The Term Test at Power-Series Boundaries

Why Check the Terms at a Boundary?

A power series has a definite interval of convergence around its center, but a boundary point requires a point-by-point test. Once a boundary point is substituted, the power series becomes an ordinary numerical series. Before using a more specialized convergence test, it is useful to check a basic necessity: the terms of any convergent series must tend to zero.

This condition is especially useful at a boundary because the powers in the general term may no longer shrink in magnitude. If the substituted terms fail to approach zero, the endpoint series diverges immediately. If they do approach zero, the test has not established convergence; it has only passed a necessary condition.

Definition: For a numerical series \(\sum_{n=1}^{\infty} u_n\), the nth-term divergence test says that if \(\lim_{n\to\infty}u_n\) does not exist or is not zero, then the series diverges. If \(\lim_{n\to\infty}u_n=0\), the test gives no conclusion about whether the series converges.

Why the Test Works

The conclusion follows directly from the definition of convergence of a series. The partial sums are \(S_N=u_1+u_2+\cdots+u_N\). If the series converges, then the sequence of partial sums has a finite limit. Each term after the first is the difference between two consecutive partial sums:

u_N=S_N-S_{N-1}.

If \(S_N\) tends to a limit \(S\), then \(S_{N-1}\) tends to that same limit. Subtracting the two limits gives zero, so \(u_N\) must tend to zero.

\lim_{N\to\infty}u_N = \lim_{N\to\infty}(S_N-S_{N-1}) = S-S = 0.

This proves the necessary condition. It also explains the test’s limitation: the argument shows that convergence requires terms to approach zero, but it does not show that this condition is sufficient.

Applying the Test to Power-Series Endpoints

Consider a real power series centered at \(c\), written as \(\sum_{n=0}^{\infty}a_n(x-c)^n\), with finite positive radius of convergence \(R\). Its real boundary points are \(x=c-R\) and \(x=c+R\). Substituting either endpoint gives a numerical series, whose nth term is the corresponding substituted expression \(a_n(x-c)^n\).

The term test is applied separately at each endpoint. At the right endpoint, the term is \(a_nR^n\); at the left endpoint, it is \(a_n(-R)^n\). The left endpoint can introduce alternating signs, but signs do not remove the necessary condition: the terms themselves must still tend to zero. In particular, if their absolute values tend to a positive number or grow without bound, the terms cannot tend to zero.

This test can establish endpoint divergence, but it cannot establish endpoint convergence. A zero limit is only a reason to continue with an appropriate test, not a final verdict.

Worked Examples

Worked Example: Nonzero Limiting Terms at Both Endpoints

Consider the power series \(\sum_{n=1}^{\infty}\frac{n}{n+1}x^n\), whose radius of convergence is \(1\). Indeed, its coefficient \(\frac{n}{n+1}\) tends to \(1\), so its nth root tends to \(1\), giving radius \(1\). Test its two boundary points.

At \(x=1\), substitution gives the series \(\sum_{n=1}^{\infty}\frac{n}{n+1}\). Its nth term has limit

\lim_{n\to\infty}\frac{n}{n+1} = \lim_{n\to\infty}\frac{1}{1+1/n} = 1.

The term limit is not zero, so the series diverges by the nth-term divergence test.

At \(x=-1\), substitution gives \(\sum_{n=1}^{\infty}(-1)^n\frac{n}{n+1}\). The terms have alternating signs, but their absolute values satisfy

\left|(-1)^n\frac{n}{n+1}\right| = \frac{n}{n+1} \longrightarrow 1.

Thus the terms do not tend to zero. The series diverges at \(x=-1\) as well. Substituting each endpoint back into the general term confirms that the two endpoint terms are respectively \(\frac{n}{n+1}\) and \((-1)^n\frac{n}{n+1}\), with neither tending to zero.

Worked Example: A Shifted Center and Nonzero Endpoint Limits

Consider \(\sum_{n=1}^{\infty}\frac{2n+3}{5n+1}\left(\frac{x-4}{2}\right)^n\). The coefficient \(\frac{2n+3}{5n+1}\) tends to \(\frac25\), so its nth root tends to \(1\); the series therefore has radius \(2\). Its boundary points are \(x=4-2=2\) and \(x=4+2=6\).

