Turning a Power-Series Point Test into a Series Test

From a Power Series to a Numerical Series

A power series is a sum whose terms depend on a variable. To decide whether it converges at one particular value, first hold that value fixed. The variable then disappears, leaving an ordinary numerical series. Tests for numerical series can be applied to those resulting terms.

For example, consider a series centered at \(a\), written in the form \(\sum_{n=0}^{\infty} c_n(x-a)^n\). At a chosen value \(x=x_0\), its terms become \(c_n(x_0-a)^n\). The convergence question at \(x_0\) is therefore the convergence question for the numerical series made from those terms.

Definition: For a power series \(\sum_{n=0}^{\infty} c_n(x-a)^n\) and a fixed real number \(x_0\), the series at \(x_0\) is the numerical series \(\sum_{n=0}^{\infty} c_n(x_0-a)^n\). The power series converges at \(x_0\) precisely when this numerical series converges.

The center \(a\) matters: the quantity being raised to the \(n\)th power is \(x_0-a\), not necessarily \(x_0\). Once the substitution is made, we can use numerical-series tests on the new terms. When absolute convergence is at issue, apply the tests to the absolute values of those terms.

Ratio and Root Tests After Substitution

For a numerical series \(\sum u_n\), the ratio test examines the limit of \(\left|u_{n+1}/u_n\right|\), when that limit exists. If the limit is less than \(1\), the series converges absolutely; if it is greater than \(1\), the series diverges. A limit equal to \(1\), or a limit that does not exist, gives no conclusion from this test. The ratio must be defined for the terms being compared.

For a fixed point \(x_0\), put \(u_n=c_n(x_0-a)^n\). Provided the terms in the denominator are nonzero, their ratio is

\left|\frac{u_{n+1}}{u_n}\right| = \left|\frac{c_{n+1}}{c_n}\right|\,|x_0-a|.

This follows by writing out two consecutive terms and cancelling the common power:

\left|\frac{c_{n+1}(x_0-a)^{n+1}}{c_n(x_0-a)^n}\right| = \left|\frac{c_{n+1}}{c_n}\right| \left|\frac{(x_0-a)^{n+1}}{(x_0-a)^n}\right| = \left|\frac{c_{n+1}}{c_n}\right|\,|x_0-a|.

The factor \(|x_0-a|\) is part of the ratio. The ratio test is applied to the substituted numerical terms, not just to the coefficients \(c_n\).

The root test examines \(\limsup_{n\to\infty}\sqrt[n]{|u_n|}\). If this value is less than \(1\), the series converges absolutely; if it is greater than \(1\), the series diverges. If it equals \(1\), the test is inconclusive. For power-series terms, the expression under the test becomes

\sqrt[n]{|c_n(x_0-a)^n|} = \sqrt[n]{|c_n|}\,|x_0-a|.

As with the ratio test, the fixed value of the variable must be included. Both tests give a conclusion about the numerical series at the chosen point; an inconclusive result does not establish either convergence or divergence.

Worked Example: Use the Ratio Test at a Fixed Point

Test the power series \(\sum_{n=1}^{\infty}\frac{n^2(x+1)^n}{4^n}\) at \(x=0\).

First substitute \(x=0\). Since \(x+1=1\), the \(n\)th term becomes

u_n=\frac{n^2(1)^n}{4^n}=\frac{n^2}{4^n}.

Now compute the ratio of consecutive absolute terms:

\frac{|u_{n+1}|}{|u_n|} = \frac{(n+1)^2/4^{n+1}}{n^2/4^n} = \frac{(n+1)^2}{n^2}\cdot\frac{4^n}{4^{n+1}} = \frac{(n+1)^2}{4n^2}.

Taking the limit gives

\lim_{n\to\infty}\frac{(n+1)^2}{4n^2} = \frac14\lim_{n\to\infty}\left(1+\frac1n\right)^2 = \frac14.

Because \(\frac14<1\), the ratio test shows that the numerical series converges absolutely. The terms at \(x=0\) are exactly \(n^2/4^n\), so this establishes convergence of the given power series at \(x=0\). The ratio limit is less than \(1\), as required for this conclusion.

Worked Example: Use the Root Test to Show Divergence

Test \(\sum_{n=1}^{\infty}\frac{2^n}{n+1}(x-1)^n\) at \(x=\frac74\).

Substitute the fixed value. The centered factor is \(\frac74-1=\frac34\), so each term simplifies to

u_n = \frac{2^n}{n+1}\left(\frac34\right)^n = \frac{(3/2)^n}{n+1}.

Take the \(n\)th root of its absolute value:

\sqrt[n]{|u_n|} = \frac{3/2}{(n+1)^{1/n}}.

Since \((n+1)^{1/n}\to 1\), it follows that

\lim_{n\to\infty}\sqrt[n]{|u_n|} = \frac32.

The limit is greater than \(1\), so the root test says that the numerical series diverges. In particular, its terms cannot tend to zero: if they did, their \(n\)th roots could not have a limit greater than \(1\). Thus the power series diverges at \(x=\frac74\). The test applies because the root limit exists and equals \(\frac32>1\).

