Every Subsequence of a Convergent Sequence Converges

Convergence Survives Every Increasing Selection

A convergent sequence settles near its limit: after some index, every term is close to that limit. Selecting terms at increasingly large indices cannot undo that eventual closeness. Earlier in this course, the theorem Subsequences of a Convergent Sequence established the precise result: if $a_n\to L$, then every subsequence of $(a_n)$ also converges to $L$. We will use that theorem, not reprove it here, and examine how to apply it in settings where the selected indices are less immediate.

The index condition matters. A subsequence has the form $(a_{n_k})_{k=0}^{\infty}$, where $n_0<n_1<n_2<\cdots$. In particular, its indices keep moving forward; it cannot repeatedly select one fixed term or return to an earlier term. The fact that these indices increase is what ensures a subsequence eventually samples only the tail where the original sequence is close to its limit.

The inheritance theorem gives more than a conclusion for one familiar choice, such as even indices. It applies to any strictly increasing index sequence: squares, prime indices, or indices chosen through several stages. It also gives a useful check on arguments: once convergence of the original sequence is known, there is no need to estimate a complicated selected sequence from scratch.

Applying the Inheritance Theorem

Worked Example: Selecting Square-Related Indices

Let $$ a_n=5+\frac{2}{n+1}. $$ Since $2/(n+1)\to0$, the limit laws give $a_n\to5$. Choose the indices $$ n_k=k^2+2k \qquad (k\in\mathbb{N}_0). $$ They are strictly increasing, because $$ n_{k+1}-n_k=(k+1)^2+2(k+1)-(k^2+2k)=2k+3>0. $$ Thus $(a_{n_k})$ is a subsequence. Substitution gives $$ a_{n_k}=5+\frac{2}{k^2+2k+1} =5+\frac{2}{(k+1)^2}. $$ The selected terms converge to $5$. This also follows immediately from the inheritance theorem, without calculating the selected formula: since $a_n\to5$, every subsequence, including this one, converges to $5$.

The calculation provides an independent check of the conclusion and shows how a selected index can simplify an expression. The theorem itself does not require such simplification; it applies even when the formula for $a_{n_k}$ is awkward.

Worked Example: A Subsequent Selection Keeps the Same Limit

Use the sequence and indices from the first example. Now select every third term of the subsequence by setting $m_j=3j+1$. These indices are strictly increasing, since $m_{j+1}-m_j=3>0$. The resulting sequence is $$ a_{n_{m_j}}=5+\frac{2}{(m_j+1)^2} =5+\frac{2}{(3j+2)^2}. $$ For every $j$, the difference from $5$ is positive and satisfies $$ \left|a_{n_{m_j}}-5\right|=\frac{2}{(3j+2)^2}. $$ As $j$ increases, the denominator grows without bound, so this difference tends to zero. Hence $a_{n_{m_j}}\to5$.

There is also a theorem-based explanation. The sequence $(a_{n_k})$ is a subsequence of $(a_n)$, and $(a_{n_{m_j}})$ is a subsequence of $(a_{n_k})$. The earlier theorem that a subsequence of a subsequence is a subsequence of the original sequence shows that this final selection is itself a subsequence of $(a_n)$. The inheritance theorem therefore gives its limit as $5$. A second round of selection does not create a new limit.

Takeaway: Once $a_n\to L$ is established, every subsequence has limit $L$, including a subsequence formed by selecting from another subsequence. The selected indices may change; the limit does not.

Finite Interleavings and Convergence

The inheritance theorem gives a one-way implication: convergence of the original sequence forces convergence of each subsequence to the same limit. There is also a useful converse when a finite collection of subsequences accounts for all the original indices. In that case, convergence along each part controls the entire sequence.

Theorem (Finite Interleaving Criterion): Suppose $\mathbb{N}_0$ is partitioned into finitely many infinite sets $E_1,\ldots,E_r$. Enumerate each set in increasing order as $E_j=\{n_k^{(j)}:k\in\mathbb{N}_0\}$. For a real number $L$, the sequence $a_n$ converges to $L$ if and only if each sequence $(a_{n_k^{(j)}})_{k=0}^{\infty}$ converges to $L$.

Proof. If $a_n\to L$, each sequence $(a_{n_k^{(j)}})$ is a subsequence of $(a_n)$, so it converges to $L$ by the theorem Subsequences of a Convergent Sequence.

For the converse, suppose each of the finitely many subsequences converges to $L$. Let $\varepsilon>0$. For each $j\in\{1,\ldots,r\}$, convergence provides an index $K_j\in\mathbb{N}_0$ such that $$ |a_{n_k^{(j)}}-L|<\varepsilon $$ whenever $k\geq K_j$. Define $$ M=\max\{n_{K_1}^{(1)},n_{K_2}^{(2)},\ldots,n_{K_r}^{(r)}\}. $$ This maximum exists because there are only finitely many real numbers in the set.

Take any $n\geq M$. Since the sets partition $\mathbb{N}_0$, $n$ belongs to exactly one set $E_j$. Write $n=n_k^{(j)}$. The enumeration of $E_j$ is strictly increasing. If $k<K_j$, then $n_k^{(j)}<n_{K_j}^{(j)}\leq M$, contradicting $n\geq M$. Therefore $k\geq K_j$, and consequently $$ |a_n-L|=|a_{n_k^{(j)}}-L|<\varepsilon. $$ This holds for every $n\geq M$. Since $\varepsilon>0$ was arbitrary, the definition of convergence gives $a_n\to L$. $\square$

Worked Example: Checking Even and Odd Terms Together

Suppose the even-indexed terms and the odd-indexed terms of $(a_n)$ both converge to $2$: $$ a_{2k}\to2 \qquad\text{and}\qquad a_{2k+1}\to2. $$ The even indices $E_1=\{0,2,4,\ldots\}$ and the odd indices $E_2=\{1,3,5,\ldots\}$ partition $\mathbb{N}_0$. The Finite Interleaving Criterion therefore gives $a_n\to2$.

