Bounded Does Not Mean Convergent
A bounded sequence stays within some fixed finite range. A convergent sequence, by contrast, has terms that eventually get as close as desired to one particular number. These conditions describe different behavior: a sequence can remain confined while continuing to jump between separated values.
This distinction matters when using the monotone convergence theorem. That theorem gives a powerful conclusion when a sequence is both monotone and bounded. If boundedness alone were enough, the monotonicity condition would be unnecessary. Alternating and periodic sequences show why it is not.
Boundedness places every term inside a fixed interval, such as \([-M,M]\). Convergence requires much more: for every positive tolerance, all sufficiently late terms must lie inside the interval \((L-\varepsilon,L+\varepsilon)\). A sequence can fit inside a fixed interval without fitting eventually inside arbitrarily small intervals around any one number.
A Useful Test for Nonconvergence
The definition of convergence gives a direct way to prove that a sequence does not converge. If a sequence converges, then any two sufficiently late terms must be close to each other. Indeed, if both are within \(\varepsilon\) of the limit, the triangle inequality says their distance is less than \(2\varepsilon\).
To justify the criterion, suppose instead that \(a_n\to L\). Apply the definition of convergence with \(\varepsilon=\delta/3\). There is an index \(N\) such that \(n\ge N\) implies \(|a_n-L|<\delta/3\). Thus, for any \(p,q\ge N\),
This contradicts the assumed existence, beyond every index, of a pair whose terms are at least \(\delta\) apart. The criterion captures the core issue: convergence rules out persistent, fixed-size jumps between late terms.
Worked Example: An Alternating Sequence
Worked Example: An Alternating Sequence
Consider \(a_n=(-1)^n\) for \(n\in\mathbb N\). Its first terms are \(-1,1,-1,1,\ldots\). We will show that it is bounded but does not converge.
For every positive integer \(n\), \((-1)^n\) is either \(-1\) or \(1\). Therefore \(|a_n|=1\), so \(|a_n|\le 1\) for all \(n\). The sequence is bounded by \(M=1\), and more precisely its terms lie in \([-1,1]\).
Suppose that \(a_n\to L\). Use the convergence definition with \(\varepsilon=1/2\). Then all sufficiently late terms must be within \(1/2\) of \(L\). There are both even and odd indices beyond any fixed index: for an even index \(p\), \(a_p=1\), and for an odd index \(q\), \(a_q=-1\). The convergence assumption would therefore imply
But the triangle inequality gives
which is impossible. Thus the sequence does not converge. This check confirms both claims: every term stays between \(-1\) and \(1\), while arbitrarily late terms still differ by \(2\).
Worked Example: A Periodic Sequence
Worked Example: A Periodic Sequence
Define a sequence by repeating the values \(2,-1,4\): for each \(n\in\mathbb N\), let \(a_n=2\) if \(n\) leaves remainder \(1\) on division by \(3\), \(a_n=-1\) if it leaves remainder \(2\), and \(a_n=4\) if it is divisible by \(3\). The sequence begins \(2,-1,4,2,-1,4,\ldots\).
Every term is one of \(-1,2,4\), so \(-1\le a_n\le4\) for all \(n\). In particular, \(|a_n|\le4\), and the sequence is bounded.
Suppose it converges to \(L\). Choose \(\varepsilon=1\). There are arbitrarily large indices \(p\) with remainder \(1\) modulo \(3\), and arbitrarily large indices \(q\) divisible by \(3\). At those indices, \(a_p=2\) and \(a_q=4\). If convergence held, sufficiently late such terms would satisfy \(|2-L|<1\) and \(|4-L|<1\). Consequently,
a contradiction. Hence the sequence is bounded but does not converge. The values remain in the interval \([-1,4]\), yet the repeated gap between \(2\) and \(4\) prevents the terms from settling near one limit.
Worked Example: Alternation with a Shrinking Term
Worked Example: Alternation with a Shrinking Term
Consider \(a_n=(-1)^n+\frac1n\). The added fraction changes with \(n\), so this sequence is not the same as a fixed repeating pattern. It is still bounded and still fails to converge.
If \(n\) is even, then \((-1)^n=1\), so \(a_n=1+\frac1n\). Since \(n\ge2\) for even \(n\), these terms satisfy \(1<a_n\le\frac32\). If \(n\) is odd, then \((-1)^n=-1\), so \(a_n=-1+\frac1n\). Since \(n\ge1\), these terms satisfy \(-1<a_n\le0\). Combining the two cases gives \(-1\le a_n\le\frac32\) for every \(n\), so the sequence is bounded.
Now suppose \(a_n\to L\), and apply the definition with \(\varepsilon=1/3\). Choose an even index \(p\) and an odd index \(q\), both sufficiently large to be past the resulting threshold. The bounds just found give \(a_p>1\) and \(a_q\le0\), so \(|a_p-a_q|>1\). But convergence would require
Since a number greater than \(1\) cannot be less than \(\frac23\), this is a contradiction. Therefore the sequence does not converge. The alternating terms remain separated even though the additional fraction varies with the index.
Where the Monotone Convergence Theorem Applies
The monotone convergence theorem for sequences states that every bounded monotone real sequence converges. A sequence is monotone if it is nondecreasing, meaning \(a_{n+1}\ge a_n\) for every \(n\), or nonincreasing, meaning \(a_{n+1}\le a_n\) for every \(n\). Boundedness and monotonicity are both required hypotheses.
For the sequence \((-1)^n\), consecutive terms alternate between \(-1\) and \(1\). The first step is an increase from \(-1\) to \(1\), and the next is a decrease from \(1\) to \(-1\). Thus it is neither nondecreasing nor nonincreasing. It is bounded, but the theorem does not apply. The same diagnosis holds for the two periodic examples: their terms repeatedly rise and fall, so they are not monotone.
There is a related point about eventual monotonicity. A sequence is eventually monotone if its terms are monotone from some index onward. The alternating sequence and the period-three sequence continue to rise and fall arbitrarily far along, so neither is eventually monotone. The sequence \((-1)^n+1/n\) also has arbitrarily late even and odd terms separated by more than \(1\), which rules out its being eventually monotone: an eventually monotone bounded sequence would converge by the monotone convergence theorem, contrary to the proof above.
Find fixed lower and upper bounds that contain every term.
Determine whether the sequence is nondecreasing or nonincreasing, at least eventually if that is the claim being tested.
Boundedness without monotonicity does not establish convergence.
A Common Mistake
A tempting but incorrect argument is: “The sequence \(a_n=(-1)^n\) is bounded between \(-1\) and \(1\). The monotone convergence theorem says bounded sequences converge. Therefore it converges.” The statement attributed to the theorem is false: the theorem requires the sequence to be monotone as well as bounded.
The correction is to check both hypotheses. Here boundedness holds, but monotonicity fails, as the consecutive terms go first up and then down repeatedly. The theorem therefore gives no conclusion. The direct argument above proves the actual conclusion: the sequence does not converge because late terms can still be \(2\) apart.
Check Your Understanding
For each question, distinguish what boundedness tells you from what would be needed to establish convergence.
- Why is \((-1)^n\) bounded, and what fixed distance separates its even-indexed and odd-indexed terms?
- State the two hypotheses of the monotone convergence theorem for sequences.
- For the sequence that repeats \(2,-1,4\), identify two values that occur at arbitrarily large indices and use their distance to explain why convergence fails.
- In the proof for \((-1)^n+1/n\), why can one choose an even and an odd index beyond the convergence threshold?
- Give a concrete error in the claim that every bounded sequence converges, and state the correction.