Positivity After a Threshold
The previous tutorial described sequences whose terms eventually pass every real bound. Eventual positivity is a more modest condition: instead of requiring terms to grow without bound, we ask only that they eventually stay above zero. The sequence may be bounded, may converge, or may continue to vary. What matters is that one index marks a point after which every term is positive.
As with divergence to infinity, the word “eventually” allows finitely many exceptions. Those early terms can be negative or zero without affecting whether the sequence is eventually positive. The threshold must, however, work for every later index, not just for infinitely many positive terms.
Every eventually positive sequence is eventually nonnegative, but the converse need not hold: a sequence that is zero from some point onward is eventually nonnegative without being eventually positive. Keeping the inequalities strict or non-strict is important, especially when using positivity in a product or when discussing whether a limit can equal zero.
Finding a Threshold
Worked Example: A Sequence with Some Negative Terms
Let \(a_n=(n-2)/(n+1)\) for \(n\in\mathbb{N}_0\). The denominator is positive for every such \(n\), since \(n+1\geq1\). Thus the sign of \(a_n\) is the sign of its numerator \(n-2\). If \(n\geq3\), then \(n-2\geq1\), so
Therefore \(a_n\) is eventually positive, with \(N=3\) as one valid threshold. It is not positive at every index: \(a_0=-2\), \(a_1=-1/2\), and \(a_2=0\). These initial terms do not prevent eventual positivity.
Worked Example: Positive Terms That Approach Zero
Define \(b_n=1/(n+2)\). Since \(n+2\geq2>0\) for every \(n\in\mathbb{N}_0\), we have \(b_n>0\) at every index, and hence the sequence is eventually positive with \(N=0\).
In fact, \(b_n\to0\). To verify this directly, let \(\varepsilon>0\). By the Archimedean property, choose \(N\in\mathbb{N}_0\) such that \(N+2>1/\varepsilon\). For every \(n\geq N\), \(n+2\geq N+2>1/\varepsilon\), and therefore
This example shows why “eventually positive” does not mean “bounded below by one fixed positive number.” The terms stay above zero, but they can get arbitrarily close to zero.
Worked Example: Positivity Despite Oscillation
Let \(c_n=2+(-1)^n/(n+1)\). Since \((-1)^n\geq-1\) and \(n+1\geq1\), we have
It follows that \(c_n\geq1>0\) for every \(n\in\mathbb{N}_0\). Thus \(c_n\) is positive at every index, even though the sign of its alternating term changes. The estimate works for both even and odd \(n\); no assumption that the alternating term is positive is needed.
Basic Operations Preserve Eventual Positivity
Once each of two sequences is positive beyond some index, their sum and product are positive beyond a common index. The two original thresholds might differ, so we use the larger one. This is a basic but useful way to handle expressions assembled from several sequences.
Proof. Since \((a_n)\) is eventually positive, there is an \(N_a\in\mathbb{N}_0\) such that \(a_n>0\) for every \(n\geq N_a\). There is also an \(N_b\in\mathbb{N}_0\) such that \(b_n>0\) for every \(n\geq N_b\). Let \(N=\max\{N_a,N_b\}\). For every \(n\geq N\), both inequalities hold, so \(a_n+b_n>0\) and \(a_nb_n>0\). This proves the sum and product assertions. If \(c>0\) and \(a_n>0\) for every \(n\geq N_a\), then \(ca_n>0\) for every \(n\geq N_a\). Hence \((ca_n)\) is eventually positive as well. \(\square\)
The requirement \(c>0\) matters for the scalar-multiple assertion: multiplication by a negative constant reverses the sign. If \(c=0\), the resulting sequence is zero and is not eventually positive. Similarly, the sum of an eventually positive sequence and an eventually nonnegative sequence is eventually positive: beyond the larger of their thresholds, it is the sum of a positive number and a nonnegative number.
Worked Example: A Product with Alternating Factors
For \(n\in\mathbb{N}_0\), define
For every \(n\geq2\), the first factor is positive: its denominator \(n+1\) is positive and its numerator \(n-1\geq1\). For every \(n\), \((-1)^n\) is either \(1\) or \(-1\), so \(3+(-1)^n\) is either \(4\) or \(2\), and in either case is positive. Consequently, for every \(n\geq2\), both factors are positive and \(p_n>0\). Thus \(p_n\) is eventually positive, with threshold \(N=2\). The alternating factor does not cause a problem because it remains positive after the added constant is included.
Eventual Comparison
An inequality can transfer eventual positivity from one sequence to another. The comparison only needs to hold on a tail, and the sequence being compared against must be at least as large as the positive sequence there.
