From Partial Fractions to Power Series
A rational function may be difficult to expand directly when its denominator is a product of several linear factors. Partial fractions separate that denominator into simpler terms. Each term can then be expanded using the geometric series, with its own convergence condition.
The key point is that the resulting power series is valid only where all of its geometric-series components converge. Thus, the factors in the denominator determine individual intervals, and their intersection determines the interval for the complete expansion.
Partial Fractions for Distinct Linear Factors
Suppose \(P(x)\) is a polynomial of degree less than \(n\), and the denominator has \(n\) distinct real roots \(r_1,\ldots,r_n\). Write \(Q(x)=\prod_{i=1}^{n}(x-r_i)\). Then the proper rational function \(P(x)/Q(x)\) has a partial-fraction decomposition
Distinctness matters: every denominator in the formula for \(A_i\) is nonzero because \(r_i\ne r_j\) whenever \(i\ne j\). If the numerator vanishes at one of the roots, that term has coefficient zero and can be omitted; in that case the corresponding factor cancels rather than giving an actual pole.
To use a power series centered at zero, each remaining pole must be nonzero. For a nonzero root \(r_i\), rewrite its term as
The geometric series \(\frac{1}{1-u}=\sum_{k=0}^{\infty}u^k\) converges when \(|u|<1\). Substituting \(u=x/r_i\) therefore gives
Each term supplies its own condition. The full partial-fraction expansion is represented by the sum of these geometric series wherever all the conditions hold. For nonzero roots with nonzero coefficients, their common interval is
This is an open interval centered at zero. If a root is at zero and is not canceled, the function has a pole at the expansion center, so it cannot have an ordinary power series there.
Why the Partial-Fraction Formula Works
The coefficients are chosen to make the decomposed expression agree with the original numerator at every root. Here is a proof for the proper rational function above.
Using the stated coefficients, consider \(S(x)=\sum_{i=1}^{n} A_i\prod_{j\ne i}(x-r_j)\). Its degree is at most \(n-1\).
At \(x=r_k\), every summand with \(i\ne k\) contains the factor \(x-r_k\) and vanishes. The remaining term is \(A_k\prod_{j\ne k}(r_k-r_j)=P(r_k)\). Thus \(S(r_k)=P(r_k)\).
The polynomial \(P-S\) has degree at most \(n-1\) and vanishes at the \(n\) distinct values \(r_1,\ldots,r_n\). A nonzero polynomial of degree at most \(n-1\) cannot have \(n\) distinct roots, so \(P-S\) is identically zero. Dividing \(P=S\) by \(\prod_i(x-r_i)\) proves the decomposition.
The convergence condition follows just as directly: the \(i\)th geometric series has ratio \(x/r_i\), so it converges exactly when \(|x/r_i|<1\). A sum of all the partial-fraction series is valid only when every ratio satisfies that inequality.
Worked Example: Two Factors with Different Convergence Intervals
Worked Example: Two Factors with Different Convergence Intervals
Expand \(f(x)=\frac{1}{(1-x)(1+2x)}\) as a power series centered at zero, and find its interval of convergence as a sum of partial-fraction expansions.
First write the partial fractions as \(\frac{A}{1-x}+\frac{B}{1+2x}\). Clearing the denominator gives
Matching the constant and linear coefficients gives \(A+B=1\) and \(2A-B=0\). The second equation gives \(B=2A\); substituting into the first gives \(3A=1\), so \(A=\frac13\) and \(B=\frac23\). Therefore
The first term is geometric with ratio \(x\), and the second with ratio \(-2x\). Thus
The intersection is \(|x|<\frac12\), or \(-\frac12<x<\frac12\). On that interval the combined expansion is
For a check, recombining the partial fractions gives \(\frac{(1/3)(1+2x)+(2/3)(1-x)}{(1-x)(1+2x)} =\frac{1}{(1-x)(1+2x)}\). Also, summing the two geometric series on their common interval recovers the same partial fractions and hence the original function. The second series cannot converge at either endpoint \(x=\pm\frac12\), because its terms there have magnitude \(\frac23\) and do not tend to zero.
Worked Example: A Negative Root and a More Distant Pole
Worked Example: A Negative Root and a More Distant Pole
Find the power-series expansion centered at zero for \(g(x)=\frac{x+5}{(x+1)(x-3)}\), and determine the interval where both partial-fraction series converge.
