When the Ratio Does Not Approach a Positive Constant
In limit comparison, a finite positive ratio gives two series terms that are comparable in size. But sometimes the ratio of terms tends to zero or grows without bound. These outcomes still give useful information: eventually one sequence of terms is smaller than the other. The direction of that inequality determines which convergence conclusions are justified.
Throughout this tutorial, consider series with terms that are positive from some index onward. A finite number of earlier terms does not affect convergence: adding or removing finitely many terms changes the partial sums by a fixed finite amount, not their long-term behavior.
These limits do not make the two series equivalent. Instead, they produce a one-sided comparison. It is essential to keep track of which series supplies the larger terms.
The Two One-Sided Rules
Recall the direct comparison principle for positive-term series. If \(0\leq a_n\leq b_n\) for all sufficiently large \(n\), then convergence of \(\sum b_n\) implies convergence of \(\sum a_n\); divergence of \(\sum a_n\) implies divergence of \(\sum b_n\). The other two possible combinations do not follow from this inequality.
| Ratio limit | Eventual size relation | Valid conclusion |
|---|---|---|
| \(\frac{a_n}{b_n}\to 0\) | \(a_n\leq b_n\) | If \(\sum b_n\) converges, then \(\sum a_n\) converges. If \(\sum a_n\) diverges, then \(\sum b_n\) diverges. |
| \(\frac{a_n}{b_n}\to\infty\) | \(a_n\geq b_n\) | If \(\sum a_n\) converges, then \(\sum b_n\) converges. If \(\sum b_n\) diverges, then \(\sum a_n\) diverges. |
The first row follows because a ratio tending to zero is eventually less than \(1\). The second follows because a ratio tending to infinity is eventually greater than \(1\). The comparison principle then gives the stated conclusions.
Why the Rules Work
Here is the full argument for both cases. The definition of a limit supplies the eventual inequalities; comparison then transfers convergence or divergence in the appropriate direction.
Since \(a_n/b_n\to 0\), there is an index \(N\) such that \(0<a_n/b_n<1\) whenever \(n\geq N\). Multiplication by \(b_n>0\) gives \(a_n<b_n\).
If \(\sum b_n\) converges, the smaller-term series \(\sum a_n\) converges. If \(\sum a_n\) diverges, the larger-term series \(\sum b_n\) must diverge.
For a ratio tending to infinity, there is an index \(N\) such that \(a_n/b_n>1\) whenever \(n\geq N\). Thus \(b_n<a_n\).
If the larger-term series \(\sum a_n\) converges, then \(\sum b_n\) converges. If the smaller-term series \(\sum b_n\) diverges, then \(\sum a_n\) diverges.
Only the terms from index \(N\) onward are needed for comparison. The finitely many terms before \(N\) cannot change whether the infinite series converges.
Worked Example: A Ratio Tending to Zero Gives Convergence
Worked Example: A Smaller Series Compared with a Convergent Series
Determine the convergence of \(\sum_{n=1}^{\infty}1/n^3\) using \(b_n=1/n^2\) as a comparison series.
Set \(a_n=1/n^3\). The terms are positive, and their ratio is
As \(n\to\infty\), \(1/n\to0\), so \(a_n/b_n\to0\). In fact, for every \(n\geq1\), \(1/n^3\leq1/n^2\). The comparison series \(\sum 1/n^2\) is a convergent p-series, because its exponent is \(2>1\). Therefore the smaller-term series \(\sum 1/n^3\) converges.
Check: The ratio tends to zero in the order \(a_n/b_n\), and the known convergent series is the denominator series \(\sum b_n\). That is exactly the valid convergence direction.
Worked Example: A Ratio Tending to Zero Gives Divergence
Worked Example: Divergence Transfers from Smaller Terms to Larger Terms
Let \(a_n=1/n\) and \(b_n=1/\sqrt n\). Use limit comparison to determine whether \(\sum b_n\) diverges.
First calculate the ratio in the specified order:
Since \(1/\sqrt n\to0\), we have \(a_n/b_n\to0\), so \(a_n\leq b_n\) for all \(n\geq1\). The series \(\sum a_n=\sum 1/n\) is the harmonic series and diverges. Because its terms are no larger than \(b_n\), the series \(\sum b_n\) must also diverge.
Check: The smaller series is divergent, which forces the larger series to diverge. The conclusion agrees with the p-series classification for \(\sum 1/n^{1/2}\), whose exponent is less than or equal to \(1\).
