When a Sequence Cannot Settle on One Limit
A sequence may look complicated when its terms are examined one at a time. A useful way to test whether it converges is to focus on selected terms whose indices follow a clear pattern. These selected terms form a subsequence.
The key fact is that a convergent sequence cannot have subsequences that settle on different finite limits. Nor can it contain a subsequence whose terms grow without bound. These observations give direct ways to prove that a sequence does not converge to any finite number.
What Is a Subsequence?
A subsequence is formed by choosing terms of a sequence in their original order, possibly skipping some terms. The chosen indices must keep increasing; terms cannot be rearranged or reused.
For example, choosing the even-numbered terms gives the subsequence \(a_2,a_4,a_6,\ldots\), which can be written as \(a_{2k}\). Choosing the odd-numbered terms gives \(a_1,a_3,a_5,\ldots\), or \(a_{2k-1}\). In each case, the indices increase in the original sequence.
The Inheritance Principle
The central result is that subsequences inherit the limit of a convergent sequence. In particular, they cannot have different limits from one another if the original sequence converges.
The reason is that a subsequence cannot keep selecting terms from the beginning of the original sequence. Its indices increase, so eventually every selected term lies past any fixed index.
If \(a_n\to L\), then for every \(\varepsilon>0\), there is a positive integer \(N\) such that \(n\ge N\) implies \(|a_n-L|<\varepsilon\).
Because \(n_1<n_2<\cdots\) are positive integers, \(n_k\ge k\) for every \(k\). Thus, if \(k\ge N\), then \(n_k\ge N\).
For every \(k\ge N\), we have \(|a_{n_k}-L|<\varepsilon\). This is exactly the definition that \(a_{n_k}\to L\).
To finish the divergence conclusion, recall that a sequence cannot have two different limits. Briefly, if a sequence had limits \(L\ne M\), choose \(\varepsilon=|L-M|/3\). Far enough along, a term would be within \(\varepsilon\) of both limits, giving \(|L-M|\le |L-a_n|+|a_n-M|<2|L-M|/3\), which is impossible. Thus, if two subsequences have distinct limits, the original sequence cannot converge: convergence would force both subsequences to have the same limit.
Worked Example: Two Subsequences Have Different Limits
Worked Example: Two Subsequences Have Different Limits
Consider \(a_n=(-1)^n\). Determine whether the sequence converges by examining its even- and odd-indexed terms.
For even indices \(n=2k\),
The even-indexed subsequence is constant at \(1\), so its limit is \(1\).
For odd indices \(n=2k-1\),
The odd-indexed subsequence is constant at \(-1\), so its limit is \(-1\). Since \(1\ne -1\), the subsequences have different finite limits. By the inheritance principle, the original sequence does not converge.
Check: If \(a_n\) had a finite limit \(L\), both subsequences would have to converge to \(L\). But the even terms converge to \(1\) and the odd terms converge to \(-1\), so no such \(L\) can exist.
Worked Example: Different Limits Without Constant Subsequences
Worked Example: Different Limits Without Constant Subsequences
Define \(a_n=1/n\) when \(n\) is even, and \(a_n=1+1/n\) when \(n\) is odd. Decide whether \((a_n)\) converges.
For \(n=2k\), the even-indexed terms are
Given any \(\varepsilon>0\), choose a positive integer \(K>1/(2\varepsilon)\). For \(k\ge K\), \(0<1/(2k)\le 1/(2K)<\varepsilon\), so \(a_{2k}\to0\).
For \(n=2k-1\), the odd-indexed terms are
Their distance from \(1\) is \(|a_{2k-1}-1|=1/(2k-1)\). Given \(\varepsilon>0\), choose a positive integer \(K>(1+1/\varepsilon)/2\). For \(k\ge K\), \(2k-1\ge 2K-1>1/\varepsilon\), so \(1/(2k-1)<\varepsilon\). Therefore \(a_{2k-1}\to1\).
