Why Normalize the Boundary Variable?
A power series centered at \(c\) is written in powers of \(x-c\), not in powers of \(x\). If its radius of convergence is \(R>0\), its boundary points are \(c-R\) and \(c+R\). The center and radius affect the algebra at those points: substituting an endpoint produces factors of \(R^n\), and the left endpoint also produces a factor of \((-1)^n\).
The normalized variable \(u=(x-c)/R\) makes both effects visible. At the right endpoint \(u=1\); at the left endpoint \(u=-1\). This does not replace testing the endpoints independently. It gives a consistent way to compute the terms that must be tested.
Rescaling the Series
Consider the power series \(\sum_{n=0}^{\infty} a_n(x-c)^n\), with radius of convergence \(R\). Define \(b_n=a_nR^n\). Since \(x-c=Ru\), each term can be rewritten as \(a_nR^nu^n=b_nu^n\). Thus the same series, expressed using the normalized variable, is \(\sum_{n=0}^{\infty}b_nu^n\).
This identity explains precisely where the radius factor belongs: it is included in \(b_n\). At the right endpoint, the terms are \(b_n\). At the left endpoint, the terms are \((-1)^nb_n\). If a series is already written in normalized form, do not multiply by \(R^n\) a second time.
Why the Endpoint Formula Works
The formula follows directly from substitution, but spelling out the algebra helps prevent sign and scale errors. At the right endpoint, \(x=c+R\), so \(x-c=R\). At the left endpoint, \(x=c-R\), so \(x-c=-R\). Raising these differences to the \(n\)th power gives the two endpoint factors.
When \(x=c+R\), \((x-c)^n=R^n\). The \(n\)th term becomes \(a_nR^n=b_n\).
When \(x=c-R\), \((x-c)^n=(-R)^n=(-1)^nR^n\). The \(n\)th term becomes \((-1)^na_nR^n=(-1)^nb_n\).
Each endpoint gives an ordinary numerical series. Its convergence must be determined from those actual terms, including any signs already present in \(a_n\).
This is also why a shifted center causes no new convergence test of its own. The center determines the endpoint coordinates; after substitution, convergence depends on the numerical endpoint terms.
Worked Example: A Shifted Center with Harmonic Boundary Terms
Worked Example: A Shifted Center with Harmonic Boundary Terms
Analyze the endpoints of \(\sum_{n=1}^{\infty}\frac{(x-4)^n}{3^nn}\). The center is \(c=4\), the radius is \(R=3\), and the normalized variable is \(u=(x-4)/3\). Since \((x-4)^n/3^n=u^n\), the series is \(\sum_{n=1}^{\infty}u^n/n\).
At the right endpoint \(x=7\), \(u=(7-4)/3=1\). The term is \(1^n/n=1/n\), so the endpoint series is \(\sum 1/n\). This is the harmonic series and diverges.
At the left endpoint \(x=1\), \(u=(1-4)/3=-1\). The term is \((-1)^n/n\), so the endpoint series is \(\sum (-1)^n/n\). The positive terms \(1/n\) decrease to zero, so the alternating-series test gives convergence. The absolute-value series is \(\sum 1/n\), which diverges; therefore convergence at \(x=1\) is conditional.
Check directly in the original expression: at \(x=1\), \((x-4)^n/3^n=(-3)^n/3^n=(-1)^n\), leaving \((-1)^n/n\). At \(x=7\), \((x-4)^n/3^n=3^n/3^n=1\), leaving \(1/n\). The two substitutions match the normalized calculation.
Coefficient Signs and Endpoint Signs
A coefficient may already contain a factor such as \((-1)^n\). At the left endpoint, that sign combines with the \((-1)^n\) from \((x-c)^n\). The product is \((-1)^n(-1)^n=1\). At the right endpoint, the endpoint itself contributes no alternating factor, so the coefficient’s sign remains. The sign pattern must therefore be computed from the full term, not inferred from the coefficient alone.
Worked Example: A Coefficient Sign That Cancels at the Left Endpoint
Consider \(\sum_{n=1}^{\infty}\frac{(-1)^n(x+2)^n}{5^n\sqrt n}\). Here \(c=-2\), \(R=5\), and \(u=(x+2)/5\). The normalized form is \(\sum_{n=1}^{\infty}(-1)^nu^n/\sqrt n\).
At the right endpoint \(x=3\), \(u=(3+2)/5=1\). The terms are \((-1)^n/\sqrt n\). Since \(1/\sqrt n\) decreases to zero, the alternating-series test gives convergence. The absolute-value series \(\sum 1/\sqrt n\) diverges because it is a \(p\)-series with \(p=1/2\). Thus this endpoint is conditionally convergent.
At the left endpoint \(x=-7\), \(u=(-7+2)/5=-1\). The terms are \((-1)^n(-1)^n/\sqrt n=1/\sqrt n\). This is the divergent \(p\)-series with \(p=1/2\), so the endpoint series diverges.
Check in the original terms: at \(x=-7\), \((x+2)^n/5^n=(-5)^n/5^n=(-1)^n\); multiplying by the coefficient sign \((-1)^n\) gives \(1\), leaving \(1/\sqrt n\). At \(x=3\), the ratio is \(5^n/5^n=1\), leaving \((-1)^n/\sqrt n\). The endpoint signs are therefore as stated.
Additional Powers and the Term Test
Not every endpoint can be settled by recognizing a harmonic or alternating series. If coefficients grow with \(n\), the first question is whether the terms tend to zero. A necessary condition for \(\sum t_n\) to converge is \(\lim_{n\to\infty}t_n=0\). If the terms fail this condition, the series diverges. Passing the term test does not itself prove convergence; it only means another test may be needed.