At \(x=2\), the scaled variable is \(\frac{2-4}{2}=-1\). The substituted nth term is

u_n = \frac{2n+3}{5n+1}(-1)^n.

Its absolute value has limit

|u_n| = \frac{2n+3}{5n+1} = \frac{2+3/n}{5+1/n} \longrightarrow \frac25.

Since the magnitude approaches \(\frac25\), not zero, the terms do not tend to zero and the endpoint series diverges.

At \(x=6\), the scaled variable is \(\frac{6-4}{2}=1\), so the nth term is \(\frac{2n+3}{5n+1}\). Directly,

\lim_{n\to\infty}\frac{2n+3}{5n+1} = \lim_{n\to\infty}\frac{2+3/n}{5+1/n} = \frac25.

This endpoint series also diverges by the term test. Checking the substitutions gives the factors \((-1)^n\) at \(x=2\) and \(1^n\) at \(x=6\); in both cases, the coefficient magnitude approaches \(\frac25\), so neither set of terms tends to zero.

Worked Example: Terms Tend to Zero, So the Test Is Inconclusive

Consider \(\sum_{n=1}^{\infty}\frac{x^n}{n}\), which has radius \(1\). At \(x=1\), the nth term is \(\frac1n\), and

\lim_{n\to\infty}\frac1n=0.

At \(x=-1\), the nth term is \(\frac{(-1)^n}{n}\), and

\left|\frac{(-1)^n}{n}\right| = \frac1n \longrightarrow 0.

At both endpoints the terms satisfy the necessary zero-limit condition. The nth-term test therefore does not decide either endpoint series. In particular, these calculations do not prove convergence; a further test would be needed to reach a conclusion. Substitution checks the results directly: the boundary factors are \(1^n\) and \((-1)^n\), and the coefficient \(1/n\) makes the term magnitudes tend to zero in both cases.

A Common Mistake: Alternation Does Not Guarantee a Zero Limit

A frequent error is to see alternating signs and assume that the terms tend to zero. For example, at \(x=-1\) in the first worked example, one might incorrectly write

(-1)^n\frac{n}{n+1}\longrightarrow 0 \qquad\text{because the signs alternate.}

This reasoning is wrong. Alternation describes the signs, not the size of the terms. The correct check uses the magnitude:

\left|(-1)^n\frac{n}{n+1}\right| = \frac{n}{n+1} \longrightarrow 1.

Since the magnitude does not tend to zero, the terms cannot tend to zero, and the series diverges. Always evaluate the limit of the actual term, or its magnitude when alternating signs make the behavior less visible.

What the Term Test Can—and Cannot—Tell You

At a boundary point, the test has a deliberately narrow role. If the substituted term has a nonzero limit, the series diverges. If the term does not have a limit, the series also diverges. For alternating terms, a magnitude tending to a positive number or to infinity is enough to show that the terms do not approach zero.

When the terms do tend to zero, the result is inconclusive. The test has not demonstrated convergence, and it has not demonstrated divergence. The endpoint must then be examined by another suitable method. Keeping this distinction clear prevents a necessary condition from being mistaken for a complete convergence test.

Takeaway: At either power-series boundary point, first substitute the endpoint into the general term. If the resulting terms fail to tend to zero, the endpoint series diverges. If they tend to zero, the nth-term test alone cannot decide the endpoint.

Check Your Understanding

For each question, distinguish carefully between a conclusion the term test proves and one it does not prove.

State the necessary condition on \(u_n\) if \(\sum_{n=1}^{\infty}u_n\) converges.
At a power-series endpoint, the substituted terms tend to \(3\). What can you conclude?
If an endpoint term has the form \((-1)^n b_n\), where \(b_n\to\frac12\), does alternation make the terms tend to zero? Explain.
If the terms at one endpoint tend to zero, what does the nth-term divergence test establish about convergence there?
Why must the left and right endpoints be tested using their own substituted terms?