \(0.8\).
At \(x=-0.4\), the distance is \(d=|-0.4+1|=0.6\), and the ratio limit is \(2(0.6)=1.2>1\). The ratio test proves divergence there; in fact the ratios are eventually greater than \(1\), so the terms do not tend to zero. Substitution gives \(1.2(n+1)/(n+2)\to1.2\), verifying the calculation. The two test points lie on opposite sides of the cutoff distance \(0.5\).
Worked Example: A Geometric Series and Its Center
Consider the power series centered at \(2\),
At \(x=4\), the distance is \(|4-2|=2\). The absolute terms are \((2/5)^n\), whose consecutive-term ratio is \(2/5<1\). The ratio test proves absolute convergence. The geometric sum is \(1/(1-2/5)=5/3\); multiplying \(5/3\) by \(1-2/5=3/5\) gives \(1\), checking the sum.
At \(x=8\), the distance is \(6\). The terms are \((6/5)^n\), with ratio \(6/5>1\); they do not tend to zero, so the series diverges. At the center \(x=2\), the series is \(1+0+0+\cdots\), so it converges. These checks agree with a radius of \(5\): \(2<5\), \(6>5\), and the center has distance zero.
What the Radius Does Not Tell You
A common mistake is to treat the boundary \(|x-a|=R\) as part of the region of convergence. The inside-and-outside theorem uses strict inequalities, and boundary behavior must be checked separately.
Worked Example: A Boundary Point Can Behave Differently
Consider \(\sum_{n=1}^{\infty}(-1)^{n-1}(x-a)^n/n\). At distance \(d<1\), the absolute terms are \(d^n/n\), bounded above by \(d^n\); the geometric series \(\sum d^n\) converges by the ratio test. At distance \(d>1\), the terms have magnitude \(d^n/n\), which does not tend to zero, so the series diverges. Thus the radius is \(R=1\).
At \(x=a+1\), the series becomes \(\sum(-1)^{n-1}/n\), which converges by the alternating-series test. At \(x=a-1\), it becomes \(-\sum 1/n\), which diverges because the harmonic series diverges. The two boundary points have the same distance from the center but different ordinary convergence behavior. This confirms that the radius theorem settles neither boundary point on its own.
Check Your Understanding
Use term magnitudes, distance comparisons, or the ratio test as appropriate.
- If a power series converges at a point \(x\) with \(|x-a|=s>0\), what can be concluded at a point with distance \(r\), where \(0<r<s\)? Explain the role of the geometric factor.
- Why does convergence at a point farther than \(R\) contradict the definition of the radius of convergence?
- For \(u_n(d)=|c_n|d^n\), how are the consecutive-term ratios at two positive distances related, when the ratios are defined?
- What does the ratio test conclude if the consecutive-term ratio tends to \(1\)?
- Does \(|x-a|=R\) guarantee convergence, divergence, or neither? Explain briefly.