Worked Example: A Ratio Tending to Infinity Gives Divergence
Worked Example: A Larger Series Compared with a Divergent Series
Let \(a_n=1/\sqrt n\) and \(b_n=1/n\). Use the ratio \(a_n/b_n\) to determine whether \(\sum a_n\) diverges.
Compute:
As \(n\to\infty\), \(\sqrt n\to\infty\). Thus \(a_n\) is eventually larger than \(b_n\) (indeed, \(a_n\geq b_n\) for \(n\geq1\)). The series \(\sum b_n=\sum 1/n\) diverges. Since the smaller-term series diverges, the larger-term series \(\sum a_n\) also diverges.
Check: This is the infinity-ratio rule: divergence of the smaller series transfers to the larger series. Again, the p-series classification confirms divergence because \(\sum 1/\sqrt n\) has exponent \(1/2\).
Worked Example: A Ratio Tending to Infinity Gives Convergence
Worked Example: Convergence of the Larger Series Controls the Smaller One
Let \(a_n=1/n^2\) and \(b_n=1/n^3\). Use the ratio to determine whether \(\sum b_n\) converges.
The ratio is
Therefore \(a_n/b_n\to\infty\), and \(b_n\leq a_n\) for every \(n\geq1\). The larger-term series \(\sum a_n=\sum1/n^2\) converges, since it is a p-series with exponent \(2>1\). The smaller-term series \(\sum b_n\) must therefore converge.
Check: The ratio tends to infinity, and the known convergent series is \(\sum a_n\), whose terms are the larger ones. The inference is in the valid direction.
What the Ratio Does Not Tell You
A one-sided comparison does not settle every possible combination of known behavior. If \(a_n/b_n\to0\), knowing that the larger series \(\sum b_n\) diverges does not determine whether the smaller series converges. Also, knowing that the smaller series \(\sum a_n\) converges does not determine whether the larger series diverges or converges.
For instance, \(a_n=1/n^2\) and \(b_n=1/n\) give \(a_n/b_n=1/n\to0\). The larger series \(\sum b_n\) diverges, while the smaller series \(\sum a_n\) converges. Thus divergence of the larger series alone cannot establish divergence of the smaller one.
Likewise, if \(a_n/b_n\to\infty\), convergence of the smaller series \(\sum b_n\) does not settle the larger series. For \(a_n=1/n\) and \(b_n=1/n^2\), the ratio is \(n\to\infty\), but \(\sum b_n\) converges while \(\sum a_n\) diverges. The direction of the inequality matters as much as the ratio limit.
Common Mistake: Reversing the Comparison
A frequent error is to see a ratio tending to zero and assume that convergence of the smaller series forces convergence of the larger one.
The conclusion is false. The ratio says \(1/n^2\leq1/n\), so the known convergent series has smaller terms, not larger terms. A smaller convergent series does not control a larger series. In fact, \(\sum1/n\) is the divergent harmonic series.
A Reliable Decision Process
When using a zero or infinite ratio, identify the ratio order and write down the term inequality before drawing a conclusion.
State which sequence is \(a_n\) and which is \(b_n\), then form \(a_n/b_n\) in that order.
Determine whether it tends to zero or infinity. Confirm that the terms are positive eventually.
A zero limit gives \(a_n\leq b_n\) eventually; an infinite limit gives \(a_n\geq b_n\) eventually.
Use convergence of the larger series or divergence of the smaller series. If neither applicable fact is known, this comparison does not settle the question.
The central lesson is simple but important: a ratio tending to zero or infinity gives a one-sided size comparison, not a two-sided equivalence. State the inequality, identify which series is known, and infer only what direct comparison allows.
Check Your Understanding
For each question, identify the direction of comparison before deciding what conclusion is justified.
- If \(a_n/b_n\to0\) and \(\sum b_n\) converges, what can be concluded about \(\sum a_n\)?
- If \(a_n/b_n\to0\) and \(\sum a_n\) diverges, what can be concluded about \(\sum b_n\)?
- If \(a_n/b_n\to\infty\) and \(\sum b_n\) diverges, what can be concluded about \(\sum a_n\)?
- If \(a_n/b_n\to\infty\) and \(\sum a_n\) converges, what can be concluded about \(\sum b_n\)?
- If \(a_n/b_n\to0\) and \(\sum b_n\) diverges, does the ratio alone determine whether \(\sum a_n\) converges? Explain briefly.
