Why Separate Even and Odd Terms?
Some sequences follow different rules depending on whether the index is even or odd. Others contain an alternating sign, so their terms switch between positive and negative values. In both situations, looking at the even-indexed terms and the odd-indexed terms separately can reveal the long-term behavior clearly.
The key is not simply to find two limits. Their relationship determines what happens to the original sequence. If both groups of terms approach the same number, the whole sequence converges to that number. If they approach different numbers, the full sequence cannot converge.
Even- and Odd-Indexed Subsequences
Throughout, the sequence is indexed by \(n\in\mathbb{N}=\{1,2,3,\dots\}\). Its even-indexed terms are \(a_2,a_4,a_6,\dots\), and its odd-indexed terms are \(a_1,a_3,a_5,\dots\). To write each group as a sequence indexed from 1, use \(k\in\mathbb{N}\): the even terms are \(a_{2k}\), and the odd terms are \(a_{2k-1}\).
For example, when \(k=1,2,3\), the even subsequence gives \(a_2,a_4,a_6\), while the odd subsequence gives \(a_1,a_3,a_5\). The index \(k\) counts the terms within a subsequence; it is not the original index of the full sequence.
A subsequence of a convergent sequence converges to the same limit. In particular, if \(a_n\to L\), then both \(a_{2k}\to L\) and \(a_{2k-1}\to L\). The useful converse also holds for these two parity subsequences: if they both converge to the same number, then the full sequence converges to that number.
The Parity-Limit Criterion
The forward direction follows from the fact that subsequences of a convergent sequence have the same limit. For completeness, the reverse direction explains why the parity test works.
For any \(\varepsilon>0\), there is an integer \(K_1\) such that \(k\ge K_1\) implies \(|a_{2k}-L|<\varepsilon\). There is also an integer \(K_2\) such that \(k\ge K_2\) implies \(|a_{2k-1}-L|<\varepsilon\).
Set \(K=\max(K_1,K_2)\). If \(n\ge 2K\) is even, then \(n=2k\) for some \(k\ge K\), so \(|a_n-L|<\varepsilon\). If \(n\ge 2K\) is odd, then \(n=2k-1\); the inequality \(2k-1\ge 2K\) gives \(k\ge K+1\), so again \(|a_n-L|<\varepsilon\).
Every integer \(n\ge 2K\) is either even or odd, and in either case \(|a_n-L|<\varepsilon\). This is precisely the definition of \(a_n\to L\).
The proof depends on covering every sufficiently large index. Even and odd indices together cover all positive integers, so controlling both groups with the same limit controls the entire sequence.
Worked Example: An Alternating Sequence That Converges
Worked Example: An Alternating Sequence That Converges
Determine whether \(a_n=\dfrac{(-1)^n}{n}\) converges, and find its limit if it does.
First substitute the even index \(n=2k\). Since \((-1)^{2k}=1\), the even subsequence is
As \(k\to\infty\), the denominator \(2k\) grows without bound, so \(a_{2k}\to 0\).
Now substitute the odd index \(n=2k-1\). Since \((-1)^{2k-1}=-1\), the odd subsequence is
As \(k\to\infty\), the denominator \(2k-1\) grows without bound, so \(a_{2k-1}\to 0\) as well. Both parity subsequences converge to the same value. By the parity-limit criterion,
Check: The first terms are \(-1,\frac12,-\frac13,\frac14,\dots\). Their magnitudes are \(1, \frac12, \frac13, \frac14,\dots\), which tend to zero; the alternating sign does not prevent the terms from approaching zero. This agrees with the two-subsequence calculation.
Worked Example: Different Parity Limits Prevent Convergence
Worked Example: Different Parity Limits Prevent Convergence
Suppose a sequence is defined by \(a_{2k}=3\) and \(a_{2k-1}=1\) for every \(k\in\mathbb{N}\). Does the full sequence converge?
The even subsequence is constant:
The odd subsequence is also constant, but at a different value:
Since \(3\ne 1\), the two parity subsequences do not have a common limit. Therefore the full sequence does not converge. Its terms are \(1,3,1,3,\dots\), so the two distinct values recur indefinitely.
Check: If the full sequence had a limit \(L\), both of its parity subsequences would have to converge to \(L\). The even subsequence forces \(L=3\), while the odd subsequence forces \(L=1\). These requirements contradict one another, confirming nonconvergence.
Worked Example: Different Formulas with the Same Limit
Worked Example: Different Formulas with the Same Limit
Consider the sequence defined, for \(k\in\mathbb{N}\), by
Find the limit of the full sequence, or show that it does not converge.
For the even terms, rewrite the numerator using the denominator:
As \(k\to\infty\), \(\frac{1}{2k+1}\to 0\), so \(a_{2k}\to 2\).
For the odd terms, use the corresponding rewrite:
As \(k\to\infty\), \(\frac{1}{3k-1}\to 0\), so \(a_{2k-1}\to 2\). Both parity subsequences approach \(2\), and therefore the full sequence converges to \(2\).
Check: The even formula is exactly \(2+\frac{1}{2k+1}\), and the odd formula is exactly \(2+\frac{1}{3k-1}\). Each added fraction tends to zero, so each subsequence approaches \(2\), as required by the conclusion.
A Common Mistake: Averaging Two Different Limits
Knowing the two parity limits does not mean that their average is the limit of the full sequence. For the sequence in the second worked example, the two limits are \(3\) and \(1\). A tempting but incorrect calculation is
This conclusion is false. The calculation averages two numbers, but convergence requires that every sufficiently late term be close to one common number. Here the even terms remain \(3\), and the odd terms remain \(1\); neither group approaches \(2\). The parity-limit criterion requires equality of the two limits, not their average.
Why the Parity Test Is Useful
The test is especially effective when a formula contains \((-1)^n\), or when the definition itself splits into even and odd cases. Substituting \(n=2k\) and \(n=2k-1\) removes the ambiguity about which terms are being examined and often simplifies the formulas.
It also prevents a common error: inferring convergence from one subsequence alone. A sequence can have an even subsequence that converges while its odd subsequence approaches another value, or does not converge at all. The full sequence includes both groups, so both must be checked.
When the two parity formulas are complicated, evaluate each limit using the appropriate sequence limit laws. Keep the original index distinction clear, simplify each subsequence, and only then compare the results. The conclusion concerns the whole sequence only after both groups have been accounted for.
Check Your Understanding
For each question, use the even- and odd-indexed subsequences to justify your conclusion.
- Write the even- and odd-indexed subsequences of \(a_n=(-1)^n\), and decide whether \(a_n\) converges.
- If \(a_{2k}\to 4\) and \(a_{2k-1}\to 4\), what is the limit of the full sequence?
- If \(a_{2k}\to -2\) and \(a_{2k-1}\to 5\), what can you conclude about the full sequence?
- A sequence has \(a_{2k}=1+\frac{1}{k}\) and \(a_{2k-1}=1-\frac{1}{k}\). Determine whether it converges and find its limit.
- Why does the average of two unequal parity-subsequence limits not establish a limit for the full sequence?