When Oscillation Prevents a Limit
A sequence can fail to approach a single value because its terms keep returning to separated values. This persistent oscillation is different from fluctuations that become smaller and smaller. When two subsequences approach different limits, the full sequence cannot converge.
This observation gives a direct way to prove divergence for some infinite series. A convergent series must have terms approaching zero. If its terms have two subsequences approaching different values, then the terms do not approach zero, so the series diverges.
Subsequences and Their Limits
A subsequence is formed by selecting terms from a sequence in their original order, without skipping backward. More precisely, if \((a_n)\) is a sequence and \(n_1<n_2<n_3<\cdots\) are positive integers, then \((a_{n_j})\) is a subsequence of \((a_n)\). For example, \(a_1,a_3,a_5,\ldots\) is the odd-indexed subsequence, and \(a_2,a_4,a_6,\ldots\) is the even-indexed subsequence.
To see why the theorem holds, suppose \(a_n\to L\). Given any \(\varepsilon>0\), there is an index \(N\) such that \(|a_n-L|<\varepsilon\) whenever \(n\ge N\). The indices in any subsequence increase, so eventually they are all at least \(N\). Thus its terms also satisfy the same inequality and converge to \(L\). In particular, two subsequences of a convergent sequence cannot have different limits.
The contrapositive is especially useful: finding two subsequences with distinct limits proves that the original sequence has no limit. It is enough to identify the subsequences, calculate their limits, and check that those limits differ.
Why This Proves Series Divergence
An infinite series \(\sum_{n=1}^{\infty}a_n\) is defined through its partial sums \(S_N=\sum_{n=1}^{N}a_n\). The series converges if the sequence of partial sums converges to a finite limit. A necessary condition is that \(a_n\to0\). Indeed, for \(n\ge2\), \(a_n=S_n-S_{n-1}\); if \(S_n\to S\), then \(S_{n-1}\to S\), so \(a_n\to S-S=0\).
This is a one-way test. Distinct subsequential limits of the terms prove divergence, but failing to find such limits does not prove convergence. The test applies because the terms fail a necessary condition for series convergence.
Worked Examples
Worked Example: Terms Alternating Between 1 and −1
Consider the series \(\sum_{n=1}^{\infty}(-1)^{n+1}\). Its term sequence is \(a_n=(-1)^{n+1}\).
For the odd-indexed terms, set \(n=2k-1\), where \(k\in\mathbb N\). Then
Thus the odd-indexed subsequence is constantly \(1\), so its limit is \(1\). For the even-indexed terms, set \(n=2k\):
The even-indexed subsequence is constantly \(-1\), so its limit is \(-1\). Since \(1\ne-1\), the term sequence has no limit and does not approach zero. The series therefore diverges.
As a check, the partial sums satisfy \(S_{2k}=0\) and \(S_{2k-1}=1\). These two subsequences of partial sums have different limits, so the partial sums do not converge. This confirms the series-divergence conclusion directly.
Worked Example: Terms Alternating Between 2 and −3
Define \(a_n=2\) when \(n\) is odd and \(a_n=-3\) when \(n\) is even. Consider \(\sum_{n=1}^{\infty}a_n\).
Every odd-indexed term equals \(2\), so \(a_{2k-1}=2\) and \(\lim_{k\to\infty}a_{2k-1}=2\). Every even-indexed term equals \(-3\), so \(a_{2k}=-3\) and \(\lim_{k\to\infty}a_{2k}=-3\). These limits differ, and hence the terms do not converge to zero. The series diverges by the necessary term condition.
We can also check what happens to the partial sums. Each consecutive odd-even pair contributes \(2+(-3)=-1\). Therefore, after \(k\) complete pairs, \(S_{2k}=-k\). One more term adds \(2\), so \(S_{2k-1}=-(k-1)+2=3-k\). In particular, \(S_{2k}=-k\) becomes unbounded below as \(k\) increases. The partial sums cannot converge, in agreement with the term test.
The first terms are \(2,-3,2,-3\), giving partial sums \(2,-1,1,-2\). These values confirm the pair calculation: each completed pair reduces the partial sum by \(1\).
Worked Example: Alternation With a Shrinking Perturbation
Consider \(a_n=(-1)^n+\frac1n\) and the series \(\sum_{n=1}^{\infty}a_n\). Its terms do not alternate between exactly two fixed numbers, but their even- and odd-indexed subsequences can still be evaluated.
For \(n=2k\),
For \(n=2k-1\),
Here the added fractions tend to zero, so the two subsequential limits are \(1\) and \(-1\). Since they differ, the terms do not approach zero and the series diverges.
Checking the first values gives \(a_1=0\), \(a_2=\frac32\), \(a_3=-\frac23\), and \(a_4=\frac54\). These agree with the formulas: the even terms are \(1+\frac{1}{2k}\), while the odd terms are \(-1+\frac{1}{2k-1}\). In particular, the initial zero term does not change the distinct limits of the two subsequences.
A General Proof for Two Different Limits
The examples use odd and even terms, but the same reasoning applies to any two subsequences. The following proof makes the argument explicit.
Let \(a_{n_j}\to L_1\) and \(a_{m_j}\to L_2\), where \(L_1\ne L_2\).
Write \(a_n\to L\) for its limit. The subsequence theorem then implies \(a_{n_j}\to L\) and \(a_{m_j}\to L\).
A real sequence cannot converge to two different limits, so \(L_1=L\) and \(L_2=L\). This gives \(L_1=L_2\), contradicting the assumption that they differ.
The contradiction shows that \(a_n\) does not converge. In particular, it cannot converge to zero, so a series with these terms diverges.
For completeness, uniqueness of limits follows directly from the triangle inequality. If a sequence had limits \(L_1\) and \(L_2\), then for every \(\varepsilon>0\), all sufficiently late terms would satisfy both \(|a_n-L_1|<\varepsilon/2\) and \(|a_n-L_2|<\varepsilon/2\). Hence
This holds for every \(\varepsilon>0\), which forces \(|L_1-L_2|=0\), and therefore \(L_1=L_2\). Thus the contradiction in the proof is valid.
Common Pitfall: Treating All Alternation as Divergence
The word “oscillating” alone does not establish that a sequence has no limit. What matters here is that the subsequences approach different values. If fluctuations shrink toward a common value, the terms may still converge. For example, the alternating signs in \((-1)^n/n\) do not give two different subsequential limits: both the even and odd terms tend to zero. This observation by itself does not determine whether a series converges; it only shows that the distinct-subsequential-limits argument is unavailable.
Another error is to confuse the sequence of terms with the sequence of partial sums. The term test concerns \(a_n\), while convergence of a series is defined by the partial sums \(S_N\). The logical link is the necessary condition: convergence of \(S_N\) forces \(a_n=S_n-S_{n-1}\to0\). Therefore, showing that the terms have distinct subsequential limits is enough to establish divergence, without needing to find a formula for every partial sum.
Check Your Understanding
For each question, identify the relevant subsequences before drawing a conclusion about the sequence or series.
- If \(a_{2k-1}=4\) and \(a_{2k}=-1\) for every \(k\in\mathbb N\), what are the two subsequential limits? What follows for \(\sum_{n=1}^{\infty}a_n\)?
- Why must every subsequence of a convergent sequence have the same limit as the full sequence?
- Suppose the odd terms of \((a_n)\) tend to \(0\) and the even terms tend to \(5\). Does \((a_n)\) converge? Does the series with terms \(a_n\) converge?
- What necessary condition for series convergence is violated when the terms have two distinct subsequential limits?
- Why does alternating sign, by itself, not prove that a term sequence has no limit?
