Nonconvergence from Subsequences

Why Look at Subsequences?

A sequence may fluctuate so much that its overall behavior is difficult to read from its formula. Instead of trying to track every term, we can examine carefully chosen terms: the first, third, and fifth terms; the terms at selected indices; or any other increasing sequence of positions. These selected terms form a subsequence.

Subsequences are useful for detecting a decisive obstruction to convergence. If one selection of terms approaches one value while another selection approaches a different value, the entire sequence cannot settle toward a single limit.

Key idea: Every subsequence of a convergent sequence must approach the same limit as the original sequence. Therefore, two subsequences with distinct limits prove that the original sequence does not converge.

Definition of a Subsequence

Let \((a_n)\) be a sequence. A subsequence is a sequence of the form \((a_{n_k})\), where \(n_1,n_2,\ldots\) are positive integer indices satisfying

n_1<n_2<n_3<\cdots.

The indices must increase strictly. This condition means that a subsequence selects terms from the original sequence in their original order, although it may skip as many terms as it likes.

For example, choosing indices \(n_k=3k\) gives the subsequence \(a_3,a_6,a_9,\ldots\). A list that repeats an index or goes backward is not a subsequence.

Recall that \(a_n\) converges to \(L\) if, for every \(\varepsilon>0\), there is an \(N\in\mathbb{N}\) such that

n\geq N \quad\Longrightarrow\quad |a_n-L|<\varepsilon.

In words, all sufficiently late terms of the sequence lie within any prescribed positive distance of \(L\).

Why Convergent Sequences Pass the Subsequence Test

Theorem: If \(a_n\to L\) and \((a_{n_k})\) is any subsequence of \((a_n)\), then \(a_{n_k}\to L\).

The reason is that a subsequence cannot keep selecting early terms forever. Its indices increase, so they eventually pass any fixed index bound.

Start with an arbitrary tolerance.
Let \(\varepsilon>0\). Since \(a_n\to L\), there is an \(N\) such that \(n\geq N\) implies \(|a_n-L|<\varepsilon\).

Use the increasing indices.
Because \(n_1<n_2<\cdots\) are positive integers, \(n_k\geq k\). Thus, whenever \(k\geq N\), we have \(n_k\geq N\).

Apply the original convergence condition.
For \(k\geq N\), the bound on the original sequence gives \(|a_{n_k}-L|<\varepsilon\). This holds for every positive \(\varepsilon\), so \(a_{n_k}\to L\).

This proves the theorem directly from the definition of convergence. The crucial point is that the same limit \(L\) works for every subsequence.

Nonconvergence test: If a sequence has two subsequences that converge to distinct finite limits, then the original sequence does not converge.

To justify this test, suppose instead that \(a_n\to L\). The theorem would force both subsequences to converge to \(L\). A sequence cannot have two different limits: if it converged to \(P\) and \(Q\), choose \(\varepsilon=|P-Q|/3\). For all sufficiently large \(n\), both \(|a_n-P|<\varepsilon\) and \(|a_n-Q|<\varepsilon\), so the triangle inequality would give

|P-Q|\leq |P-a_n|+|a_n-Q| <\frac{2}{3}|P-Q|,

which is impossible when \(P\ne Q\). Consequently, two distinct subsequential limits contradict the assumption that the original sequence converges.

Worked Examples

Worked Example: An Oscillation in the Logarithm of the Index

Show that \(a_n=\sin(\log n)\) does not converge.

We choose indices near exponentials whose logarithms are angles where sine equals \(1\) or \(-1\). For \(k\in\mathbb{N}\), define

u_k=\left\lceil e^{2\pi k+\pi/2}\right\rceil, \qquad v_k=\left\lceil e^{2\pi k+3\pi/2}\right\rceil.

Here \(\lceil x\rceil\) is the least integer greater than or equal to \(x\). Both index sequences are strictly increasing: their underlying exponential quantities are multiplied by \(e^{2\pi}>1\) when \(k\) increases by one, and the quantities are already much larger than one. Thus these choices produce subsequences.

Set \(T_k=e^{2\pi k+\pi/2}\). Since \(T_k\leq u_k<T_k+1\),

0\leq \log u_k-\left(2\pi k+\frac{\pi}{2}\right) =\log\left(\frac{u_k}{T_k}\right) <\log\left(1+\frac{1}{T_k}\right)\longrightarrow 0.

Write this difference as \(\delta_k\). Then \(\delta_k\to0\), and periodicity of sine gives

a_{u_k} =\sin\left(2\pi k+\frac{\pi}{2}+\delta_k\right) =\cos(\delta_k)\longrightarrow 1.

For the other subsequence, the same rounding estimate gives \(\log v_k-(2\pi k+3\pi/2)\to0\). Therefore, using the identity \(\sin(3\pi/2+\delta)=-\cos(\delta)\),

a_{v_k} =\sin(\log v_k)\longrightarrow -1.

The two subsequences have distinct limits, \(1\) and \(-1\). Hence \((\sin(\log n))\) does not converge.

