Power series: radius, endpoints and certified approximation

Learning goals and prerequisites

Recognize a power series as a family of numerical series indexed by an input, determine its convergence interval, and bound a finite approximation on a stated input range. Prerequisites are sequences, finite and infinite geometric sums, polynomial algebra and real completeness. No termwise differentiation or integration is needed for the core arguments.

A power series centered at a real number $a$ is

$$\sum_{n=0}^{\infty}c_n(x-a)^n,$$

where the coefficients $c_n$ are fixed real numbers. For each selected $x$, its partial sums form a numerical sequence. The sum is a function only on inputs for which that sequence converges. At $x=a$, the value is $c_0$; the $n=0$ term is the constant polynomial $c_0$, not an ambiguous numerical evaluation of $0^0$.

A finite polynomial is a terminating power series and converges everywhere. An infinite power series need not converge anywhere except its center, so a formula cannot be assigned a domain merely by looking at each individual term.

Proof bridge: finite sums, geometric limits and comparison

In the geometric identities below, $a,r$ may be real or complex; put $\rho=|r|$. Signed remainders are understood as complex remainders when appropriate. For complex sums, convergence means convergence in modulus.

We take the Archimedean property and completeness of the real numbers as basic properties of $\mathbb R$. The Archimedean property says that for every real $M$ there is a positive integer $N\in\mathbb Z_{\ge1}$ with $N>M$. Completeness says that every nonempty real set bounded above has a least upper bound. Induction is used only for finite integer arguments. No infinite cancellation or rearrangement is assumed.

Geometric powers tend to zero. Let $0\le\rho<1$. If $\rho=0$, then $\rho^n=0$ for $n\ge1$. Otherwise put $t=1/\rho-1>0$. The inequality $(1+t)^n\ge1+nt$ follows by induction: multiplication of $1+nt$ by $1+t$ gives $1+(n+1)t+nt^2\ge1+(n+1)t$. Therefore $0\le\rho^n\le1/(1+nt)$. For any $\varepsilon>0$, the Archimedean property supplies $N$ with $1+Nt>1/\varepsilon$; then $n\ge N$ gives $\rho^n<\varepsilon$. Thus $r^n\to0$ whenever $|r|<1$.

For $|r|<1$ and the finite geometric sum $S_N=a(1-r^N)/(1-r)$, direct subtraction gives

$$\left|S_N-\frac a{1-r}\right|=\frac{|a|\,|r|^N}{|1-r|}\longrightarrow0.$$

This proves the infinite sum and its exact signed remainder $a r^N/(1-r)$ directly from the definition of convergence, including negative $r$. If $a=0$, every partial sum is zero for every real or complex $r$. If $a\ne0$ and $|r|\ge1$, the terms have magnitude at least $|a|$ and do not tend to zero.

Why convergent series have terms tending to zero. If $S_n\to S$, then

$$|a_n|=|S_n-S_{n-1}|\le|S_n-S|+|S_{n-1}-S|.$$

For any $\varepsilon>0$, both terms on the right are less than $\varepsilon/2$ for all sufficiently large $n$. Hence $a_n\to0$. Also, every convergent real sequence is bounded: eventually its distance from its limit is less than one, and its finitely many earlier values are bounded. These facts justify the geometric-series divergence assertion above.

Harmonic divergence by finite blocks. For each integer $k\ge1$, the finite block $n=2^{k-1}+1,\ldots,2^k$ has $2^{k-1}$ terms, each at least $1/2^k$. Consequently

$$\sum_{n=1}^{2^K}\frac1n =1+\sum_{k=1}^K\sum_{n=2^{k-1}+1}^{2^k}\frac1n \ge1+\frac K2.$$

These partial sums are unbounded, so the harmonic series cannot converge, although $1/n\to0$ by the Archimedean property. All grouping above is finite; it does not appeal to a theorem about rearranging infinite series.

Comparison and absolute convergence. First, a nondecreasing real sequence $(s_n)$ bounded above converges to $L=\sup\{s_n:n\ge1\}$. Indeed, for each $\varepsilon>0$ some $s_N>L-\varepsilon$, or else $L-\varepsilon$ would be an upper bound; for $n\ge N$, $L-\varepsilon<s_n\le L$.

If $0\le u_n\le b_n$ from some index onward and $\sum b_n$ converges with $b_n\ge0$, the tail partial sums of the $u_n$ are nondecreasing and bounded by a fixed finite initial contribution plus a bound for the $b_n$ partial sums. The preceding supremum argument proves that $\sum u_n$ converges. For real $a_n$, if $\sum|a_n|$ converges, apply this argument to $a_n^+=(|a_n|+a_n)/2$ and $a_n^-=(|a_n|-a_n)/2$, each between zero and $|a_n|$. Their partial sums converge, so their difference $\sum_{n=1}^N a_n$ converges: the error in the difference is at most the sum of the two individual errors. For complex $a_n$, apply the real result to real and imaginary parts, whose absolute values are at most $|a_n|$. If these component sums have limits $U,V$, the complex partial sums tend to $U+iV$ because $|z-(U+iV)|\le|\operatorname{Re}z-U|+|\operatorname{Im}z-V|$. This proves absolute convergence implies convergence, and justifies comparison with a convergent geometric majorant. A finite number of initial terms does not affect convergence because it only adds a fixed finite constant to later partial sums.

