Why Domain Checks Matter
A power series is not automatically equal to its associated algebraic expression everywhere that expression makes sense. The series represents the function only at inputs where the series converges. Substitution, multiplication, and division do not remove that requirement.
The distinction matters because algebraic simplification can produce a formula defined at points where one of the original series diverges. At such points, the simplified formula may have a value, but the claimed identity involving the original series is not justified.
Convergence Regions and Substitution
A power series centered at \(c\) has the form \(\sum_{n=0}^{\infty} a_n(u-c)^n\). Its convergence region is the set of values of \(u\) for which this series converges. If its radius of convergence is \(R\), it converges absolutely when \(|u-c|<R\), diverges when \(|u-c|>R\), and requires a separate test at either endpoint when \(R\) is finite.
Suppose the series represents a function \(F(u)\) on a convergence region \(C\), and we substitute \(u=g(x)\). The substituted series is \(\sum_{n=0}^{\infty}a_n(g(x)-c)^n\). For this series to represent \(F(g(x))\) by direct substitution, we need \(g(x)\in C\). In the interior of a radius-\(R\) interval, this means \(|g(x)-c|<R\).
The restriction is on the value being substituted into the series, not merely on whether \(g(x)\) itself is defined. A formula for \(g\) may exist for many \(x\)-values while producing values outside \(C\).
Worked Example: Substituting a Rational Function into a Geometric Series
Use the geometric series \(\sum_{n=0}^{\infty}u^n=1/(1-u)\), which converges when \(|u|<1\). Substitute \(u=2x/(1+x^2)\). Determine where the substituted series is valid, and compare that region with the simplified rational expression.
For every real \(x\), \(1+x^2>0\), so the substituted expression is defined. The required convergence condition is
Since \(1+x^2>0\), this is equivalent to \(2|x|<1+x^2\). Subtracting \(2|x|\) gives
The inequality is strict exactly when \(|x|\ne 1\). Thus the substituted geometric series converges for every real \(x\) except \(x=-1\) and \(x=1\). Within that region, its sum is
At \(x=-1\), the rational expression on the right is defined and equals \(2/4=1/2\). But the substituted series has terms \((-1)^n\). Its terms do not tend to zero, and its partial sums alternate between \(1\) and \(0\); therefore the series diverges. The rational expression has a value there, but it is not the sum of the substituted series there. At \(x=1\), the rational expression is undefined as well as the series divergent.
Check at an interior point, \(x=0\): the substituted series is \(\sum_{n=0}^{\infty}0^n=1\), and the rational expression gives \((1+0)/(0-1)^2=1\). The series and formula agree where the convergence condition applies.
Products: Use the Common Convergence Region
Suppose two power series represent \(F(x)\) and \(G(x)\). To multiply their series as representations of those functions, first require that each series converges at the chosen \(x\). A simple safe setting is a point in the interior of both radii of convergence: there each series converges absolutely.
If \(\sum_{n=0}^{\infty}a_nx^n\) and \(\sum_{n=0}^{\infty}b_nx^n\) converge absolutely, their Cauchy product has coefficients \(c_n=\sum_{k=0}^{n}a_kb_{n-k}\), and its sum is the product of the two sums. Absolute convergence is what permits the terms in the product to be regrouped by powers of \(x\). At endpoints, do not assume this conclusion from the radii alone: check convergence there and verify that the hypotheses for the product operation hold.
Worked Example: The Product Has a Smaller Convergence Region Than the Formula Suggests
Multiply the geometric series \(\sum_{n=0}^{\infty}(x/2)^n\) and \(\sum_{n=0}^{\infty}(x/3)^n\). The first converges for \(|x|<2\), and the second for \(|x|<3\). Their common interior region is therefore \(|x|<2\).
For the Cauchy product, the coefficient of \(x^n\) is
The coefficient formula follows by summing the finite geometric sequence with first term \((1/3)^n\) and ratio \(3/2\); the displayed difference-of-powers form gives the same finite sum. Thus the product series is
At \(x=1\), which lies in the common region, the original series sum to \(2\) and \(3/2\), respectively, so their product is \(3\). The coefficient series gives
This agrees with the product. The rational expression obtained by multiplying the geometric sums is \(1/((1-x/2)(1-x/3))\), which is defined at \(x=5/2\). But the coefficient series does not converge there: its \(n\)th term is
which does not tend to zero. In contrast, the rational expression has the finite value
That value belongs to the rational formula, not to the divergent product series at \(x=5/2\). The product series is guaranteed to represent the product on the common region \(|x|<2\), not wherever the rational expression happens to be defined.
