A subsequence keeps some of the terms of a sequence, in their original order. Subsequences are the main tool for showing that a sequence does not converge, and they lead to the most important existence theorem about sequences in the next topic. They build on limits of sequences.
Key idea
A subsequence of $(a_n)$ is $(a_{n_k})$ where $n_1 < n_2 < n_3 < \cdots$ are positive integers. If $a_n \to L$, every subsequence tends to $L$. So if two subsequences have different limits, $(a_n)$ diverges. A subsequential limit is a number some subsequence converges to.
Why it works
Strictly increasing indices satisfy $n_k \ge k$ (induction: $n_1 \ge 1$, and $n_{k+1} > n_k \ge k$). If $a_n \to L$ and $\varepsilon > 0$, take $N$ with $|a_n - L| < \varepsilon$ for $n \ge N$. For $k \ge N$ we have $n_k \ge k \ge N$, so $|a_{n_k} - L| < \varepsilon$. $\blacksquare$ The proof uses the increasing indices in an essential way: they guarantee that the subsequence moves out along the sequence.
Worked examples
Worked example: Two subsequences
Show that $a_n = (-1)^{n}\frac{2n}{n + 1}$ diverges.
- Along even $n$: $a_{2k} = \frac{4k}{2k + 1} \to 2$.
- Along odd $n$: $a_{2k+1} = -\frac{2(2k + 1)}{2k + 2} \to -2$.
Answer: Two subsequences with different limits, so $(a_n)$ diverges.
Worked example: All subsequential limits
Find the set of subsequential limits of $a_n = \cos\left(\frac{n\pi}{3}\right)$.
- The values repeat with period $6$: $\frac{1}{2}, -\frac{1}{2}, -1, -\frac{1}{2}, \frac{1}{2}, 1$.
- Each value occurs infinitely often, so each is the limit of a constant subsequence.
- Any subsequence must eventually stay near one value (they are a fixed distance apart), so there are no others.
Answer: $\left\{-1, -\frac{1}{2}, \frac{1}{2}, 1\right\}$
Worked example: A subsequence along a pattern
Let $a_n = (-1)^{n}\frac{3n + 1}{n + 2}$. Find the limit of $a_{4k+3}$ as $k \to \infty$, and decide whether $a_{3k}$ converges.
- The index $4k + 3$ is always odd, so $a_{4k+3} = -\frac{3(4k + 3) + 1}{(4k + 3) + 2} \to -3$.
- The index $3k$ is even for even $k$ and odd for odd $k$, so $a_{3k}$ itself has subsequences tending to $3$ and to $-3$: it diverges.
Answer: $a_{4k+3} \to -3$; $(a_{3k})$ diverges.
The set of subsequential limits tells a lot. A bounded sequence converges exactly when this set has one element. In the next topic you will see that it is never empty for a bounded sequence, and two topics later its largest and smallest elements get names: lim sup and lim inf.
Common mistake
Reordering is not a subsequence. $a_2, a_1, a_4, a_3, \ldots$ uses every term but in a different order, so it is not a subsequence, and the theorem says nothing about it. The indices must strictly increase: $n_k = k^{2}$ and $n_k = 2k + 1$ are fine; $n_k = \lceil \frac{k}{2} \rceil$ (which repeats indices) is not. A second slip: one subsequence converging does not make the whole sequence converge.