1. Why convergence spreads inward

Lemma: bounded terms give geometric control

Suppose the power series converges at $x_0\ne a$. Then its terms are bounded: for some $M>0$, $|c_n(x_0-a)^n|\le M$ for all $n$. If $|x-a|<|x_0-a|$, then

$$|c_n(x-a)^n|\le M\left(\frac{|x-a|}{|x_0-a|}\right)^n,$$

so the series converges absolutely by geometric comparison.

Proof. Terms of a convergent numerical series tend to zero and are therefore bounded, including their finite initial segment. Factor $(x-a)^n$ as $(x_0-a)^n[(x-a)/(x_0-a)]^n$. The resulting geometric ratio has magnitude below one, proving the comparison. Absolute convergence in turn makes signed tails Cauchy by the triangle inequality, so the original series converges as well.

Theorem: radius of convergence

There is an extended number $R\in[0,\infty]$ such that the series converges absolutely for $|x-a|<R$ and diverges for $|x-a|>R$. If $0<R<\infty$, the two endpoints $a-R,a+R$ require separate tests.

Proof. Let $R$ be the supremum of the distances $|x-a|$ at which the series converges, with the center supplying distance zero. If a distance $r<R$ is given, the definition of supremum supplies a convergent point at distance greater than $r$; the lemma then gives absolute convergence at every point of distance $r$. A point farther than $R$ cannot be a convergent point by the upper-bound property. The same argument covers an unbounded set of convergence distances, with $R=\infty$, and the case where only distance zero occurs, with $R=0$. The argument uses a strict comparison of distances and therefore does not decide either endpoint at distance exactly $R$.

The interval can consequently be open, closed, or closed at only one endpoint. Having the same distance from the center does not force the same signed endpoint behavior.

2. A complete endpoint example

Consider $\sum_{n=1}^{\infty}(x-2)^n/n$. For $|x-2|<1$, its absolute terms are bounded by $|x-2|^n$, so it converges. If $|x-2|>1$, write $\rho=|x-2|>1$. The terms $\rho^n/n$ do not approach zero: their successive ratio is $\rho n/(n+1)$, eventually bounded below by some number greater than one. Thus the radius is one.

At $x=3$, the series is harmonic and diverges. At $x=1$, it is the alternating harmonic series and converges, but not absolutely. To justify the alternating conclusion, first negate it to obtain the positive-start alternating harmonic series. For that negated series, even partial sums increase and odd partial sums decrease; each even sum is at most each later odd sum, and the paired difference tends to zero. Real completeness gives a common limit. Negating back preserves convergence. This is the usual alternating argument for positive magnitudes decreasing to zero, not a permission to use arbitrary alternating signs. The convergence interval is therefore $[1,3)$.

For comparison, $\sum_{n\ge1}(x-2)^n/n^2$ converges absolutely at both endpoints. The numerical series $\sum1/n^2$ converges because for $n\ge2$, $1/n^2\le1/[n(n-1)]=1/(n-1)-1/n$, a telescoping majorant. Thus its interval is $[1,3]$, despite the same radius one.

3. Functions from finite geometric identities

For $|x|<1$, the finite formula $\sum_{n=0}^N x^n=(1-x^{N+1})/(1-x)$ gives

$$\frac1{1-x}=\sum_{n=0}^{\infty}x^n, \qquad \frac1{3-2x}=\frac13\sum_{n=0}^{\infty}\left(\frac{2x}{3}\right)^n \quad\text{when }|x|<\frac32.$$

The rational functions may be defined beyond these intervals, but these particular series representations are not. At $x=2$, for example, $1/(3-2x)=-1$ exists, while the geometric representation has ratio $4/3$ and diverges. A function's formula domain and one expansion's convergence domain are distinct.

Substitution can also change the shape of the powers. The series

$$\sum_{n=0}^{\infty}\frac{(x-1)^{2n}}{4^n}$$

is geometric with ratio $(x-1)^2/4$. It converges exactly for $|x-1|<2$, so its interval is $(-1,3)$, and its sum there is $1/[1-(x-1)^2/4]$. At either endpoint the terms are one, violating the zero-term necessary condition.

4. Approximation on a whole input range

Suppose convergence at distance $s>0$ gives $|c_n|s^n\le M$. For a fixed $0<r<s$, every input with $|x-a|\le r$ has omitted tail after degree $N$ bounded by

$$\left|\sum_{n=N+1}^{\infty}c_n(x-a)^n\right| \le\frac{M(r/s)^{N+1}}{1-r/s}.$$

The same bound works throughout the smaller closed interval. It is a certified approximation because it controls all omitted terms together, independently of the chosen input inside that interval.