To see the threshold argument explicitly, fix $\varepsilon>0$. There are $K_1,K_2\in\mathbb{N}_0$ such that $$ |a_{2k}-2|<\varepsilon\quad(k\geq K_1), \qquad |a_{2k+1}-2|<\varepsilon\quad(k\geq K_2). $$ Set $M=\max\{2K_1,2K_2+1\}$. For $n\geq M$, if $n=2k$, then $n\geq2K_1$, so $k\geq K_1$ and $|a_n-2|<\varepsilon$. If $n=2k+1$, then $n\geq2K_2+1$, so $k\geq K_2$ and again $|a_n-2|<\varepsilon$. Every index is even or odd, so the estimate holds for all $n\geq M$.

The finite hypothesis is essential to this proof: it lets us choose one threshold that works across all parts by taking a maximum. For infinitely many interleaved subsequences, the individual thresholds need not have a finite common bound. Thus, convergence on each part of an infinite partition does not by itself permit the same argument.

A Further-Subsequence Test for Convergence

A convergent sequence passes its limit to every subsequence. A related question reverses the direction: what if every subsequence has some further subsequence that converges to a specified $L$? This condition is strong enough to force the original sequence to converge to $L$. The proof explains a common strategy in analysis: if convergence fails, select infinitely many terms that stay a fixed positive distance away.

Theorem (Further-Subsequence Criterion for Convergence): Let $(a_n)$ be a real sequence and $L\in\mathbb{R}$. If every subsequence of $(a_n)$ has a further subsequence converging to $L$, then $a_n\to L$.

Proof. We prove the contrapositive. Suppose $a_n$ does not converge to $L$. By the epsilon-$N$ definition, there is an $\varepsilon_0>0$ such that for every $N\in\mathbb{N}_0$, some $n\geq N$ satisfies $$ |a_n-L|\geq\varepsilon_0. $$ Choose such an index $n_0$ for $N=0$. Having chosen $n_k$, apply the same condition with $N=n_k+1$ to choose $n_{k+1}>n_k$ satisfying $$ |a_{n_{k+1}}-L|\geq\varepsilon_0. $$ This constructs a subsequence $(a_{n_k})$ whose every term has distance at least $\varepsilon_0$ from $L$.

No further subsequence of $(a_{n_k})$ can converge to $L$. Indeed, every term in any such further subsequence still satisfies the same lower bound. If it converged to $L$, the definition of convergence with tolerance $\varepsilon_0$ would require all sufficiently late terms to satisfy $|a_{n_k}-L|<\varepsilon_0$, a contradiction. We have found a subsequence with no further subsequence converging to $L$, contrary to the stated hypothesis. Therefore $a_n\to L$. $\square$

Worked Example: Why One Convergent Subsequence Is Not Enough

Consider $$ a_n=(-1)^n. $$ The even-indexed subsequence is constant: $$ a_{2k}=(-1)^{2k}=1, $$ so it converges to $1$. But the full sequence does not converge to $1$: for every $N$, there is an odd $n\geq N$, and at each such index $$ |a_n-1|=|-1-1|=2. $$ Thus the epsilon-$N$ condition fails for $L=1$ and, for example, $\varepsilon=1$.

This does not conflict with the inheritance theorem. That theorem starts with convergence of the full sequence and concludes convergence of every subsequence. A single convergent subsequence gives no such conclusion about the original sequence. The Further-Subsequence Criterion uses a stronger hypothesis: every subsequence must have a further subsequence converging to the specified $L$.

How to Use the Inheritance Principle

When the sequence is already known to converge, use the inheritance theorem as a shortcut. Verify that the chosen indices are strictly increasing, identify the selected terms as a subsequence, and carry over the original limit. If there are nested selections, the theorem that a subsequence of a subsequence is a subsequence of the original sequence keeps the reasoning organized.

When the goal is to establish convergence from selected pieces, first check that those pieces cover all indices and that there are only finitely many of them. The Finite Interleaving Criterion then combines their thresholds. When the hypothesis instead concerns further subsequences, test convergence by contradiction: failure supplies a fixed tolerance and infinitely many later terms violating it, which can be assembled into a subsequence.

These uses highlight the direction of each argument. Convergence of the whole sequence guarantees convergence along every selection. Convergence along a finite partition can be combined to recover convergence of the whole. A lone convergent subsequence, however, is not enough; it may simply select the terms that behave well while other terms remain far from its limit.

Check Your Understanding

Use the inheritance theorem and the results proved in this tutorial to answer the following questions.

If $a_n\to L$, what is the limit of a subsequence whose indices are strictly increasing?
Why does selecting a further subsequence from an already selected subsequence preserve the original limit?
In the Finite Interleaving Criterion, where does the finiteness of the partition enter the proof?
If the even-indexed and odd-indexed terms both converge to $L$, what can be concluded about the full sequence, and why?
What fixed-distance property can be used to construct a subsequence when $a_n$ fails to converge to $L$?
Why does the existence of one subsequence converging to $L$ not imply that the original sequence converges to $L$?