Direct Comparison for Endpoint Series

Why Direct Comparison Helps at an Endpoint

Finding a power series' radius of convergence identifies where its endpoint tests are needed, but it does not decide convergence at those endpoints. After substitution, the question is whether the resulting numerical series converges. Direct comparison answers that question by relating its terms to a familiar series whose behavior is already known.

The method is useful when the substituted terms have an evident upper or lower bound involving a harmonic series, a \(p\)-series, or a geometric series. The direction of the bound matters: an upper bound by a convergent series can establish convergence, while a lower bound by a divergent series can establish divergence.

Definition: Suppose \(a_n\ge 0\) and \(b_n\ge 0\) for all sufficiently large \(n\). If \(a_n\le b_n\) eventually and \(\sum b_n\) converges, direct comparison implies that \(\sum a_n\) converges. If \(b_n\le a_n\) eventually and \(\sum b_n\) diverges, direct comparison implies that \(\sum a_n\) diverges. Changing finitely many terms does not affect convergence or divergence.

Benchmark Series and the Direction of Comparison

The standard benchmarks used here are the harmonic series, the \(p\)-series, and the geometric series. The \(p\)-series \(\sum_{n=1}^{\infty}1/n^p\) converges when \(p>1\) and diverges when \(p\le 1\). The harmonic series is its case \(p=1\). A geometric series \(\sum_{n=0}^{\infty}r^n\) converges when \(|r|<1\) and diverges when \(|r|\ge 1\).

For endpoint series, the terms being compared must be nonnegative. If substitution produces signs that change, direct comparison is generally applied to the absolute values. Convergence of the series of absolute values proves absolute convergence, and hence convergence of the original series. But divergence of the absolute-value series alone does not prove that a signed series diverges; an alternating series, for instance, can converge without converging absolutely.

Substitute an endpoint.
Write the resulting numerical series explicitly, including its signs and starting index.

Identify nonnegative terms.
If the terms change sign, first decide whether the aim is to test absolute convergence using their magnitudes.

Choose a benchmark and the needed direction.
For convergence, seek an upper bound by a convergent series. For divergence, seek a lower bound by a divergent series.

Verify the bound on a tail.
Check the inequality for all sufficiently large indices; a finite number of exceptions does not change the verdict.

Why the Comparison Rules Work

The comparison rules follow from the behavior of partial sums. For a nonnegative series \(\sum a_n\), its partial sums \(S_N=\sum_{n=1}^{N}a_n\) are nondecreasing. A nondecreasing sequence of partial sums converges if it is bounded above.

Suppose \(0\le a_n\le b_n\) for all \(n\ge N_0\), and \(\sum b_n\) converges. For \(N\ge N_0\),

\sum_{n=1}^{N}a_n \le \sum_{n=1}^{N_0-1}a_n+\sum_{n=N_0}^{N}b_n \le \sum_{n=1}^{N_0-1}a_n+\sum_{n=N_0}^{\infty}b_n.

The right-hand side is a finite constant independent of \(N\). Since \(a_n\ge 0\) for \(n\ge N_0\), the partial sums are nondecreasing from index \(N_0-1\) onward, and the bound shows they are bounded above, so they converge. (Finitely many initial terms change the partial sums only by a fixed constant.) This proves the convergence rule.

For the divergence rule, suppose \(0\le b_n\le a_n\) eventually and \(\sum b_n\) diverges. The partial sums of \(\sum a_n\) dominate those of \(\sum b_n\), apart from at most a fixed finite difference from the initial terms. Since the partial sums of \(\sum b_n\) grow without bound, so do the partial sums of \(\sum a_n\). Therefore \(\sum a_n\) diverges. These arguments also show why the inequalities cannot be reversed.

Worked Endpoint Examples

Worked Example: Convergence at Both Endpoints by a \(p\)-Series

Consider the power series

\sum_{n=1}^{\infty}\frac{(x-1)^n}{3^n n(n+1)}.

Its radius is \(3\): the \(n\)th root of the absolute value of its coefficient is

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

The endpoints are \(x=1-3=-2\) and \(x=1+3=4\). At \(x=4\), substitution gives

\frac{(4-1)^n}{3^n n(n+1)} = \frac{3^n}{3^n n(n+1)} = \frac{1}{n(n+1)}.

Since \(n(n+1)\ge n^2\) for \(n\ge1\),

0\le\frac{1}{n(n+1)}\le\frac{1}{n^2}.

The comparison series \(\sum 1/n^2\) converges because \(p=2>1\). Therefore the endpoint series at \(x=4\) converges. At \(x=-2\), substitution gives

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

Its absolute-value terms are again \(1/[n(n+1)]\), bounded above by \(1/n^2\). The endpoint series at \(x=-2\) therefore converges absolutely. The endpoint verdicts are consistent with the substitutions: the right endpoint gives positive terms, and the left endpoint gives alternating terms whose absolute values pass the same comparison.