Proof. Since \((a_n)\) is eventually positive, choose \(N_1\in\mathbb{N}_0\) such that \(a_n>0\) for every \(n\geq N_1\). Let \(N=\max\{N_0,N_1\}\). For every \(n\geq N\), the comparison and positivity both apply, giving
Thus \(b_n>0\) for every \(n\geq N\), which is exactly the definition of eventual positivity. \(\square\)
The direction of comparison cannot be reversed without further information. A sequence lying below an eventually positive sequence could still be negative. For example, \(a_n=1\) and \(b_n=-1\) satisfy \(b_n\leq a_n\) at every index, but \(b_n\) is not eventually positive. The theorem uses the lower sequence as the positive one and the upper sequence as the one whose positivity is concluded.
What Eventual Positivity Says About a Limit
Eventual positivity constrains the limit of a convergent sequence, but it does not force that limit to be strictly positive. The following result rules out negative limits. Its proof makes the contradiction explicit: convergence to a negative number would eventually place the terms below zero, in conflict with eventual positivity.
Proof. Suppose, to the contrary, that \(L<0\). Then \(\varepsilon=-L/2\) is positive. Since \(a_n\to L\), there is an \(N_1\in\mathbb{N}_0\) such that \(|a_n-L|<-L/2\) whenever \(n\geq N_1\). In particular,
for every \(n\geq N_1\). But eventual positivity gives an \(N_2\in\mathbb{N}_0\) such that \(a_n>0\) for every \(n\geq N_2\). For any \(n\geq\max\{N_1,N_2\}\), the same term would have to satisfy both \(a_n<0\) and \(a_n>0\), an impossibility. Therefore \(L\) cannot be negative, and \(L\geq0\). \(\square\)
There is a useful converse when the limit is strictly positive. By Eventual Sign Stability, if \(a_n\to L\) with \(L\neq0\), then the terms eventually have the sign of \(L\). In particular, if \(L>0\), the sequence is eventually positive. Together with the theorem just proved, this gives a precise statement: a convergent eventually positive sequence has a nonnegative limit, and a convergent sequence with a positive limit is eventually positive. The zero-limit case is different and allows either sign behavior.
Worked Example: Two Sequences with Limit Zero
The sequence \(u_n=1/(n+2)\) is positive at every index and converges to zero, as verified earlier. In contrast, define \(v_n=(-1)^n/(n+1)\). For even \(n\), \(v_n>0\); for odd \(n\), \(v_n<0\), so \(v_n\) is not eventually positive. Nevertheless, \(v_n\to0\): for every \(\varepsilon>0\), choose \(N\in\mathbb{N}_0\) with \(N+1>1/\varepsilon\). If \(n\geq N\), then
The Absolute-Value Criterion for Convergence to Zero therefore gives \(v_n\to0\). These examples show why the conclusion of the theorem is \(L\geq0\), not \(L>0\): a zero limit is compatible with eventual positivity, but it does not imply eventual positivity.
Common Pitfalls
Eventual positivity is a statement about the sign of every term on a tail. It is not enough that positive terms occur infinitely often. For example, \(w_n=(-1)^n\) has positive terms at every even index, however large the index, but it also has negative terms at every odd index. No threshold makes all later terms positive.
It is also important not to confuse eventual positivity with a uniform positive lower bound. The sequence \(1/(n+2)\) is eventually positive, but given any \(c>0\), its terms eventually become smaller than \(c\). Eventual positivity only supplies the lower bound zero on a tail; it does not supply a positive number that all terms on that tail exceed.
Finally, finite initial exceptions are permitted, but infinitely many exceptions are not. When checking a candidate threshold, verify the required inequality for every index at or beyond it. If the argument combines two eventual conditions, take the larger of their thresholds so that both conditions hold simultaneously. This simple step is what allows the sum, product, and comparison results to apply without overlooking any indices.
Check Your Understanding
Use the definitions and results in this tutorial to answer the following questions.
- What quantifiers define an eventually positive sequence, and how does the definition differ from eventual nonnegativity?
- Why is the product of two eventually positive sequences eventually positive even when their thresholds differ?
- If \(a_n\leq b_n\) for all sufficiently large \(n\), and \(a_n\) is eventually positive, what follows about \(b_n\)?
- Why can a convergent, eventually positive sequence not have a negative limit?
- Does eventual positivity imply a strictly positive limit? Give an example or explain why not.
- Why does having positive terms at infinitely many indices not by itself establish eventual positivity?