Check: The estimates show that the rounding changes each target logarithm by a quantity tending to zero. The corresponding sine values therefore approach the claimed values, rather than merely matching them exactly.

Worked Example: A Cosine Oscillation with a Vanishing Term

Show that \(b_n=\cos(\log n)+2/n\) does not converge.

For \(k\in\mathbb{N}\), choose \(p_k=\lfloor e^{2\pi k}\rfloor\) and \(q_k=\lfloor e^{(2k+1)\pi}\rfloor\), where \(\lfloor x\rfloor\) is the greatest integer less than or equal to \(x\). Both index sequences strictly increase: the exponential targets increase by factors greater than one, and the gaps between successive targets exceed one.

If \(T=e^{2\pi k}\), then \(T-1<p_k\leq T\), so \(p_k/T\to1\). Continuity of the logarithm at \(1\) gives \(\log(p_k/T)\to0\), and hence \(\log p_k-2\pi k\to0\). Also \(2/p_k\to0\). Thus

b_{p_k} =\cos(\log p_k)+\frac{2}{p_k} \longrightarrow 1+0=1.

Applying the same floor estimate to \(q_k\) gives \(\log q_k-(2k+1)\pi\to0\), and \(2/q_k\to0\). Since \(\cos((2k+1)\pi)=-1\),

b_{q_k} =\cos(\log q_k)+\frac{2}{q_k} \longrightarrow -1+0=-1.

These are distinct subsequential limits, so \((b_n)\) does not converge. The added term \(2/n\) tends to zero and does not remove the persistent oscillation.

Check: Along the first subsequence the cosine term tends to \(1\), and along the second it tends to \(-1\); in both cases the extra term tends to zero.

Worked Example: Oscillation in the Square Root of the Index

Show that \(c_n=\sin(\sqrt n)\) does not converge.

For \(k\in\mathbb{N}\), let \(r_k=\lfloor(2\pi k+\pi/2)^2\rfloor\). These indices are positive and strictly increasing because the squared targets increase by more than one between successive \(k\). Put \(t_k=2\pi k+\pi/2\). The floor definition gives \(0\leq t_k^2-r_k<1\), and therefore

0\leq t_k-\sqrt{r_k} =\frac{t_k^2-r_k}{t_k+\sqrt{r_k}} <\frac{1}{t_k+\sqrt{r_k}} \longrightarrow 0.

Thus \(\sqrt{r_k}-t_k\to0\), so continuity and periodicity of sine imply \(c_{r_k}\to1\). Now define \(s_k=\lfloor(2\pi k+3\pi/2)^2\rfloor\). The same calculation, with \(w_k=2\pi k+3\pi/2\), gives \(w_k-\sqrt{s_k}\to0\). Hence

c_{s_k} =\sin(\sqrt{s_k})\longrightarrow \sin\left(2\pi k+\frac{3\pi}{2}\right)=-1,

where the displayed limiting statement means that the difference between \(\sqrt{s_k}\) and its target angle tends to zero. Since the two subsequences approach \(1\) and \(-1\), the original sequence does not converge.

Check: The square-root estimate follows by multiplying \(t_k-\sqrt{r_k}\) by \(t_k+\sqrt{r_k}\); its numerator is less than one and its denominator grows without bound.

A Common Mistake

Finding one convergent subsequence is not enough to prove that the original sequence converges. For \(a_n=\sin(\log n)\), the calculation above shows that \(a_{u_k}\to1\). The incorrect conclusion would be: “A subsequence tends to \(1\), so \(a_n\to1\).” The theorem proved earlier only says that if the original sequence converges to \(L\), then every subsequence converges to \(L\). It does not say that one convergent subsequence forces convergence of the whole sequence.

In this example the second subsequence, \(a_{v_k}\to-1\), directly exposes the error. The two subsequences cannot both have the same limit required of every subsequence of a convergent sequence.

Use the test in the right direction: Distinct subsequential limits prove nonconvergence. A single convergent subsequence, or several subsequences with the same limit, does not by itself prove convergence of the full sequence.

When the Test Is Useful

The test is especially effective when a sequence repeatedly visits different regions of the real line. Rather than estimate all terms at once, identify two index patterns that keep the terms near different values. The subsequences must be genuine selections in increasing index order, and their limits must actually be computed or justified. Once those conditions are met, no analysis of the remaining terms is needed.

This method gives a conclusive proof, not just numerical evidence. A finite list of terms can suggest oscillation, but it cannot establish that the pattern persists indefinitely. Subsequence limits do establish persistent behavior: infinitely late terms along one selection approach one value, while infinitely late terms along another selection approach a different value.

Check Your Understanding

For each question, explain how the subsequence principle applies.

What two conditions must the indices \(n_k\) satisfy for \(a_{n_k}\) to be a subsequence?
Why must every subsequence of a convergent sequence have the same limit as the original sequence?
If two subsequences approach \(4\) and \(-2\), what can you conclude about the original sequence?
Why does one subsequence converging to \(L\) fail to prove that the original sequence converges to \(L\)?
In the example \(b_n=\cos(\log n)+2/n\), which part of the formula vanishes along both selected subsequences, and which part produces the distinct limits?