For a more specific example, on $|x|\le r<1$,

$$\left|\sum_{n=N+1}^{\infty}\frac{x^n}{n+1}\right| \le\frac{r^{N+1}}{(N+2)(1-r)},$$

since $1/(n+1)\le1/(N+2)$ throughout that tail. Near an endpoint with $r$ close to one, the denominator $1-r$ explains why a small first omitted term may still leave a sizeable total error. Claims about derivatives or integrals of the summed function require additional theorems, supplied later in analysis rather than assumed here.

5. Radii zero and infinity are genuine cases

For $\sum_{n\ge0}n!x^n$, any nonzero input has successive absolute-term ratio $(n+1)|x|$, eventually at least two. Terms cannot tend to zero, so the radius is zero. At the center the series has only its constant contribution.

For $\sum_{n\ge0}x^n/n!$, first take $x=0$: the constant term is one and every higher-degree polynomial term is zero, so the series converges. If $x\ne0$, its absolute terms are positive and their successive ratio is $|x|/(n+1)$, eventually at most $1/2$. From that index onward, repeated comparison bounds the absolute tail by a convergent geometric tail. Hence its radius is infinity. Identifying its sum as the exponential and proving the exponential's laws is a separate construction; the convergence statement itself does not assume that identity.

Graded practice and diagnostics

1. Centered substitution. Find the convergence interval and sum of $\sum_{n\ge0}[(x+2)/5]^n$, checking both endpoints.
2. Endpoint proof. Determine the interval of $\sum_{n\ge1}(x+1)^n/n^2$ and justify the endpoint tests without citing a ratio equal to one as decisive.
3. Domain distinction. Explain why the rational function $1/(1-x)$ can be defined at $x=2$ even though its displayed geometric expansion cannot represent it there.
4. Certified tail. For $\sum_{n\ge0}x^n/(n+1)$ on $|x|\le1/2$, give an explicit error bound after degree $N$ and prove it tends to zero.
5. Radius diagnosis. Prove that $\sum_{n\ge0}n!(x-3)^n$ has radius zero and identify its sum at the center.
6. Proof repair. A student says convergence at $x=a+s$ proves convergence at both $a\pm s$ and all farther points. State exactly what the inward lemma proves, and provide an endpoint counterexample to the stronger assertion.

Full solutions

1. The ratio has magnitude below one when $-5<x+2<5$, or $-7<x<3$. At $x=3$ the terms are one; at $x=-7$ they alternate between one and minus one and still do not tend to zero. Both endpoints diverge. Inside, the sum is $1/[1-(x+2)/5]=5/(3-x)$.
2. Interior convergence holds for $|x+1|<1$ by geometric comparison, and outside that distance put $\rho=|x+1|>1$. The successive absolute-term ratio is $\rho n^2/(n+1)^2$, eventually at least $q=(1+\rho)/2>1$. Iteration bounds these positive terms below by a positive geometric sequence growing without bound, so they do not tend to zero. At $x=0$ and $x=-2$, the absolute series is $\sum1/n^2$, convergent by the telescoping bound above. The full interval is $[-2,0]$. A ratio tending to one alone would have been inconclusive.
3. The function value is $-1$ at two because its denominator is nonzero. Its power-series partial sums there are $1+2+\cdots+2^N=2^{N+1}-1$, diverging. Equality to a power-series sum was proved only where the limiting geometric remainder vanishes, namely $|x|<1$.
4. The bound is $(1/2)^{N+1}/[(N+2)(1-1/2)]=2^{-N}/(N+2)$. It tends to zero since it is at most $2^{-N}$. The derivation bounds the full absolute tail uniformly over the stated interval, including negative inputs where cancellation could give a smaller actual error.
5. At any $x\ne3$, the magnitude ratio is $(n+1)|x-3|$, eventually greater than two, so the terms fail the necessary zero-term condition. Only the center converges; its value is the constant term $0!=1$. Thus the radius is zero.
6. Convergence at distance $s>0$ implies absolute convergence at every strictly smaller distance, not at the opposite endpoint or farther away. For $\sum_{n\ge1}(-1)^n x^n/n$, the input $x=1$ gives a convergent alternating harmonic series, but $x=-1$ gives the divergent positive harmonic series. Inputs with $|x|>1$ have terms not tending to zero. The strict inequality in geometric comparison is indispensable.

Summary and terms

A power series assigns a numerical series to each input. Its center is the input around which powers are taken, and its radius separates interior absolute convergence from exterior divergence. An endpoint test is independent of the interior test. A uniform tail bound on a range certifies every input in that smaller range without asserting any unproved interchange of derivatives or integrals.