Write \(g(x)=\frac{A}{x+1}+\frac{B}{x-3}\). Clearing denominators yields \(x+5=A(x-3)+B(x+1)=(A+B)x+(-3A+B)\). Thus \(A+B=1\) and \(-3A+B=5\). Subtracting the first equation from the second gives \(-4A=4\), so \(A=-1\) and \(B=2\). Hence
The first geometric ratio is \(-x\), so its condition is \(|x|<1\). The second ratio is \(x/3\), giving \(|x|<3\). Their intersection is \(-1<x<1\), and the expansion is
Check the decomposition by recombining its fractions: \(\frac{-(x-3)+2(x+1)}{(x+1)(x-3)} =\frac{x+5}{(x+1)(x-3)}\). For instance, its constant series coefficient is \(-1-\frac23=-\frac53\), which agrees with \(g(0)=\frac{5}{(1)(-3)}=-\frac53\). At \(x=1\) and \(x=-1\), the first geometric series has terms whose magnitude is \(1\), so neither endpoint is included.
Worked Example: Three Factors and the Nearest Pole
Worked Example: Three Factors and the Nearest Pole
Expand \(h(x)=\frac{1}{(x-1)(x-2)(x+2)}\) about zero and identify the common interval.
Use \(h(x)=\frac{A}{x-1}+\frac{B}{x-2}+\frac{C}{x+2}\). Evaluating after clearing denominators at the three roots gives
Consequently,
The three geometric conditions are \(|x|<1\), \(|x|<2\), and \(|x|<2\). Their intersection is \(-1<x<1\). Therefore
To check the partial fractions, put them over the common denominator. Their numerator is \(-\frac13(x-2)(x+2)+\frac14(x-1)(x+2)+\frac1{12}(x-1)(x-2)\). The coefficients of \(x^2\) add to \(-\frac13+\frac14+\frac1{12}=0\); those of \(x\) add to \(0+\frac14-\frac14=0\); and the constants add to \(\frac43-\frac12+\frac16=1\). Thus the numerator is exactly \(1\). The nearest poles are at \(1\) and \(-1\), so the first geometric series fails to converge at either endpoint; the common interval is open.
A Common Mistake: Using Only One Convergence Condition
A frequent error is to find the convergence interval for one partial fraction and treat it as the interval for the entire expansion. For example, in the first worked example, the term \(\frac{1/3}{1-x}\) converges for \(|x|<1\), but the other term converges only for \(|x|<\frac12\). Claiming that the full expansion converges throughout \((-1,1)\) ignores that restriction.
Do not infer endpoint convergence from the interval of convergence of a different summand. At an endpoint of a geometric series' interval, its terms have constant nonzero magnitude, so they fail the necessary condition that the terms of a convergent series tend to zero.
Why This Method Is Useful
Partial fractions turn a rational function with several distinct linear factors into a finite sum of geometric-series expansions. This makes the coefficients accessible and makes the convergence restriction visible: each pole limits the distance from zero over which its corresponding geometric expansion works. The nearest uncanceled pole sets the common radius for these summands.
For a proper rational function with distinct nonzero real roots \(r_i\), the practical procedure is therefore:
- Find the partial-fraction coefficients.
- Rewrite each term in the form of a constant times \(1/(1-x/r_i)\).
- Expand using the geometric series and record \(|x|<|r_i|\) for that term.
- Intersect the conditions before stating the combined power series.
Keeping the conditions attached to their individual terms prevents a correct algebraic decomposition from becoming an incorrect claim about where its series represents the function.
Check Your Understanding
For each question, keep the convergence condition for every summand separate until you take the intersection.
- For \(1/((x-2)(x+4))\), what are the convergence conditions of the two geometric expansions centered at zero, and what is their intersection?
- Find the partial-fraction coefficients for \(1/((x-1)(x-3))\), then state the common interval for its geometric expansions.
- Why does a genuine pole at zero prevent an ordinary power series centered at zero?
- At an endpoint where a geometric-series ratio has absolute value \(1\), what necessary condition for convergence fails?
- For a proper rational function with distinct nonzero real roots, which root determines the common radius of convergence when all partial-fraction coefficients are nonzero?