Worked Example: Harmonic Lower Bound Proves Divergence

Consider

\sum_{n=1}^{\infty} \left(\frac{x}{2}\right)^n \frac{n+1}{n(n+2)}.

The coefficient root has limit \(1/2\), since \(\bigl((n+1)/(n(n+2))\bigr)^{1/n}\to1\). The radius is therefore \(2\), so the endpoints are \(x=2\) and \(x=-2\). At \(x=2\), the terms are

\left(\frac{2}{2}\right)^n\frac{n+1}{n(n+2)} = \frac{n+1}{n(n+2)}.

For every \(n\ge1\), the inequality \((n+1)/(n+2)\ge1/2\) follows from \(2n+2\ge n+2\). Dividing by \(n>0\) gives

\frac{n+1}{n(n+2)}\ge\frac{1}{2n}.

The lower comparison series \(\sum 1/(2n)\) diverges, since it is one-half the harmonic series. Thus the endpoint series at \(x=2\) diverges. Its terms nevertheless tend to zero: also \((n+1)/(n+2)<1\), so

0\le\frac{n+1}{n(n+2)}<\frac{1}{n}\longrightarrow0.

The term test is inconclusive here; it is the lower comparison that proves divergence.

At \(x=-2\), the endpoint series is

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

The series of absolute values is the divergent positive series just considered, so this endpoint is not absolutely convergent. The terms \(a_n=(n+1)/(n(n+2))\) decrease to zero: indeed,

a_n=\frac12\left(\frac1n+\frac1{n+2}\right),

and both summands decrease to zero as \(n\) increases. The alternating-series test therefore gives convergence at \(x=-2\), but not absolute convergence. Direct comparison establishes the failure of absolute convergence; it does not by itself establish convergence of this signed series.

Worked Example: Comparison with a Geometric Series at an Endpoint

Consider

\sum_{n=1}^{\infty} \left(\frac{x}{5}\right)^n\left(1+\frac{1}{n}\right).

The coefficient root tends to \(1/5\), so the radius is \(5\). At the right endpoint \(x=5\), substitution gives the positive terms

\left(\frac{5}{5}\right)^n\left(1+\frac1n\right) = 1+\frac1n \ge 1.

The comparison series \(\sum_{n=1}^{\infty}1\) is geometric with ratio \(1\), and it diverges. The lower comparison proves that the endpoint series at \(x=5\) diverges. At the left endpoint \(x=-5\), the terms are

(-1)^n\left(1+\frac1n\right).

Their magnitudes are at least \(1\), so they do not tend to zero. The necessary term test therefore proves divergence at \(x=-5\) as well. This example shows that a geometric comparison with ratio \(1\) is a useful divergence benchmark, even though it is not a convergent geometric series.

A Common Comparison Error

A common mistake is to use an inequality in the wrong direction. For example, \(1/n^2\le1/n\), and \(\sum 1/n\) diverges. It would be incorrect to conclude from these facts that \(\sum 1/n^2\) diverges. An upper bound by a divergent series gives no convergence or divergence conclusion.

The correct comparison is with the convergent \(p\)-series itself: the terms \(1/n^2\) equal those of \(\sum 1/n^2\), which converges because \(p=2>1\). More generally, keep the two valid implications distinct: an upper bound by a convergent series proves convergence; a lower bound by a divergent series proves divergence.

Using Comparison in Endpoint Analysis

At a power-series boundary, first substitute the actual endpoint into the full term. The center and radius determine which values to substitute, but the endpoint terms determine which comparison is appropriate. In particular, do not compare the coefficient sequence alone if the endpoint substitution changes its factors or signs.

Direct comparison is especially effective when elementary inequalities reveal a clear benchmark: a bound by \(1/n^p\) for \(p>1\) can prove convergence, while a lower bound by \(1/n^p\) for \(p\le1\) can prove divergence. A lower bound by \(r^n\) with \(r\ge1\) also proves divergence. For series with changing signs, apply these arguments to absolute values when deciding absolute convergence, and use a suitable convergence test for the signed series if absolute convergence fails.

Check Your Understanding

For each question, identify the direction of comparison needed before stating a conclusion.

If \(0\le a_n\le 1/n^3\) for all sufficiently large \(n\), what does direct comparison imply about \(\sum a_n\)?
If \(a_n\ge 1/(4n)\) for all sufficiently large \(n\), what does direct comparison imply about \(\sum a_n\)?
At an endpoint, the terms are \((-1)^n a_n\), where \(a_n\ge0\), and \(\sum a_n\) diverges. Does this alone prove that \(\sum (-1)^n a_n\) diverges?
Why does \(0\le a_n\le 1/n\) fail to prove convergence, even though \(a_n\) is bounded above by a familiar series term?
What must be checked before applying a direct comparison to a power-series endpoint?