Introduction
A comparison between terms can settle whether a series diverges without requiring an exact formula for its partial sums. The key idea is to use a series already known to diverge as a lower bound: if the terms of another series are eventually at least as large, its partial sums cannot remain bounded.
The direction of the inequality matters. A larger nonnegative term sequence can inherit divergence from a smaller divergent one. The comparison only needs to hold eventually, so finitely many initial terms do not affect the conclusion. They do, however, need to be handled correctly when writing the partial-sum argument.
Eventual Comparison and Divergence
A statement holds eventually if it holds for every index from some fixed index onward. Thus, saying \(0\le b_n\le a_n\) eventually means that there is an integer \(N\) such that the inequalities hold for every \(n\ge N\).
For a series \(\sum_{n=1}^{\infty}c_n\), its \(m\)-th partial sum is \(S_m=\sum_{n=1}^{m}c_n\). When the terms are nonnegative eventually, divergence means that these partial sums are unbounded above; equivalently, they tend to \(+\infty\). Any finite initial segment contributes only a fixed constant, so it cannot prevent this eventual growth.
In the rule below, the comparison conditions are required only from some index onward. The series being compared may have finitely many negative terms before that index.
The comparison series must diverge in the direction of unbounded partial sums. In typical applications, \(b_n\ge0\) eventually, and its divergent tail has partial sums tending to \(+\infty\). The theorem does not say that a larger series converges whenever a smaller one converges; that is a different kind of comparison conclusion.
Why the Rule Works
From the index \(N\) onward, each term \(a_n\) contributes at least as much as the corresponding term \(b_n\). Summing these inequalities gives a lower bound for the tail of the \(a_n\)-series. The full partial sum also includes the fixed initial segment of \(a_n\), which may be negative. Keeping this finite contribution visible is essential.
For every \(n\ge N\), the hypotheses give \(a_n\ge b_n\ge0\).
For \(m\ge N\), adding the inequalities gives \(\sum_{n=N}^{m}a_n\ge\sum_{n=N}^{m}b_n\).
The full partial sum is \(S_m^a=\sum_{n=1}^{N-1}a_n+\sum_{n=N}^{m}a_n\), so \(S_m^a\ge\sum_{n=1}^{N-1}a_n+\sum_{n=N}^{m}b_n\).
The first term on the right is a fixed finite constant. Since the divergent nonnegative tail of \(\sum b_n\) has unbounded partial sums, the right-hand side tends to \(+\infty\). Thus \(S_m^a\to+\infty\), and \(\sum a_n\) diverges.
This proves the rule. Notice the role of each condition: the inequality \(a_n\ge b_n\) transfers the lower bound, \(b_n\ge0\) makes the comparison terms nonnegative, and divergence of \(\sum b_n\) supplies an unbounded tail.
Worked Applications
Worked Example: A Rational Term Compared with the Harmonic Series
Determine whether \(\sum_{n=1}^{\infty}\frac{n}{n^2+1}\) diverges. Compare its terms with \(b_n=\frac{1}{2n}\).
For \(n\ge1\), both denominators in the following difference are positive:
The last inequality holds because \(n^2-1\ge0\) for every integer \(n\ge1\). Therefore \(0\le\frac{1}{2n}\le\frac{n}{n^2+1}\) for every \(n\ge1\).
The comparison series is a constant multiple of the harmonic series:
The harmonic series diverges, and multiplying its terms by the positive constant \(\frac12\) does not make its partial sums bounded. Direct comparison therefore proves that \(\sum_{n=1}^{\infty}\frac{n}{n^2+1}\) diverges. The term inequality above verifies the required lower bound for every index, and the comparison series is divergent.
Worked Example: A Comparison That Begins After the First Term
Determine whether \(\sum_{n=1}^{\infty}\frac{3n+1}{n^2+4n}\) diverges. The denominator is positive for \(n\ge1\); use \(b_n=\frac1n\) and check the comparison from \(n=2\).
