Is it possible to have a convergent subsequence of a divergent sequence?

Question

Is it possible to have a convergent subsequence of a divergent sequence? Thanks!

Answer 1

Sure. Consider $0, 1, 0, 1, 0, 1, \dots$

Furthermore, the Bolzano-Weierstrass Theorem says that every bounded sequence has a convergent subsequence.

Answer 2

It depends on your definition of divergence: If you mean non-convergent, then the answer is yes; If you mean that the sequence "goes to infinity", than the answer is no.

Answer 3

Another example: Let $(x_n)=\sin(\frac{n\pi}{2})$. Obviously $(x_n)$ is a non-convergent sequence, buy if you consider the subsecuence: $(x_n)_{n=2k}$, with $k\in \mathbb{N}$, then $(x_n)_{n=2k}\to 0$.

Answer 4

Another example:

By Fundamental arithmetic theorem we know that for every $n>1$: $$n=p_{1}^{\alpha_{1}}\times p_{2}^{\alpha_{2}}\times\cdots \times p_{n}^{\alpha_{n}}$$ As there are finitely many primes in the factorization, we can choose the smallest:

$$p^{(n)}=\min\{p_{i}: i=1,\cdots, n\}$$

So you can define the sequence:

$$a_{n}= p^{(n)}$$

which is clearly divergent. But for any prime number $p$, you can extract a subsequence of $a_n$ that converges to p as follows:

Consider $N_{p}=\{n\in \mathbb{N}: p^{(n)}=p\}$.

The subsequence of $a_{n}$ defined over this set is the constant sequence $p$ that converges to $p$.

Answer 5

Consider the sequence $(-1)^n$. You can argue that it has a convergent subsequence in two ways:

(1) It is bounded, so by the Bolzano Weirstrass Theorem, it has a convergent subsequence.

(2) Observe that the subsequence, $(1,1,1,1,1,1,1...)$ is convergent (since it is constant).