Power series and intervals of convergence

One expression gives many numerical series

Start with 1+t+t²+⋯. At t=1/2 its partial sums approach 2. At t=−1/2 they approach 2/3. At t=1 the totals grow without bound; at t=−1 they alternate between 1 and 0. The expression is the same, but choosing t chooses a different numerical series.

A power series lets a variable play this role. Its first task is to define where an infinite sum exists. A finite polynomial can be evaluated everywhere, but its infinite continuation may converge only near one centre. A formula obtained by summing the series carries that convergence domain with it.

Preparation: use sequences, partial sums and convergence tests, especially geometric comparison, absolute convergence and the alternating test. Read the centre, radius and endpoint examples first. The later calculus sections use differentiation, integration and the fundamental theorem of calculus; the optional general radius formula introduces a precise upper-limit idea.

Separate the centre, coefficients and input

A real power series centred at a has the form Σₙ₌₀∞ cₙ(x−a)ⁿ, with fixed real coefficients cₙ and real input x. Its degree-N partial sum is P_N(x)=c₀+c₁(x−a)+⋯+c_N(x−a)ᴺ. It contains N+1 indexed terms, some of which may be zero. The number N is a highest index, not automatically a term count.

At x=a every positive power vanishes. The constant term c₀ remains, with the zeroth power defined as 1 in this notation, including when its base is zero. Thus every power series converges at its centre. This says nothing yet about a nearby nonzero displacement.

For example set t=(x+1)/3. Then Σₙ₌₀∞tⁿ is a power series centred at a=−1, with coefficients cₙ=3⁻ⁿ. The input x=1/2 gives t=1/2; the input x=−5/2 gives t=−1/2. Do not confuse the displacement x+1, the scaled displacement t and the coefficient 3⁻ⁿ.

A function formula comes with an interval

For t≠1, finite multiplication and subtraction give P_N=1+t+⋯+tᴺ=(1−tᴺ⁺¹)/(1−t). If |t|<1, the remaining power tends to zero. Hence Σₙ₌₀∞tⁿ=1/(1−t). The exact signed remainder after degree N is tᴺ⁺¹/(1−t).

With t=(x+1)/3, this gives 3/(2−x) on |x+1|<3, or −4<x<2. The rational function is defined at many other inputs, but the displayed power series diverges there. In particular, at x=−4 its terms are (−1)ⁿ and do not tend to zero. At x=2 its terms are 1.

For a fixed |t|<1, |R_N|=|t|ᴺ⁺¹/|1−t|≤|t|ᴺ⁺¹/(1−|t|). On a whole set |t|≤r<1 the same upper bound rᴺ⁺¹/(1−r) works for every input. For r=1/2 and N=7 this bound is 1/128. At positive t=1/2 it is attained; at negative t=−1/2 the exact absolute error is smaller, 1/384.

Approximation deteriorates near the boundary if N is fixed. For any fixed N, the positive-input error tᴺ⁺¹/(1−t) grows without bound as t approaches 1 from below. Increasing N gives convergence at each fixed interior input, but does not give one small error on the entire open interval at once.

Convergence at one distance controls every smaller distance

Suppose Σcₙdⁿ converges for a nonzero displacement d from the centre. Its individual terms form a bounded sequence, so choose M>0 with |cₙdⁿ|≤M for every n. For any displacement y with |y|<|d|, |cₙyⁿ|≤M(|y|/|d|)ⁿ. A geometric comparison proves absolute convergence at a+y. This works even if convergence at a+d was conditional.

Let D be the set of nonnegative distances |x−a| at which the series converges, and let R be its supremum, allowing R=∞. The centre gives 0∈D. If 0<r<R, there is some d∈D with d>r; otherwise r would be an upper bound below the supremum. The preceding comparison proves absolute convergence at both displacements of magnitude r. If a distance exceeds R it cannot belong to D.

Therefore exactly three regimes are possible: R=0 gives convergence only at a; 0<R<∞ gives absolute convergence for |x−a|<R and divergence for |x−a|>R; R=∞ gives absolute convergence at every real input. This R is the radius of convergence. At the two boundary points of a finite positive radius, the comparison ratio becomes 1 and the argument makes no decision.

The interval of convergence is the actual set of real inputs where the sum exists. For a finite positive radius it includes the interior and may include either, both or neither endpoint. R=0 gives the single-point set {a}; R=∞ gives the whole real line. A radius and a fully checked interval answer different questions.

