From a Geometric Series to a Reciprocal
A linear denominator can often be rewritten so that its reciprocal is a constant multiple of the geometric-series function. The key is to express the denominator in terms of the distance from a chosen center. That makes the center and the convergence condition visible, rather than leaving them hidden in a substitution.
The geometric series formula is
The condition \(|r|<1\) is essential: it is exactly where the infinite sum equals the stated reciprocal. For an affine expression in \(x\), the ratio \(r\) will depend on \(x\), so the condition becomes an inequality describing where the power series converges.
Normalize the Denominator at the Center
Consider \(f(x)=1/(a-bx)\), where \(a,b\) are real constants and \(b\ne 0\). Choose a real center \(x_0\) at which the denominator is nonzero, so \(a-bx_0\ne 0\). Write \(x=x_0+(x-x_0)\), then separate the denominator into its value at the center and its change from that value:
Factoring out \(a-bx_0\) puts the reciprocal into geometric-series form:
Here the geometric ratio is \(\dfrac{b}{a-bx_0}(x-x_0)\). Applying the geometric-series formula gives the following result.
The first few terms make the coefficient pattern explicit:
If the center is \(x_0=0\), this reduces to the familiar expansion \(\frac{1}{a}\sum_{n=0}^{\infty}(b x/a)^n\), valid when \(|bx/a|<1\), provided \(a\ne 0\). Choosing a different center changes both the powers \((x-x_0)^n\) and the coefficients.
Why the Formula Works
The derivation uses only the geometric-series identity. To verify that the coefficients and condition follow, define \(D=a-bx_0\), which is nonzero by assumption, and set \(r=\frac{b}{D}(x-x_0)\). Then \(a-bx=D(1-r)\). If \(|r|<1\), the geometric series sums to \(1/(1-r)\), and the resulting expression is exactly the reciprocal of the original denominator.
Set \(D=a-bx_0\). The assumption \(D\ne 0\) ensures the reciprocal is defined at \(x_0\).
Rewrite \(a-bx=D\left(1-\frac{b}{D}(x-x_0)\right)\), so \(r=\frac{b}{D}(x-x_0)\).
The expansion is valid precisely when \(|r|<1\), or \(|x-x_0|<|D|/|b|\).
The \(n\)th term is \(\frac{1}{D}r^n=\frac{b^n}{D^{n+1}}(x-x_0)^n\).
This also justifies the radius: solving \(\left|\frac{b(x-x_0)}{D}\right|<1\) gives \(|x-x_0|<|D|/|b|\). For real \(x\), this is the open interval centered at \(x_0\) with radius \(R=|D|/|b|\).
Worked Examples
Worked Example: Expand \(1/(6-3x)\) about zero
First factor out the constant term in a way that leaves \(1\) in the denominator’s first factor:
The ratio is \(r=x/2\), so the expansion converges when \(|x/2|<1\), or \(|x|<2\). Thus
To check it, sum the geometric series within its convergence interval:
The endpoints are not included. At \(x=2\), the original denominator is zero. At \(x=-2\), the ratio is \(-1\), so the terms \(\frac{1}{6}(-1)^n\) do not approach zero; the series diverges even though the function itself is defined there.
Worked Example: Expand \(1/(3+2x)\) about \(x_0=1\)
At the chosen center, the denominator is \(3+2(1)=5\). Writing \(x=1+(x-1)\) gives
To match \(1-r\), take \(r=-\frac{2}{5}(x-1)\). The convergence condition is \(\left|\frac{2}{5}(x-1)\right|<1\), or \(|x-1|<5/2\). Therefore
For a check, sum the geometric series on this interval:
The center is \(1\), and the radius is \(5/2\). The corresponding interval is \((-3/2,7/2)\). The left endpoint makes the denominator zero. At the right endpoint, the ratio is \(-1\), so the series terms do not approach zero, and the series diverges there as well.
Worked Example: Expand \(1/(5-2x)\) about \(x_0=2\)
The denominator at the center is \(5-2(2)=1\). Using \(x=2+(x-2)\),
The geometric ratio is \(r=2(x-2)\). Hence \(|r|<1\) means \(|x-2|<1/2\), and the expansion is
Check the result by summing the series:
As a numerical check inside the interval, take \(x=9/4\). Then \(x-2=1/4\), so the ratio is \(1/2\), and the series sums to \(\frac{1}{1-1/2}=2\). Direct substitution gives \(\frac{1}{5-2(9/4)}=\frac{1}{1/2}=2\), in agreement.
The Center Determines the Convergence Inequality
The quantity controlling convergence is not just the coefficient of \(x\). It is the ratio formed by comparing the change in the denominator to its value at the center. For \(1/(a-bx)\), the denominator changes by \(-b(x-x_0)\) from its center value \(a-bx_0\). The geometric condition compares the magnitudes of these two quantities.
The zero of the denominator is \(x=a/b\). Since \(a-bx_0=b(a/b-x_0)\), the radius can also be written \(R=|x_0-a/b|\): it is the distance from the center to the pole. This identifies the radius but does not make either endpoint part of the convergence interval. At each endpoint the geometric ratio has absolute value \(1\), so the terms fail to approach zero.
If \(b=0\), the function is constant \(1/a\) when \(a\ne 0\); the geometric expansion is unnecessary. If \(a-bx_0=0\), the proposed center is a point where the reciprocal is undefined, so this expansion about that center does not apply.
A Common Scaling Mistake
A frequent error is to substitute into the geometric series but forget the factor pulled out of the denominator. For example, writing
drops the factor \(1/6\). At \(x=0\), this incorrect expression gives \(1\), while the original function gives \(1/6\). The correction is to factor the denominator completely:
Another useful check is to sum the proposed geometric series and verify that it simplifies back to the original reciprocal. That check catches a missing scale factor as well as an incorrect sign in the ratio.
Why This Representation Is Useful
Shifting and scaling the geometric series turns a simple reciprocal into a power series with a chosen center. The center determines the powers, the value of the denominator there determines the coefficient scale, and the slope determines how quickly the geometric ratio changes with \(x\). Keeping these roles separate makes the calculation easier to check and makes the convergence condition explicit.
When a power series is needed near a particular point, it is therefore not enough to substitute an affine expression mechanically. First evaluate the denominator at the chosen center, then divide the change from that value by the center value. The resulting ratio gives both the expansion and its domain of convergence.
Check Your Understanding
Use the geometric-series condition carefully, and check endpoints separately where relevant.
- For \(1/(4-2x)\), find the expansion centered at zero and its radius of convergence.
- For \(1/(2+3x)\), expand about \(x_0=1\). State the center and convergence interval.
- Explain why \(1/(a-bx)\) cannot be expanded by this method about \(x_0=a/b\) when \(b\ne 0\).
- For \(1/(1+x)\) centered at zero, determine whether the series converges at \(x=1\) and at \(x=-1\). Explain each result.
- In an affine-denominator expansion, what information determines the constant factor, and what information determines the geometric ratio?
