Why the Inequality Direction Matters
A series can be difficult to classify directly, even when its terms are positive. A useful strategy is to compare its terms with those of a series whose convergence or divergence is already known. The comparison is only useful, however, when the inequality points in the direction that supports the conclusion.
The key idea is that the partial sums of a nonnegative-term series can only increase. If the terms are no larger than those of a convergent series, the partial sums are bounded above. If the terms are at least as large as those of a divergent series, the partial sums grow without bound.
The Direct Comparison Tests
Suppose \(a_n\) and \(b_n\) are nonnegative for all sufficiently large \(n\). A comparison need only hold from some index onward: changing or omitting finitely many terms does not change whether a series converges or diverges.
- If \(0\leq a_n\leq b_n\) for all sufficiently large \(n\), and \(\sum b_n\) converges, then \(\sum a_n\) converges.
- If \(0\leq b_n\leq a_n\) for all sufficiently large \(n\), and \(\sum b_n\) diverges, then \(\sum a_n\) diverges.
These are two different tests, not two versions of the same inequality. To prove convergence, place the series of interest below a convergent series. To prove divergence, place it above a divergent series.
Why the Tests Work
The reasoning comes from comparing partial sums. First consider the convergence test, and suppose the termwise inequality holds for \(n\geq N_0\). For any \(N\geq N_0\),
Because \(\sum b_n\) converges and its terms are nonnegative, its partial sums are bounded above. The right-hand side is therefore bounded independently of \(N\). The partial sums of \(\sum a_n\) are nondecreasing, because each added term is nonnegative, and they are bounded above. Thus they approach a finite limit, so \(\sum a_n\) converges.
For the divergence test, if \(b_n\leq a_n\) eventually, then for sufficiently large \(N\), the partial sum of \(a_n\) is at least a fixed finite initial sum plus the corresponding tail sum of \(b_n\). Since \(\sum b_n\) diverges and has nonnegative terms, its partial sums grow without bound. The partial sums of \(\sum a_n\) must also grow without bound, so \(\sum a_n\) diverges.
Find a convergent comparison series that is larger than the series being tested.
Find a divergent comparison series that is smaller than the series being tested.
Explain that the termwise inequality and the known comparison result meet the conditions of the relevant test.
Worked Examples
Worked Example: Comparing with a Convergent p-Series
Determine whether \(\sum_{n=1}^{\infty}\frac{1}{n^2+4n+5}\) converges or diverges.
For every \(n\geq1\), \(4n+5\geq0\), so \(n^2+4n+5\geq n^2>0\). Taking reciprocals of positive quantities reverses the inequality:
The comparison series \(\sum_{n=1}^{\infty}\frac{1}{n^2}\) is a convergent p-series, since its exponent is \(p=2>1\). The terms of the given series are nonnegative and no larger than its terms. By the direct comparison test, the given series converges.
Check: The denominator inequality holds for every \(n\geq1\), and both sides are positive, so the reciprocal inequality has the required direction. The comparison series is convergent, which is the condition needed to conclude convergence.
Worked Example: Comparing Below with the Harmonic Series
Determine whether \(\sum_{n=1}^{\infty}\frac{2n+3}{n^2+4n+3}\) converges or diverges.
The terms are positive for \(n\geq1\). To compare them with \(1/n\), check the inequality for \(n\geq3\). All denominators are positive, so
Subtracting the right side from the left gives
At \(n=3\), this expression is \(9-3-3=3>0\). For \(n\geq3\), it increases as \(n\) increases, since its value at \(n+1\) minus its value at \(n\) is \(2n>0\). Thus the inequality holds for every \(n\geq3\).
The harmonic series \(\sum_{n=1}^{\infty}\frac{1}{n}\) diverges. The given terms are at least as large as its terms from \(n=3\) onward. By the divergence part of the direct comparison test, the given series diverges.
Check: The comparison is only needed for \(n\geq3\), so the first two terms do not affect the conclusion. The given terms are above the divergent comparison terms, as required.
Worked Example: Comparing with a Geometric Series
Determine whether \(\sum_{n=1}^{\infty}\frac{n+2}{n^2 3^n}\) converges or diverges.
For \(n\geq1\), \(n+2\leq3n\), because this inequality is equivalent to \(2\leq2n\). Also, \(1/n\leq1\). Therefore,
The comparison series is geometric:
Its common ratio is \(1/3\), whose absolute value is less than \(1\), so it converges. In fact, its sum is \(3\cdot\frac{1/3}{1-1/3}=3/2\). The given series has nonnegative terms bounded above by this convergent geometric series, so it converges by direct comparison.
Check: The inequality chain holds for every \(n\geq1\), and the final comparison series has ratio \(1/3\). The comparison therefore establishes convergence.
Common Pitfall: An Upper Bound Does Not Prove Divergence
Consider \(\sum_{n=1}^{\infty}\frac{n}{n^2+1}\). A tempting calculation is
which is correct because \(n^2+1\geq n^2\). But the harmonic series on the right diverges. Being bounded above by a divergent series does not establish that the series on the left diverges. The convergence test cannot be used because its comparison series must converge.
Instead, compare in the other direction. For \(n\geq1\),
The inequality is true for every \(n\geq1\), and \(\sum_{n=1}^{\infty}\frac{1}{2n}\) diverges because it is one-half of the harmonic series. The given series is therefore larger than a divergent series and diverges by direct comparison.
Writing a Complete Comparison Justification
A clear justification does more than state that one series is “like” another. Show the inequality, explain why it holds on the relevant range, identify whether the comparison series converges or diverges, and connect those facts to the correct part of the test.
Before concluding, check that the terms are nonnegative where the comparison is used. If the inequality holds only for \(n\geq N_0\), say so; finitely many initial terms do not affect the classification. Finally, verify that the known comparison series has the behavior you need. A familiar series is useful only when its convergence or divergence supports the intended conclusion.
Check Your Understanding
For each question, identify the comparison inequality and the direct comparison conclusion it permits.
- If \(0\leq a_n\leq b_n\) eventually and \(\sum b_n\) converges, what can you conclude about \(\sum a_n\)?
- If \(0\leq b_n\leq a_n\) eventually and \(\sum b_n\) diverges, what can you conclude about \(\sum a_n\)?
- Why does \(0\leq a_n\leq b_n\) with \(\sum b_n\) divergent not, by itself, prove that \(\sum a_n\) diverges?
- For \(n\geq1\), compare \(\frac{1}{n^2+1}\) with \(\frac{1}{n^2}\). What does direct comparison establish?
- Does a comparison inequality need to hold for every term, or can it hold only from some index onward? Explain briefly.