Apply a numerical test after fixing the input

For an eventually nonzero coefficient sequence, suppose L=lim|cₙ₊₁/cₙ| exists, possibly 0 or +∞. At a fixed x≠a the magnitude ratio of the terms is |x−a||cₙ₊₁/cₙ|. If 0<L<∞, the ratio test proves convergence for |x−a|<1/L and divergence for |x−a|>1/L, so R=1/L. The endpoints still need separate tests.

If L=0 the series converges absolutely at every fixed input, hence R=∞. If L=+∞ it diverges at every x≠a, hence R=0. Handle x=a directly rather than multiplying an infinite limit by zero or dividing by a zero term. If coefficients vanish infinitely often or the coefficient ratio has no limit, this convenient ratio formula may be unavailable while the power series still has a well-defined radius.

For Σₙ₌₀∞(x−2)ⁿ/n!, with 0!=1, each fixed noncentral input has term ratio |x−2|/(n+1)→0. Thus R=∞. For Σₙ₌₀∞ n!(x−2)ⁿ, that ratio is (n+1)|x−2|→∞ at every x≠2. Thus R=0. Neither result depends on the first few partial sums looking small.

The same radius allows different endpoint answers

Keep t=(x+1)/3 and consider Σₙ₌₁∞tⁿ/n. At t≠0, the term-ratio limit is |t|; at t=0 the series is zero. Therefore its radius in x is 3, with absolute convergence on (−4,2). At x=−4 it becomes Σ(−1)ⁿ/n, which converges conditionally. At x=2 it becomes the divergent harmonic series. The full interval is [−4,2).

Replacing 1/n with 1/n² keeps the same ratio limit, but both endpoints now converge absolutely by comparison with Σ1/n². Replacing the coefficient 1/n with (−1)ⁿ/n reverses which endpoint is conditional and which diverges. The original geometric coefficients include neither endpoint. These are four different endpoint patterns with exactly the same centre and radius.

Endpoint tests use the actual substituted numerical series, including its sign and starting index. An alternating endpoint needs decreasing magnitudes tending to zero. An absolute-convergence claim needs a convergent absolute series. A ratio limit of 1 supplies neither conclusion.

Missing powers do not remove the convergence question

For Σₙ₌₀∞(x−1)²ⁿ/4ⁿ, treat [(x−1)²/4]ⁿ as a geometric term. It converges exactly when |x−1|<2; at either endpoint every term is 1. The centre is 1, the radius is 2, and the interval is (−1,3). As a power series in x−1 it has zero coefficients at every odd power, so an ordinary consecutive-coefficient ratio would repeatedly divide by zero.

When Q=lim|cₙ|¹ᐟⁿ exists for n≥1, the root test instead gives R=1/Q, using R=∞ for Q=0 and R=0 for Q=∞. Zero coefficients are allowed. A limit need not exist, so this version is a useful method rather than the most general formula. The radius theorem above did not assume either a ratio or a root limit.

Optional: an upper limit handles irregular coefficients

Let uₙ=|cₙ|¹ᐟⁿ for n≥1. Define Q=limsup uₙ as the limit of the nonincreasing extended upper bounds sup{uₙ:n≥m} as m grows. Q records the largest size that recurs arbitrarily late, not an average of the roots. It may be 0, finite positive or +∞.

If |y|Q<1 with Q finite, choose q>Q with |y|q<1. Eventually uₙ≤q, so |cₙyⁿ|≤(|y|q)ⁿ and geometric comparison proves absolute convergence. This also handles Q=0 for every finite y by choosing a sufficiently small positive q.

If |y|Q>1, choose b with 1/|y|<b<Q. Infinitely many roots exceed b; otherwise the upper limit would be at most b. At those indices |cₙyⁿ|>(|y|b)ⁿ, so the terms do not approach zero. For Q=∞ the same argument works for every nonzero y. The centre is handled separately.

Thus the general Cauchy–Hadamard formula is R=1/Q with the stated extended conventions. In the missing-odd-powers example, the coefficient roots are 1/2 at positive even indices and 0 at odd indices. Their ordinary limit fails, but their upper limit is 1/2, giving R=2. The boundary case |y|Q=1 remains undecided.

A smaller closed interval admits one error control

Pointwise convergence fixes x first and then chooses a cutoff. Uniform convergence on a set means that, for each ε>0, one cutoff makes |f(x)−P_N(x)|<ε for every input in the set whenever N is beyond that cutoff. It controls all those inputs together.

For a power series of positive radius R, choose 0<r<s<R. Convergence at a+s bounds all |cₙsⁿ| by a constant M>0. For every |x−a|≤r, the entire omitted absolute tail after degree N is at most M(r/s)ᴺ⁺¹/(1−r/s). This bound tends to zero independently of x. Therefore power series converge uniformly on every smaller closed interval [a−r,a+r].