Worked Example: Growing Coefficients at Both Boundaries
Consider \(\sum_{n=1}^{\infty}(-1)^n(n+1)\left(\frac{x-3}{2}\right)^n\). The center is \(c=3\), the radius is \(R=2\), and \(u=(x-3)/2\). Its normalized terms are \((-1)^n(n+1)u^n\).
At the right endpoint \(x=5\), \(u=(5-3)/2=1\). The \(n\)th term is \((-1)^n(n+1)\), whose absolute value is \(n+1\), not a quantity tending to zero. In particular, \(\lvert(-1)^n(n+1)\rvert=n+1\to\infty\). The series diverges by the term test.
At the left endpoint \(x=1\), \(u=(1-3)/2=-1\). The \(n\)th term is \((-1)^n(n+1)(-1)^n=n+1\), which also fails to tend to zero. This endpoint series diverges as well.
Check the signs in the original expression: at \(x=1\), \((x-3)/2=-1\), so its \(n\)th power contributes \((-1)^n\); the separate coefficient sign contributes another \((-1)^n\), and their product is \(1\). At \(x=5\), the power factor is \(1\), so the coefficient sign remains. Both computed terms agree with the direct substitutions.
When Both Endpoints Converge Absolutely
Endpoint signs do not guarantee that the two endpoints have different outcomes. If the coefficient magnitudes already produce a convergent absolute-value series, multiplication by \((-1)^n\) at the left endpoint does not change absolute convergence. This is useful to check explicitly rather than assuming the left endpoint must be alternating convergence.
Worked Example: Absolute Convergence at Both Endpoints
Consider \(\sum_{n=1}^{\infty}\frac{(-1)^n}{n^2}\left(\frac{x+1}{3}\right)^n\). The center is \(c=-1\), the radius is \(R=3\), and \(u=(x+1)/3\).
At the right endpoint \(x=2\), \(u=1\), so the terms are \((-1)^n/n^2\). Their absolute values sum to \(\sum 1/n^2\), which converges because it is a \(p\)-series with \(p=2>1\). Hence this endpoint converges absolutely.
At the left endpoint \(x=-4\), \(u=-1\), so the terms are \((-1)^n(-1)^n/n^2=1/n^2\). This is again a convergent \(p\)-series with \(p=2\), so this endpoint also converges absolutely.
Check the substitutions in the original factor: at \(x=2\), \((x+1)/3=1\), leaving the coefficient \((-1)^n/n^2\). At \(x=-4\), \((x+1)/3=-1\); its \(n\)th power cancels the coefficient sign, leaving \(1/n^2\). The absolute convergence verdict is the same at both endpoints.
A Common Scaling Error
When \(R\) is not \(1\), replacing the endpoint factor \((x-c)^n\) by only a sign loses the scale \(R^n\). For example, at \(x=3\) in a series centered at \(c=7\) with \(R=4\), the exact numerator is \((3-7)^n=(-4)^n\), not merely \((-1)^n\). If the term is \((x-7)^n/(4^nn)\), the powers of \(4\) cancel:
The following calculation is incorrect because it replaces \((3-7)^n\) by \((-1)^n\) but leaves the denominator \(4^n n\) unchanged:
The correction is to preserve the entire endpoint difference before simplifying: \((3-7)^n=(-4)^n\). Then the factor \(4^n\) cancels the denominator’s \(4^n\). In general, first compute \(x-c\), raise that whole value to the \(n\)th power, and only then simplify against any radius factors.
A Reliable Endpoint Routine
For a series centered at a shifted point, use the same order of operations each time. Keep the coefficient’s sign separate until after substituting the endpoint, and distinguish the actual endpoint series from a shortcut based only on its appearance.
Write \(c\) and \(R\), with \(R>0\), and identify the endpoints \(c-R\) and \(c+R\).
Set \(u=(x-c)/R\), or rewrite the original terms using \(b_n=a_nR^n\).
Use \(u=1\) at the right endpoint and \(u=-1\) at the left endpoint. Combine all signs and powers before naming the resulting series.
Check whether its terms tend to zero, then apply an appropriate convergence test. For absolute convergence, test the series of absolute values.
Why This View Is Useful
Normalization separates the geometry of the power series from the arithmetic of its coefficients. The center tells where the series is anchored; the radius sets the scale; the normalized boundary values reduce to \(1\) and \(-1\). This makes endpoint signs easier to track, especially when a coefficient already has an alternating factor or when powers of the radius appear in both numerator and denominator.
The central practical rule is simple: evaluate the full \(n\)th term at each endpoint. Do not decide that a series alternates merely because its coefficient contains \((-1)^n\); the left endpoint may cancel that sign. Do not discard \(R^n\) merely because the endpoint is a distance \(R\) from the center; that factor may cancel another \(R^n\) and change the boundary terms substantially.
Check Your Understanding
For each question, write the endpoint terms explicitly before deciding what convergence test applies.
- For \(\sum_{n=1}^{\infty}\frac{(x-5)^n}{2^nn}\), what are the numerical series at \(x=7\) and \(x=3\)?
- For \(\sum_{n=1}^{\infty}\frac{(-1)^n(x-1)^n}{4^n n^2}\), which sign pattern appears at each endpoint, and are the endpoints absolutely convergent?
- Why does the left-endpoint term of a normalized series \(\sum b_nu^n\) equal \((-1)^nb_n\)?
- At \(x=c-R\), simplify \(a_n(x-c)^n\) in terms of \(a_n\), \(R\), and \(n\).
- What conclusion follows if the terms of an endpoint series fail to tend to zero?