The two subsequences have limits \(0\) and \(1\), respectively. Since these limits differ, the original sequence does not converge.
Check: Each subsequence limit follows directly from its distance to the proposed limit: the even terms approach \(0\), and the odd terms approach \(1\). A single limit for the full sequence would have to be both values, which is impossible.
An Unbounded Subsequence Also Rules Out Convergence
Another useful test is to find a subsequence whose terms are unbounded in absolute value. A sequence converging to a finite limit must be bounded, and the same is true of each of its subsequences. Therefore, an unbounded subsequence rules out finite convergence.
Here is why finite convergence guarantees boundedness. If \(a_n\to L\), use the definition with \(\varepsilon=1\). There is an index \(N\) such that \(n\ge N\) implies \(|a_n-L|<1\). The triangle inequality then gives
The terms before \(N\) form a finite set, so their absolute values have a finite maximum. Together, the finite initial segment and the bounded tail show that the whole sequence is bounded. Every subsequence is then bounded as well, since its terms are among the terms of the original sequence.
Worked Example: An Unbounded Subsequence
Worked Example: An Unbounded Subsequence
Define \(a_n=0\) for odd \(n\), and \(a_n=n\) for even \(n\). Determine whether the sequence converges to a finite limit.
The even-indexed subsequence has terms
To check that this subsequence is unbounded, let \(M>0\) be any proposed upper bound. Choose a positive integer \(k>M/2\). Then \(a_{2k}=2k>M\). Since this can be done for every \(M>0\), the even-indexed subsequence is unbounded.
If the original sequence converged to a finite limit, the even-indexed subsequence would also converge to that limit and hence be bounded. That contradicts the calculation above. Therefore, \((a_n)\) does not converge to a finite limit.
Check: The odd terms are all \(0\), but that does not prevent the even terms \(2,4,6,\ldots\) from becoming arbitrarily large. Convergence requires control of all sufficiently late terms, not just one selected pattern of terms.
A Common Mistake: Averaging Different Subsequences
Suppose a sequence has one subsequence tending to \(0\) and another tending to \(1\). A tempting but incorrect argument is to average those limits and declare that the original sequence tends to \(1/2\). Subsequence limits do not combine that way.
For a concrete check, let \(a_n=0\) for odd \(n\) and \(a_n=1\) for even \(n\). The two subsequences are constant, with limits \(0\) and \(1\). If the full sequence converged to \(1/2\), then eventually every term would be within \(1/4\) of \(1/2\). But every term is either \(0\) or \(1\), and
Thus no term is within \(1/4\) of \(1/2\), let alone every sufficiently late term. The correction is to use the subsequences as evidence against convergence: their different limits prove there is no single finite limit for the full sequence.
How to Use the Tests Carefully
These methods give sufficient evidence of divergence. To use them, first define the subsequence by a valid increasing index pattern, then calculate its limit or show that it is unbounded. Keep the conclusion precise:
- If two subsequences have different finite limits, the original sequence does not converge.
- If a subsequence is unbounded in absolute value, the original sequence does not converge to a finite limit.
- If the subsequences you examine have the same limit, that fact alone does not prove that the original sequence converges.
The last point matters. A few selected patterns may miss other terms that behave differently. Subsequence evidence is decisive when it reveals incompatible behavior, but conclusions must match what the evidence actually establishes.
Check Your Understanding
Use the subsequence criteria to answer each question. Give a reason for each conclusion.
- For \(a_n=3+(-1)^n\), find the limits of the even- and odd-indexed subsequences. What do they imply about \((a_n)\)?
- A sequence has a subsequence converging to \(4\) and another converging to \(4\). Does this establish that the full sequence converges? Explain.
- Suppose \(a_{3k}=k^2\) for every positive integer \(k\). What does this tell you about whether \((a_n)\) can converge to a finite limit?
- Why must the indices defining a subsequence be strictly increasing?
- What is the error in claiming that a sequence with subsequence limits \(0\) and \(1\) must converge to their average?