The sum is continuous in the open convergence interval. Near any interior input, first make the two series tails small using the shared bound, then use continuity of the finite polynomial P_N to control its change. The triangle inequality combines the two tail errors and the polynomial change. This proves continuity without assuming arbitrary pointwise limits preserve it.

This conclusion is local to closed intervals strictly inside the radius. The geometric example already shows why it cannot automatically be extended to the whole open interval or its endpoints. A finite graph and a few small errors do not establish uniform convergence on a larger set.

Differentiate and integrate where the tails are controlled

If f(x)=Σₙ₌₀∞cₙ(x−a)ⁿ has radius R>0, then for |x−a|<R, f′(x)=Σₙ₌₁∞n cₙ(x−a)ⁿ⁻¹. Also the integral of f from a to x is Σₙ₌₀∞cₙ(x−a)ⁿ⁺¹/(n+1). Both resulting power series have the same radius R, though their endpoint behaviour can change. The integral from the centre fixes its constant to zero; an indefinite integral adds an arbitrary constant.

Here is why the limit operations are justified. Choose 0<r<s<R and |cₙsⁿ|≤M as above. On |x−a|≤r, the derivative terms have magnitudes at most (M/s)n(r/s)ⁿ⁻¹. This numerical series converges by the ratio test. Its tails give a uniform bound on the derivative-polynomial tails. The derivative polynomials therefore converge uniformly to a continuous function g on that closed interval.

Every finite polynomial satisfies P_N(x)=c₀ plus the integral of P_N′ from a to x. A uniform error δ in the derivative changes this integral by at most rδ. Passing to the limit gives f(x)=c₀ plus the integral of g. The fundamental theorem yields f′=g. The original uniform tail bound similarly justifies integrating f term by term. No general rule permits differentiating every pointwise-convergent function sequence.

The preceding estimates show that the new radii are at least R. They cannot be larger: if the derivative series converged in a larger neighbourhood, integrate it term by term there to recover the original coefficients, contradicting the original radius. If the integrated series had a larger radius, differentiating it there would give the same contradiction. Repeating the argument shows that the sum is infinitely differentiable in the interior.

A shared radius does not preserve endpoint inclusion

Starting from Σₙ₌₀∞xⁿ=1/(1−x) on |x|<1, differentiation gives Σₙ₌₁∞n xⁿ⁻¹=1/(1−x)² there. At x=1 the derivative-series terms are n; at x=−1 their magnitudes are n. Both endpoints diverge, just as the original endpoints did.

Integration from 0 to x gives Σₙ₌₁∞xⁿ/n=−log(1−x) for |x|<1. At x=−1 the series converges conditionally; at x=1 it diverges. Thus integration has changed the interval from (−1,1) to [−1,1) while leaving the radius 1. Differentiating this latter series removes the included endpoint again.

The interior theorem alone does not identify the sum at a newly included endpoint with the limiting function value there. For this example that identification can be proved directly: for t∈[0,1], 1/(1+t)=Σₖ₌₀ᴺ(−t)ᵏ+(−t)ᴺ⁺¹/(1+t). Integrate this finite identity from 0 to 1. The remainder has magnitude at most 1/(N+2), so log 2=Σₖ₌₀∞(−1)ᵏ/(k+1). Consequently Σₙ₌₁∞(−1)ⁿ/n=−log 2. The endpoint equality now has its own proof.

Combining series guarantees a common interior, sometimes more

Power series about the same centre can be added or subtracted coefficient by coefficient wherever both converge. Scalar multiplication is equally direct. On a common open interior they converge absolutely, so their product is given by the Cauchy coefficients eₙ=Σₖ₌₀ⁿcₖdₙ₋ₖ. The absolute double sum is bounded by the product of the two finite absolute totals; truncating a sufficiently large finite rectangle leaves an arbitrarily small tail. Grouping by total degree therefore gives the product sum.

The common interior is guaranteed, but it need not be the largest possible interval for the resulting coefficients. The two series Σxⁿ and −Σxⁿ each have radius 1; their coefficientwise sum is the zero series, with infinite radius. This does not make the two separate divergent sums meaningful outside their domains. The combined coefficient series is a new object there.

Substituting an expression h(x) into a known series is valid at each input where h(x) lies in its proved convergence set. For example 1/(1+t²)=Σₙ₌₀∞(−1)ⁿt²ⁿ for |t|<1. At t=±1 its terms alternate without tending to zero, so neither endpoint is included. Multiply by 2t to get 2t/(1+t²)=Σₙ₌₀∞2(−1)ⁿt²ⁿ⁺¹ on the same interval. At an input where a prefactor is zero, check the resulting series directly rather than assuming multiplication preserves divergence.