Quotients: Check the Denominator Twice
For a quotient of represented functions, two checks are needed. Both source series must converge at the input, and the function represented by the denominator series must be nonzero there. The nonzero condition makes the quotient defined. If a power series for the quotient is also claimed, that new series must itself be shown to converge at the input.
These are distinct questions. The quotient of two function values can be defined at a point where a source series no longer represents one of those values. Conversely, convergence of both source series does not make a quotient valid if the denominator sum is zero.
Worked Example: A Quotient Whose Algebraic Formula Extends Beyond Its Source Series
Consider the two series
The first converges for \(|x|<2\) and represents \(1/(1-x/2)\). The second converges for \(|x|<1\) and represents \(1/(1+x)\). Both series therefore represent their functions on the common region \(|x|<1\). In that region, \(1/(1+x)\ne 0\), so the quotient of the represented functions is
At \(x=1/2\), the first series has ratio \(1/4\), so its sum is \(1/(1-1/4)=4/3\). The second has ratio \(-1/2\), so its sum is \(1/(1+1/2)=2/3\). Their quotient is \((4/3)/(2/3)=2\). The simplified expression also gives
At \(x=3/2\), the simplified rational expression is defined and equals \((5/2)/(1/4)=10\). But the series for \(B(3/2)\) has terms \((-3/2)^n\), which do not tend to zero, so that source series diverges. Consequently, the quotient identity obtained from the two source series is not justified at \(x=3/2\). The value \(10\) can be obtained from the rational expression or its separately derived expansion, but not by evaluating the divergent source series for \(B\).
A Reliable Check Before Using a Series Identity
The following sequence of checks separates convergence issues from later algebra. It is especially useful when a substitution, product, or quotient is followed by simplification.
Write down each series being used and the region where it represents its function.
For a substitution \(u=g(x)\), check that \(g(x)\) lies in the source convergence region. For products and quotients, take the intersection of the source regions.
For a product, justify the coefficient multiplication, using absolute convergence in the interior. For a quotient, also check that the denominator sum is nonzero.
If an operation produces a new coefficient series, verify its convergence where you use it. Do not infer its full convergence region from the simplified function alone.
A radius gives absolute convergence in the interior, not an automatic answer at the endpoints. Test the relevant series at each endpoint.
Why Algebra Alone Cannot Validate a Series Identity
The geometric-series identity illustrates the issue in its simplest form:
The right-hand side is an algebraic expression, while the left-hand side is an infinite sum. The algebraic expression may be defined for some \(u\)-values outside the series’ convergence region, but that does not make the infinite sum converge there. A derivation that substitutes, multiplies, or divides series must retain the domain conditions that justified each step.
A common mistake is to simplify first and then declare the resulting rational expression to be the sum wherever it is defined. For example, one might substitute \(u=2x/(1+x^2)\) and claim the geometric series equals \((1+x^2)/(x-1)^2\) at every \(x\ne1\). This claim includes \(x=-1\), where the rational expression equals \(1/2\), even though the series becomes \(1-1+1-1+\cdots\) and diverges. The correction is to carry the original condition \(|u|<1\) through the substitution; it excludes both \(x=-1\) and \(x=1\).
Check Your Understanding
For each question, identify the convergence condition before deciding what the series represents.
- For the geometric series \(\sum_{n=0}^{\infty}u^n\), what condition must a substituted value \(u=g(x)\) satisfy for the standard sum formula to apply?
- Two power series have radii \(3\) and \(5\). What is their common interior region for multiplying them by the Cauchy product?
- Why does the fact that a rational product formula is defined at a point not establish convergence of its product series there?
- What two conditions must hold before taking the quotient of functions represented by power series at an input?
- Why must endpoints of a power series’ convergence interval be checked separately?