From a Geometric Series to a Function Expansion
A rational function may have a power-series representation even when its formula is not a polynomial. A useful starting point is the geometric series: by replacing its variable with an expression involving \(x\), we can produce expansions for functions such as \(1/(1+x)\) and \(1/(4-x^2)\).
The key is to keep two questions separate. First, the substitution gives a formal pattern of terms. Second, the geometric-series convergence condition tells us which values of \(x\) make the infinite sum equal the function. A formula that looks correct algebraically need not converge for every \(x\) where the function is defined.
Why the Geometric-Series Formula Works
For a finite number of terms, multiplication by \(1-r\) makes the intermediate powers cancel. For \(r\ne 1\), the sum of the first \(N+1\) terms is therefore
If \(|r|<1\), then \(r^{N+1}\) approaches \(0\) as \(N\) increases. Taking the limit of the finite-sum formula gives
This also explains why the condition matters. If \(r=1\), the terms do not approach zero. If \(r=-1\), they alternate between \(1\) and \(-1\), so they do not approach zero either. If \(|r|>1\), the terms \(r^n\) do not approach zero. In all these cases, the series diverges: its terms fail the necessary condition that they approach zero.
After substituting an expression \(g(x)\) for \(r\), the direct convergence condition becomes \(|g(x)|<1\). Solve that inequality to find where the formula applies, then test any boundary values separately.
A Reliable Substitution Method
Rewrite the function so its variable-dependent factor has the form \(1-r\), possibly after factoring out a constant.
Replace each occurrence of \(r^n\) by the corresponding expression in \(x\), keeping any outside factor.
Use \(|r|<1\) to find the interior of the convergence interval.
Substitute each endpoint into the series and decide whether its terms approach zero and whether the series converges.
At a convenient interior value, sum the resulting geometric series and verify that it matches the rational function.
Worked Example: Substituting \(x^2\)
Worked Example: Expand \(1/(1+x^2)\)
Start with \(1/(1-r)=\sum_{n=0}^{\infty}r^n\), valid when \(|r|<1\). Set \(r=-x^2\), since \(1-r=1-(-x^2)=1+x^2\). Therefore, wherever \(|-x^2|<1\),
The convergence condition reduces to \(|x|^2<1\), or \(-1<x<1\). At \(x=1\), the series is \(1-1+1-1+\cdots\); at \(x=-1\), it is also \(1-1+1-1+\cdots\), because \(x^2=1\). In both cases the terms do not approach zero, so both endpoints are excluded. The interval of convergence is \((-1,1)\).
Check the sum at \(x=1/2\). The function value is \(1/(1+(1/2)^2)=1/(5/4)=4/5\). The series has ratio \(-1/4\), so its sum is \(1/(1-(-1/4))=1/(5/4)=4/5\). The series and the rational function agree at this interior value.
Worked Example: Substituting \(-x\)
Worked Example: Expand \(1/(1+x)\)
Choose \(r=-x\). Then \(1-r=1-(-x)=1+x\), so the geometric formula gives
provided that \(|-x|<1\), which is equivalent to \(-1<x<1\). At \(x=1\), the terms are \(1,-1,1,-1,\ldots\), and they do not approach zero. At \(x=-1\), the terms are \(1,1,1,\ldots\), and they also do not approach zero. Thus the interval is \((-1,1)\). Notice that the function itself is undefined at \(-1\), while it is defined at \(1\); nevertheless, the series diverges at both endpoints.
For a check, take \(x=1/2\). The series becomes \(1-1/2+1/4-1/8+\cdots\), with ratio \(-1/2\). Its sum is \(1/(1+1/2)=2/3\). Direct substitution gives \(1/(1+1/2)=2/3\) as well.
Worked Example: Scaling the Substitution
Worked Example: Expand \(1/(3-x)\)
Factor \(3\) from the denominator to create the form \(1-r\):
Set \(r=x/3\). Applying the geometric series and retaining the outside factor gives
The condition is \(|x/3|<1\), or \(-3<x<3\). At \(x=3\), the function is undefined and every term of the series equals \(1/3\), so the series diverges. At \(x=-3\), the terms are \((-1)^n/3\), which do not approach zero; the series diverges there too. The interval of convergence is \((-3,3)\).
Check at \(x=1\). The series is \(1/3+1/9+1/27+\cdots\), with ratio \(1/3\), so its sum is \((1/3)/(1-1/3)=(1/3)/(2/3)=1/2\). The original function gives \(1/(3-1)=1/2\), confirming the expansion at this value.
Reading the Convergence Interval Carefully
The examples illustrate two important distinctions. First, the function's domain and the series' interval of convergence are not the same thing. For instance, \(1/(1+x^2)\) is defined for every real \(x\), but the geometric expansion derived above converges only for \(-1<x<1\). The series represents the function on its interval of convergence; it does not automatically represent it everywhere the function exists.
Second, solving the strict inequality \(|g(x)|<1\) identifies the interior where the geometric-series formula applies. The boundary values satisfy \(|g(x)|=1\), not \(|g(x)|<1\), so the formula alone gives no conclusion about them. Substitute each endpoint into the actual series. For the direct geometric substitutions in these examples, the resulting terms fail to approach zero, so the endpoints are excluded.
A Common Mistake
A frequent error is to forget that the convergence condition applies to the substituted expression, not automatically to \(x\). For example, after substituting \(r=-x^2\), it would be incorrect to write “\(|x|<1\) because \(|r|<1\)” without showing the step connecting them. The condition is
Another error is to treat a boundary value as convergent because the rational function has a finite value there. For \(1/(1+x)\) at \(x=1\), the function equals \(1/2\), but the series \(1-1+1-1+\cdots\) diverges because its terms do not approach zero. A function value does not establish convergence of a particular series representation.
Why These Substitutions Matter
Geometric substitutions turn a familiar formula into a practical way to recognize power series for rational functions. The powers in the expansion reveal useful structure: substituting \(x^2\) produces only even powers, while substituting \(-x\) produces alternating signs. Factoring a constant first can also change the interval of convergence, as the expansion of \(1/(3-x)\) shows.
For AP problems, a disciplined solution makes the reasoning visible: show the algebraic rewrite, identify the substituted quantity, derive the inequality, test the endpoints, and check the resulting sum when a convenient value is available. These steps prevent sign errors and keep the claimed interval tied to the actual convergence condition.
Check Your Understanding
For each question, make clear which substitution is used and how the convergence condition is obtained.
- Use the geometric series to write \(1/(1-x^2)\) as a power series and determine its interval of convergence, including endpoint tests.
- Rewrite \(1/(2+x)\) as a geometric-series expansion. What condition must \(x\) satisfy for the expansion to converge?
- For the expansion of \(1/(4-x)\), determine both the interval of convergence and the behavior at each endpoint.
- Why must the endpoints be tested separately after solving \(|g(x)|<1\)?
- Give an example from this tutorial where the rational function is defined at an endpoint but its geometric-series expansion diverges there.