A represented function determines the coefficients

Repeated term-by-term differentiation inside a positive radius gives f⁽ⁿ⁾(a)=n!cₙ, including f(a)=c₀. Thus if a function has a power-series representation near a, its coefficients are uniquely cₙ=f⁽ⁿ⁾(a)/n!. Two convergent power series with the same centre representing the same function on a neighbourhood cannot have different coefficients.

The converse needs care. Having derivatives of every order does not by itself prove equality with the resulting infinite series. The next lesson constructs Taylor polynomials and checks when their remainders tend to zero. For now, a representation has been established either by an actual convergence argument and identity or by valid operations on an already represented function.

Finish with a domain and a justified approximation

A complete power-series solution records the centre and index convention, determines the radius, tests each finite endpoint, and states where a claimed function identity is proved. For an approximation, specify the degree or number of terms and give a remainder bound valid at that input or on the intended whole interval.

Keep four claims separate: the finite polynomial is defined, the infinite series converges, its sum equals a stated function, and the chosen truncation meets an error target. The geometric identity proves all four on its stated interior. A ratio test alone generally proves only convergence and divergence regions.

Use the explorer to compare interval patterns and finite polynomial errors. Predict an endpoint classification or the effect of moving toward the boundary before revealing the calculation. The proof supplies the infinitely many cases that a finite display cannot show.

Worked examples

Try it yourself

1. What is the centre of Σcₙ(x+5)ⁿ?

   Show the answer and reasoning

   −5.

   Write x+5=x−(−5).

2. What is the sum at x=a of Σₙ₌₀∞cₙ(x−a)ⁿ?

   Show the answer and reasoning

   c₀.

   All positive powers vanish, and the constant term remains.

3. How many indexed terms are in P₆=Σₙ₌₀⁶cₙ(x−a)ⁿ?

   Show the answer and reasoning

   Seven.

   The indices are 0 through 6; some coefficients may be zero.

4. A power series centred at 2 converges at 7. What is guaranteed at −2?

   Show the answer and reasoning

   Absolute convergence.

   Its distance from the centre is 4, smaller than 5. The geometric comparison applies even if convergence at 7 was conditional.

5. Does convergence at 7 for a series centred at 2 guarantee convergence at −3?

   Show the answer and reasoning

   No.

   The two points are equally far from the centre; boundary behaviours can differ.

6. Find the interval of Σₙ₌₁∞[(x+1)/3]ⁿ/n².

   Show the answer and reasoning

   [−4,2].

   The interior is absolute and both boundary absolute series are Σ1/n².

7. If the eventual coefficient ratio tends to zero, what is the radius?

   Show the answer and reasoning

   Infinite.

   At each fixed noncentral input the term ratio tends to zero; the centre is immediate.

8. Does a failed coefficient-ratio limit mean there is no radius?

   Show the answer and reasoning

   No.

   The radius theorem does not require that limit; geometric regrouping, roots or the upper-limit formula may determine it.

9. Does differentiating a power series preserve every included endpoint?

   Show the answer and reasoning

   No.

   Σxⁿ/n includes −1, but its differentiated geometric series does not.

10. Does pointwise convergence on (−1,1) of Σxⁿ give a uniform small truncation error on that whole interval?

    Show the answer and reasoning

    No.

    For any fixed degree, the error grows without bound as x approaches 1 from below.

11. Can adding two coefficient series increase their radius beyond either original radius?

    Show the answer and reasoning

    Yes.

    Σxⁿ plus −Σxⁿ gives the zero coefficient series, which converges everywhere. This does not assign separate divergent sums outside the original domains.

12. If a function has a convergent power series near a, how are its coefficients recovered?

    Show the answer and reasoning

    cₙ=f⁽ⁿ⁾(a)/n!.

    Repeated term-by-term differentiation is valid inside the positive radius. Existence of all derivatives alone does not prove a representation.

What to carry forward

• Every power series converges at its centre, where only its constant term remains.
• The radius controls absolute convergence in the interior and divergence outside; each finite endpoint needs its own test.
• A coefficient ratio or root formula is a method with hypotheses, not a prerequisite for a radius to exist.
• The general reciprocal-upper-limit formula handles missing or irregular coefficients.
• A finite polynomial can be defined where its infinite continuation diverges. Function identities retain their proved domains.
• Power series have uniform tail control on smaller closed intervals, enabling valid term-by-term calculus in the interior.
• Differentiation and integration preserve the radius but can change endpoint inclusion.
• A power-series representation has unique derivative coefficients; smoothness alone is not a representation proof.