Absolute Value and Distance Between Terms
The previous tutorial established when a geometric sequence converges, using the epsilon-N definition and, in some cases, the sizes of the terms. Absolute value is central to that definition: \(a_n\to L\) means that the distance \(|a_n-L|\) becomes smaller than every positive tolerance. This tutorial examines how absolute value itself behaves under limits.
The key estimate compares the absolute values of two real numbers with the distance between them. Once it is proved, the main limit result follows directly from the definition of convergence. The result is useful in both directions of practice: it can turn a limit of a sequence into a limit of its absolute values, and it can show convergence to zero by bounding the absolute values of the terms.
The Reverse Triangle Inequality
The triangle inequality gives an upper bound for the absolute value of a sum. A related estimate says that taking absolute values cannot change two numbers by more than their original distance. This is called the reverse triangle inequality.
Proof. Write \(x=(x-y)+y\). By the Triangle Inequality established earlier in the course,
Subtracting \(|y|\) gives \(|x|-|y|\leq|x-y|\). Interchange \(x\) and \(y\) in the same argument. Since \(|y-x|=|x-y|\), this gives \(|y|-|x|\leq|x-y|\). Together, the two inequalities say that the number \(|x|-|y|\) lies between \(-|x-y|\) and \(|x-y|\). Therefore,
This proves the reverse triangle inequality. \(\square\)
The estimate is called “reverse” because it compares a difference of absolute values to the absolute value of a difference. It does not say that the two sides are equal. For example, if \(x=3\) and \(y=-3\), then \(\bigl||x|-|y|\bigr|=0\), while \(|x-y|=6\). The inequality holds, but equality is not required.
Worked Example: Comparing Absolute Values Directly
Take \(x=-7\) and \(y=2\). The absolute values differ by
whereas the distance between \(x\) and \(y\) is
Thus \(5\leq9\), as the reverse triangle inequality asserts. The estimate is informative even though these particular values do not give equality: the change in absolute value is no greater than the distance between the original numbers.
Absolute Values Preserve Limits
Suppose the terms \(a_n\) approach \(L\). To prove that their absolute values approach \(|L|\), we need to control \(\bigl||a_n|-|L|\bigr|\). Apply the reverse triangle inequality with \(x=a_n\) and \(y=L\). It bounds that quantity by \(|a_n-L|\), precisely the error already controlled by the definition of \(a_n\to L\).
Proof. Let \(\varepsilon>0\). Since \(a_n\to L\), the epsilon-N definition gives an \(N\in\mathbb{N}_0\) such that \(|a_n-L|<\varepsilon\) for every \(n\geq N\). For every such \(n\), the reverse triangle inequality gives
This is the epsilon-N condition for \(|a_n|\to|L|\). Therefore the absolute-value sequence converges to \(|L|\). \(\square\)
The proof works without requiring the terms to be positive, or to have the same sign as their limit. Absolute value may change the signs of individual terms, but the reverse triangle inequality ensures that this operation does not amplify their distance from the corresponding absolute value of the limit.
Worked Example: A Sequence Approaching a Negative Number
Define \(a_n=-4+\dfrac{3}{n+1}\) for \(n\in\mathbb{N}_0\). First verify convergence to \(-4\). For every \(n\in\mathbb{N}_0\),
Given \(\varepsilon>0\), choose \(N\in\mathbb{N}_0\) so that \(N+1>3/\varepsilon\), which is possible by the Archimedean property. If \(n\geq N\), then \(n+1\geq N+1>3/\varepsilon\), and hence
Thus \(a_n\to-4\). The theorem on absolute values preserving limits now gives \(|a_n|\to|-4|=4\). Notice that the limit of the absolute values is positive, even though the original sequence converges to a negative number.
Worked Example: Alternating Terms That Converge to Zero
Let \(b_n=\dfrac{(-1)^n}{n+2}\). Its signs alternate, but
Given \(\varepsilon>0\), choose \(N\in\mathbb{N}_0\) with \(N+2>1/\varepsilon\). For every \(n\geq N\),
Therefore \(b_n\to0\). Its absolute-value sequence is \(1/(n+2)\), which also converges to zero. In this case, the absolute values make the size of the error especially easy to calculate.
