From a Geometric Series to a Function Series
A geometric series can do more than sum a list of numerical terms. When its common ratio depends on \(x\), it can represent a function as a power series wherever the geometric series converges. The central task is to substitute carefully and then translate the convergence condition into a condition on \(x\).
The basic formula, for a real number \(r\) satisfying \(|r|<1\), is
If an expression \(g(x)\) takes the place of \(r\), the corresponding formula is
The condition is essential. Substitution does not make the geometric series converge for every input where the function on the right happens to be defined. First require \(|g(x)|<1\); then, if necessary, examine any boundary values separately.
Why the Substitution Works
The geometric-series formula comes from the finite partial sum. If \(r\ne 1\), then multiplying the sum of the first \(N+1\) terms by \(1-r\) cancels all intermediate powers:
Therefore
When \(|r|<1\), the powers \(r^{N+1}\) approach zero as \(N\) increases. Taking the limit of the finite sums gives \(1/(1-r)\). For a fixed value of \(x\), set \(r=g(x)\). If \(|g(x)|<1\), the same argument applies to that numerical ratio, and the sum is \(1/[1-g(x)]\). Thus substitution works because, at each eligible \(x\), it is precisely the original geometric-series argument with a particular value of \(r\).
Worked Example: A Series for \(1/(1+x^2)\)
Worked Example: A Series for \(1/(1+x^2)\)
Rewrite the function using the geometric-series formula, and find the full convergence interval.
We want the denominator to have the form \(1-r\). Since \(1+x^2=1-(-x^2)\), choose \(r=-x^2\). Substitution gives
This equality follows from the geometric-series formula only when
We must test the endpoints because the strict inequality does not include them. At \(x=1\), the terms are \((-1)^n\), which do not approach zero, so the series diverges. At \(x=-1\), the terms are also \((-1)^n\), so the series diverges there as well.
Check the representation at \(x=\tfrac12\). The series becomes
Its common ratio is \(-\tfrac14\), so its sum is
The function value agrees:
Therefore the series represents the function on the interval \((-1,1)\), with neither endpoint included.
Transforming the Convergence Condition
After substitution, the convergence question is an inequality involving the substituted expression, not automatically the familiar interval \((-1,1)\) for \(x\). The original geometric series requires \(-1<r<1\). If \(r=g(x)\), solve \(-1<g(x)<1\) for \(x\), taking care with signs, squares, and any domain restrictions.
The set of eligible \(x\)-values can be an interval, several intervals, or another set, depending on \(g\). Even if the function \(1/[1-g(x)]\) is defined at a point, the series still needs to be tested there: the formula for the sum was derived under the geometric-series convergence condition.
Express the function in the form \(1/[1-g(x)]\), including any constant factor if one is needed.
Replace every occurrence of \(r\) in \(\sum_{n=0}^{\infty}r^n\) by \(g(x)\).
Find the \(x\)-values satisfying \(|g(x)|<1\).
At each candidate endpoint, inspect the actual terms; the strict inequality alone gives no endpoint conclusion.
Worked Example: A Series for \(1/(1-3x)\)
Worked Example: A Series for \(1/(1-3x)\)
Find a geometric-series representation and determine exactly where it converges.
Here the denominator already has the form \(1-r\), with \(r=3x\). Thus
The geometric condition becomes
At \(x=\tfrac13\), the \(n\)th term \((3x)^n\) equals \(1\). At \(x=-\tfrac13\), it equals \((-1)^n\). In both cases the terms fail to approach zero, so the series diverges at both endpoints.
For a check, choose \(x=\tfrac16\). The substituted ratio is \(3x=\tfrac12\), so the series sums to
The function gives the same value:
Consequently, the power series represents the function for \(-\tfrac13<x<\tfrac13\), and not at either endpoint.
Worked Example: A Shifted Geometric Ratio
Worked Example: A Shifted Geometric Ratio
Represent \(1/(2-2x)\) by a geometric series centered at \(x=\tfrac12\), and determine where the representation is valid.
Rewrite the denominator as
Thus the substituted ratio is \(g(x)=2x-1\), giving
The condition is
At \(x=0\), the terms are \((-1)^n\), and at \(x=1\), they are \(1\). Neither sequence of terms approaches zero, so both endpoints are excluded.
Check at \(x=\tfrac14\). The common ratio is \(2(\tfrac14)-1=-\tfrac12\), so
Direct evaluation confirms the result:
The convergence interval is therefore \((0,1)\). This example shows that the interval for \(x\) depends on the substituted expression: it need not be centered at zero.
A Common Mistake: Dropping the Absolute Value
A frequent error is to solve only one side of the geometric condition. For \(1/(1-3x)\), it is incorrect to say that convergence requires merely \(3x<1\), and therefore \(x<\tfrac13\). That inequality would include \(x=-1\), even though the terms there are \((-3)^n\), which do not approach zero.
The correct requirement includes both sides:
Another mistake is to assume that equality at an endpoint automatically gives convergence. At \(x=-\tfrac13\), the ratio is \(-1\), so the terms alternate between \(1\) and \(-1\); they do not approach zero. A series can converge only if its terms approach zero, so this endpoint must be rejected.
Why the Result Matters
Substitution turns one known series into representations for many functions, but the condition travels with the substitution. The resulting series describes the function only at inputs where the substituted ratio has magnitude less than one, except for any boundary cases that must be assessed separately. This distinction matters when using a series to calculate approximate values: an expansion that is algebraically correct as a pattern is not a valid function representation outside its convergence set.
For each substitution, keep the three checks distinct:
- Confirm that the denominator has been matched to \(1-g(x)\) correctly.
- Solve the full inequality \(|g(x)|<1\), rather than just one-sided inequality.
- Test boundary values using the terms of the series, not only the function’s formula.
Check Your Understanding
For each question, state the substituted ratio and justify the convergence conditions.
- Use substitution to write a geometric series for \(1/(1+x^4)\). Find its convergence interval and test both endpoints.
- For the series \(\sum_{n=0}^{\infty}(2-4x)^n\), solve the geometric convergence condition for \(x\).
- At \(x=-\tfrac13\), why does the series for \(1/(1-3x)\) diverge even though the function is defined there?
- Write \(1/(3+x)\) as a constant multiple of a geometric series, and find the condition on \(x\) under which the representation applies.
- Explain why \(|g(x)|<1\) must be checked after substituting \(g(x)\), rather than assuming the original condition \(|r|<1\) is already a condition on \(x\).
