Comparisons on a Tail
The previous tutorial showed that a bounded difference can make two sequences share the same boundedness status. A different way to compare sequences is to ask whether the absolute value of one is at most a fixed multiple of the absolute value of the other, at least from some index onward. Such a comparison need not say that the sequences agree or that their difference is bounded. It can still transfer useful information about boundedness.
As with eventual boundedness, the comparison is allowed to begin after a threshold. The constant in the comparison, however, must be fixed: it cannot increase with the index.
The absolute values make this definition useful for sequences with terms of either sign. It is a different kind of condition from an inequality such as \(a_n\leq b_n\): the order inequality alone does not control the magnitude of either sequence. The constant \(C\) also matters. For instance, a bound with a different constant for each \(n\) is not the fixed-multiple comparison defined here.
When an Upper Comparison Transfers Boundedness
If \((b_n)\) is bounded, then a fixed multiple of its absolute values is bounded too. An eventual magnitude comparison therefore bounds the tail of \((a_n)\). The finitely many terms before the comparison begins can be included by using the Tail Boundedness Is Equivalent to Boundedness theorem from the previous tutorial.
Proof. Suppose \((b_n)\) is bounded. Choose \(B\geq0\) such that \(|b_n|\leq B\) for every \(n\in\mathbb{N}_0\). For every \(n\geq N\), the assumed comparison gives
Thus \((a_n)\) is eventually bounded, so the Tail Boundedness Is Equivalent to Boundedness theorem implies that it is bounded. The final assertion is the contrapositive: if \((b_n)\) were bounded, the result just proved would force \((a_n)\) to be bounded. Therefore, if \((a_n)\) is unbounded, \((b_n)\) cannot be bounded. \(\square\)
The theorem has a particular direction. An upper comparison lets boundedness pass from \(b_n\) to \(a_n\); it also lets unboundedness pass from \(a_n\) to \(b_n\), by the contrapositive. The reverse conclusions do not follow from this inequality alone. In particular, an unbounded sequence can dominate a bounded one.
Worked Example: Bounding an Alternating Sequence by a Constant
Define \(a_n=(-1)^n(n+1)/(n+2)\) and \(b_n=1\) for \(n\in\mathbb{N}_0\). Since \(n+1\leq n+2\), and \(n+2>0\),
for every \(n\). Thus the magnitude comparison holds with \(N=0\) and \(C=1\). The sequence \((b_n)\) is bounded by \(1\), so the theorem shows that \((a_n)\) is bounded. In fact, the displayed inequality directly gives \(|a_n|\leq1\) for every \(n\). The alternating signs do not interfere with the comparison because it is made using absolute values.
Two-Sided Comparisons
An upper comparison alone need not settle whether the two sequences are both bounded or both unbounded. To transfer boundedness in both directions, use comparisons in both directions. Equivalently, require the magnitudes to be within fixed positive multiples of one another on a shared tail.
Proof. Choose \(N,c,C\) as in the definition. Suppose first that \((b_n)\) is bounded, and choose \(B\geq0\) such that \(|b_n|\leq B\) for every \(n\). For \(n\geq N\),
So \((a_n)\) is eventually bounded and hence bounded by the Tail Boundedness Is Equivalent to Boundedness theorem.
Conversely, suppose \((a_n)\) is bounded, with \(|a_n|\leq A\) for every \(n\), where \(A\geq0\). Since \(c>0\), the lower comparison implies that for every \(n\geq N\),
Thus \((b_n)\) is eventually bounded and therefore bounded. We have proved both implications. \(\square\)
The strict positivity of \(c\) is necessary for this argument: it allows the lower comparison to be rearranged to bound \(|b_n|\). If \(c=0\), the inequality \(0\leq|a_n|\) gives no information about \(b_n\).