Comparison at a Fixed Point

Not every point test requires a ratio or root calculation. The direct comparison test is useful when the absolute values of the substituted terms can be bounded by the terms of a known convergent series. In particular, if \(0\le |u_n|\le v_n\) for all sufficiently large \(n\), and \(\sum v_n\) converges, then \(\sum |u_n|\) converges. Consequently, \(\sum u_n\) converges absolutely.

The comparison must be made after substitution. A bound that is true for the coefficients alone may not bound the actual terms, because the factor \((x_0-a)^n\) also contributes.

Worked Example: Compare the Substituted Terms with a Geometric Series

Test \(\sum_{n=1}^{\infty}\frac{(x-2)^n}{n^2}\) at \(x=\frac52\).

Substitution gives \(x-2=\frac12\), so the terms are

u_n=\frac{(1/2)^n}{n^2}.

For every integer \(n\ge 1\), \(n^2\ge 1\), and therefore

0\le |u_n| = \frac{(1/2)^n}{n^2} \le \left(\frac12\right)^n.

The comparison series is geometric, with first term \(\frac12\) and ratio \(\frac12\). Its sum is finite:

\sum_{n=1}^{\infty}\left(\frac12\right)^n = \frac{1/2}{1-1/2} = 1.

By direct comparison, \(\sum |u_n|\) converges. Thus the substituted numerical series, and hence the power series at \(x=\frac52\), converges absolutely. The inequality holds term by term and the comparison series converges, so the conditions for direct comparison are satisfied.

A Reliable Point-Test Procedure

At a chosen point, keep the substitution and the series test as separate steps. This makes it easier to see whether a factor or sign has been lost and makes the conditions of the test explicit.

Fix the point.
Write down the value \(x_0\) and the center \(a\) of the power series.

Substitute into the full term.
Replace every occurrence of \(x\), including in \((x-a)^n\), to obtain \(u_n=c_n(x_0-a)^n\).

Choose a numerical-series test.
Use the ratio or root test when its limit is manageable, or comparison when a useful term-by-term bound is available.

Check the test’s condition.
Compare the computed limit with \(1\), or verify both the inequality and the convergence of the comparison series.

State the pointwise conclusion.
Say whether the power series converges or diverges at the particular value tested. An inconclusive test does not justify either claim.

If \(x_0=a\), the factor \((x_0-a)^n\) is zero for every \(n\ge1\), so the series at that point reduces to its constant term \(c_0\). Evaluate this case directly rather than forming a ratio that may have zero in its denominator.

A Common Substitution Error

A frequent mistake is to replace \(x\) by the chosen value but forget that the power series is centered at \(a\). For example, consider \(\sum_{n=1}^{\infty}\frac{(x-2)^n}{n3^n}\) at \(x=6\). The centered factor is \(6-2=4\), so the correct terms are

u_n=\frac{4^n}{n3^n}=\frac{(4/3)^n}{n}.

The ratio of consecutive absolute terms is

\frac{|u_{n+1}|}{|u_n|} = \frac{(4/3)^{n+1}}{n+1}\cdot\frac{n}{(4/3)^n} = \frac43\cdot\frac{n}{n+1} \longrightarrow \frac43.

Since \(\frac43>1\), the ratio test shows divergence. A mistaken substitution that drops \((x-2)^n\) would instead produce \(1/(n3^n)\), whose ratio limit is \(1/3\) and whose series converges. That is a different numerical series and gives the wrong conclusion for the original point. Always substitute the entire centered factor before applying a test.

What a Point Test Does—and Does Not—Establish

A successful ratio, root, or comparison test answers a question at the fixed value used in the substitution. It does not, by itself, classify every other value of \(x\). For instance, a ratio limit of \(\frac14\) in a point test establishes convergence for that substituted series; it is not a claim that the same limit holds at every point.

There is also a distinction between a decisive and an inconclusive result. A ratio or root limit below \(1\) gives absolute convergence, and a limit above \(1\) gives divergence. A limit equal to \(1\) does not settle the question. In that situation, the terms at the point are still a numerical series, but another suitable test is needed. The conclusion should match exactly what the chosen test proves.

Key takeaway: To test a power series at a fixed point, substitute that value into every term first. Then apply a numerical-series test to the resulting terms, checking the test’s hypotheses and stating only the pointwise conclusion it supports.

Check Your Understanding

For each question, begin by writing the numerical terms obtained at the specified point.

What numerical series results from \(\sum_{n=1}^{\infty} n(x-3)^n\) at \(x=4\)?
For \(\sum_{n=1}^{\infty}\frac{(x+1)^n}{5^n}\) at \(x=1\), what is the ratio of consecutive absolute terms and what does the ratio test conclude?
What does a root-test limit of \(1\) tell you about convergence?
At \(x=\frac12\), how can you compare the terms of \(\sum_{n=1}^{\infty}\frac{x^n}{n}\) with a geometric series?
Why should the series at its center be evaluated directly rather than tested by a ratio that divides by a zero term?