A Useful Special Case: Convergence to Zero
For convergence to zero, the absolute-value criterion has a converse as well. This is because the distance from \(|a_n|\) to zero is exactly \(|a_n|\), which is also the distance from \(a_n\) to zero. Consequently, bounding the terms in absolute value is not merely a convenient sufficient condition: it is precisely the epsilon-N requirement for convergence to zero.
Proof. Suppose first that \(a_n\to0\). The theorem on absolute values preserving limits gives \(|a_n|\to|0|=0\). Conversely, suppose \(|a_n|\to0\). Given \(\varepsilon>0\), there is an \(N\in\mathbb{N}_0\) such that \(\bigl||a_n|-0\bigr|<\varepsilon\) for every \(n\geq N\). Since \(|a_n|\geq0\), this inequality is \(|a_n|<\varepsilon\). But \(|a_n-0|=|a_n|\), so \(|a_n-0|<\varepsilon\) for every \(n\geq N\). This is exactly \(a_n\to0\). \(\square\)
This criterion often simplifies a proof: instead of tracking the signs of the terms, it is enough to find an upper bound on \(|a_n|\) that tends to zero. It also explains why an alternating sequence can converge to zero. Alternation concerns sign, while convergence to zero depends on the size of the terms.
What Absolute Values Do Not Tell Us
For a general limit \(L\), the theorem on absolute values preserving limits cannot simply be reversed. Knowing that \(|a_n|\) converges does not determine whether \(a_n\) converges. Absolute values discard sign information, and that information may be essential.
Worked Example: Convergent Absolute Values but Divergent Terms
Let \(c_n=5(-1)^n\). For every \(n\in\mathbb{N}_0\), \(|c_n|=5\), so the absolute-value sequence is constant and converges to \(5\). The original sequence does not converge. Indeed, its even and odd terms are
If \(c_n\) converged to some \(L\), the earlier theorem on subsequences of a convergent sequence would imply that both the even-indexed and odd-indexed subsequences converge to \(L\). Those subsequences are constant, with limits \(5\) and \(-5\), respectively. The uniqueness of limits would then require \(5=-5\), which is false. Hence \((c_n)\) does not converge, despite the convergence of \((|c_n|)\).
The special case \(a_n\to0\) avoids this problem: if the absolute values tend to zero, all signs are harmless because positive and negative terms alike lie within a small distance of zero. For a nonzero limit, signs can matter. The sequence in the example keeps switching between two distinct values, while its absolute values cannot register that switching.
A common pitfall is to claim that \(a_n\to L\) exactly when \(|a_n|\to|L|\). The forward implication is the Absolute Values Preserve Limits theorem; the reverse implication is false in general. Another useful check is to distinguish the two expressions \(|a_n-L|\) and \(\bigl||a_n|-|L|\bigr|\). The first measures the error in the original sequence, and the second measures the error in its absolute-value sequence. The reverse triangle inequality says the second error is no larger than the first, which proves the forward implication—but it does not say the first error is controlled by the second.
If \(a_n\to L\), apply the reverse triangle inequality to obtain \(\bigl||a_n|-|L|\bigr|\leq|a_n-L|\).
Find an eventual bound \(|a_n|<\varepsilon\); signs do not need to be analyzed separately.
Use this implication safely when the relevant limit is zero. For a nonzero limit, convergence of absolute values alone may lose essential sign information.
Check Your Understanding
Use the reverse triangle inequality and the epsilon-N definition to explain each conclusion.
- State the reverse triangle inequality for \(x,y\in\mathbb{R}\).
- If \(a_n\to-6\), what must \(|a_n|\) converge to?
- Why does \(|a_n|\to0\) imply \(a_n\to0\)?
- For \(c_n=5(-1)^n\), what does the absolute-value sequence converge to, and why does \((c_n)\) fail to converge?
- Which direction of the statement “\(a_n\to L\) if and only if \(|a_n|\to|L|\)” is always valid, and why is the converse not always valid?