Worked Example: Comparing a Perturbed Quadratic with a Quadratic
Let \(b_n=(n+2)^2\) and \(a_n=(n+2)^2+3(-1)^n\). Because \(|3(-1)^n|=3\), the triangle inequality gives
Here \(a_n\) is positive: since \(b_n\geq4\), \(a_n\geq b_n-3\geq1\). Also \(b_n\geq4\), so \(3\leq(3/4)b_n\). Consequently,
For a lower estimate, the reverse triangle inequality gives \(|a_n|\geq b_n-3\). Since \(b_n\geq4\),
because this last inequality is equivalent to \((3/4)b_n\geq3\). Thus \((1/4)|b_n|\leq|a_n|\leq(7/4)|b_n|\) for every \(n\). The two sequences are eventually comparable in magnitude. The sequence \(b_n=(n+2)^2\) is unbounded, since its values exceed any fixed real bound for sufficiently large \(n\). The theorem therefore shows that \(a_n\) is unbounded as well. The alternating perturbation changes the terms but does not change their boundedness status.
Why a One-Sided Order Inequality Is Not Enough
It is important to distinguish a magnitude comparison from an order comparison. The inequality \(a_n\leq b_n\) does not imply \(|a_n|\leq|b_n|\). For example, a large negative value of \(a_n\) can be less than a small positive value of \(b_n\), even though its absolute value is much larger. When terms are known to be nonnegative, an order comparison can give a magnitude comparison, but without that sign information it may not.
Worked Example: An Unbounded Upper Sequence Does Not Force Unboundedness Below
Set \(a_n=1\) and \(b_n=n+1\). Both sequences are nonnegative, and
for every \(n\in\mathbb{N}_0\). The sequence \((b_n)\) is unbounded: for any \(R\geq0\), choose an integer \(n>R\); then \(b_n=n+1>R\). Yet \((a_n)\) is bounded, since \(|a_n|=1\) for every \(n\). Thus the fact that an unbounded sequence lies above a sequence does not force the lower sequence to be unbounded. The upper comparison theorem cannot be applied to conclude otherwise: it requires the sequence being bounded to be the one on the right side of the magnitude inequality.
For nonnegative sequences, a lower comparison can transfer unboundedness upward: if \(0\leq b_n\leq a_n\) on a tail and \(b_n\) is unbounded, then \(a_n\) is unbounded. Indeed, a bound on \(a_n\) would also bound \(b_n\) on that tail, and the finite initial segment of \(b_n\) could then be included to give a global bound, a contradiction. The direction of the inequality determines which sequence’s boundedness can be transferred.
A Practical Comparison Strategy
When using comparisons, first identify the conclusion sought. If the goal is to prove that \((a_n)\) is bounded, look for an eventual upper bound on \(|a_n|\) involving a sequence already known to be bounded. If the goal is to prove that \((b_n)\) is unbounded, one useful route is to show that \(|a_n|\leq C|b_n|\) on a tail for some fixed \(C\), and then prove that \((a_n)\) is unbounded. If boundedness status must be equivalent, seek both an upper and a positive lower comparison.
For sequences that may change sign, compare \(|a_n|\) and \(|b_n|\) rather than relying on their order.
Write down the constants and the index from which the comparison holds. If two estimates begin at different indices, use the larger threshold.
An upper magnitude bound transfers boundedness from the comparison sequence to the sequence below it. A positive lower bound is needed to transfer boundedness back.
These comparisons concern boundedness, not convergence. Two sequences can both be unbounded without having the same limiting behavior, and two bounded sequences need not have the same limit or even converge. A magnitude comparison is useful only for conclusions that its inequalities actually support. In particular, a single inequality should not be treated as though it were a two-sided estimate.
Check Your Understanding
Use the definitions and theorems in this tutorial to answer the following questions.
- What fixed quantities must exist for \((a_n)\) to be eventually dominated in magnitude by \((b_n)\)?
- If \(|a_n|\leq C|b_n|\) eventually and \((b_n)\) is bounded, what can be concluded about \((a_n)\)?
- Why does the same upper comparison imply that unboundedness of \((a_n)\) forces unboundedness of \((b_n)\)?
- Which part of a two-sided magnitude comparison is used to transfer boundedness from \((a_n)\) to \((b_n)\), and why must its constant be positive?
- Can \(0\leq a_n\leq b_n\) hold for every \(n\) when \((a_n)\) is bounded and \((b_n)\) is unbounded? Give an example.