For \(n\ge2\), the difference is
The denominator is positive, and \(2n-3\ge1>0\) for \(n\ge2\). Also \(\frac1n>0\), so \(0\le\frac1n\le\frac{3n+1}{n^2+4n}\) for every \(n\ge2\). The comparison series \(\sum_{n=2}^{\infty}\frac1n\) diverges: removing its first term from the harmonic series changes its partial sums by only a fixed constant. The direct comparison rule applies with \(N=2\), proving that the given series diverges. The first term does not need to satisfy the comparison; the inequality has been verified for every index in the required tail.
Worked Example: A Negative Initial Term Does Not Change the Conclusion
Define \(a_1=-10\) and \(a_n=\frac1n\) for \(n\ge2\). Determine whether \(\sum_{n=1}^{\infty}a_n\) diverges, using \(b_n=\frac1n\).
For every \(n\ge2\), \(0\le b_n=a_n\). The comparison series \(\sum_{n=1}^{\infty}\frac1n\) diverges, so direct comparison applies from \(N=2\). To see explicitly how the initial term enters, for \(m\ge2\) the partial sum is
where \(H_m=\sum_{n=1}^{m}\frac1n\). Since \(H_m\to+\infty\), \(S_m=H_m-11\to+\infty\), so the series diverges. As a check on the finite initial contribution, \(S_2=-10+\frac12=-\frac{19}{2}\), whereas the comparison tail through index \(2\) is only \(\sum_{n=2}^{2}b_n=\frac12\). The full partial sum is not at least that tail sum; it is the tail sum plus the fixed initial term \(-10\).
A Common Error: Dropping the Finite Initial Sum
A tempting but incorrect line is to claim that the comparison \(a_n\ge b_n\) for \(n\ge N\) implies
This is not justified: the left-hand side includes the terms \(a_1,\ldots,a_{N-1}\), which might have a negative sum. In the previous example, \(N=2\) and \(m=2\) give \(S_2=-\frac{19}{2}\), while the comparison tail equals \(\frac12\). Thus the displayed inequality would assert \(-\frac{19}{2}\ge\frac12\), which is false.
The correct inequality retains the finite initial sum:
The initial sum on the right is a fixed finite constant; it does not depend on \(m\). The divergent tail of \(\sum b_n\) eventually exceeds any fixed negative amount, so adding that constant still gives quantities tending to \(+\infty\). This establishes divergence without making a false claim about the full partial sum being larger than the tail alone.
Using the Test Reliably
When applying direct comparison for divergence, keep the logical direction and the eventual condition in view. A useful checklist is:
- Choose a comparison series whose divergence is already known.
- Verify that the comparison terms are nonnegative on the indices used.
- Show that the terms of the target series are at least as large from some index \(N\) onward.
- Conclude divergence for the target series; do not claim that a finite initial segment must satisfy the inequality.
The known divergent comparisons in this unit include harmonic and other divergent \(p\)-series. For instance, \(\sum 1/n^p\) diverges when \(p\le1\). The comparison itself must still be verified: recognizing the general shape of a term is not a substitute for establishing the required inequality.
The result is useful when exact summation is difficult but a simple divergent lower bound is available. It proves divergence from below, rather than trying to bound the terms above. In particular, a comparison with a divergent series is informative here only when the target terms are eventually no smaller than the comparison terms.
Check Your Understanding
Use the theorem and the proof carefully. In particular, distinguish a tail sum from a full partial sum.
- State the eventual inequalities required for direct comparison for divergence.
- Why is a finite initial segment allowed to contain negative terms?
- If \(0\le\frac{1}{n+2}\le a_n\) for all \(n\ge5\), what additional fact about the comparison series is needed to conclude that \(\sum a_n\) diverges?
- Explain why \(\sum_{n=1}^{m}a_n\ge\sum_{n=N}^{m}b_n\) may fail even when \(a_n\ge b_n\ge0\) for every \(n\ge N\).
- For the series in the first worked example, identify the index from which the termwise comparison holds and the divergent comparison series.
