$(a_n)_{n \in \mathbb{N}}$ be a sequence with the convergent subsequences $(a_{2n}), (a_{2n+1})$ and $(a_{3n})$. Is $(a_n)$ then convergent? Proof or counter-example.

My idea with this question was to show the following equation as counter-example.$$a_n = \cases{ 50 - \frac{1}{n} \text{for 2n}\\-100-\frac{1}{n} \text{for 2n+1}}$$

Of course, $(a_{2n}), (a_{2n+1})$ and $(a_{3n})$ are convergent, but is the sequence divergent? The definition of divergence is like this $$\forall a \in \mathbb{R}\exists \epsilon \gt 0 \forall n_0 \in \mathbb{N} \exists n \ge n_0: |a_n - a| \ge \epsilon$$

Now, $|a_n - a| \ge 0$, so if $|a_n - a| \gt 0$ then there's obviously an $\epsilon$ such that the equation is true. But what if it's 0? And, BTW, is it actually divergent or should I look for another example or try to proof that such a sequence is always convergent? Thank you for your help!

  • $\begingroup$ Your sequence is divergent, but the subsequence $(a_{3n})$ doesn't converge, so it doesn't give you a counter-example. $\endgroup$
    – mdp
    May 3, 2013 at 10:29
  • 2
    $\begingroup$ In your example, $(a_{3n})$ is not convergent. For the general case, note that $(a_{3n})$ has infinitely many terms in common with $(a_{2n})$ and infinitely many terms in common with $(a_{2n+1})$ ... $\endgroup$ May 3, 2013 at 10:30

2 Answers 2


Let $(a_{2n}) \to L$ and $(a_{3n}) \to H$ then we can show $H=L$ by using the fact that a convergent sequence is also Cauchy.

So as $(a_{2n})$ converges it is also cauchy and we can take $N_0$ such that for all $m \ge n \ge N_0$ we have that $|a_{2n} - a_{2m}| < \epsilon_0$. In particular take $m=3n$. Then we have that: $|a_{2n} - a_{3*2n}| < {\epsilon \over 2} \tag{1}$

Also as we know that $(a_{3n})$ goes to $H$ we can say that $|a_{6n} - H| < {\epsilon \over 2} \tag{2}$ for $n \ge N_1$.

So we can then ask what is $|a_{2n} - H|$? If we take $n \ge \max (N_0,N_1)$ then using $(1),(2)$ and the triangle inequality we get:

$$|a_{2n} - H| \le |a_{2n}-a_{6n}| + |H-a_{6n}| < {\epsilon \over 2 } + {\epsilon \over 2} = \epsilon$$

And so all the even numbers tend to the same number as the multiples of 3. An analogous argument for the odd numbers will prove the claim.


A subsequence of a convergent sequence converges to the same limit.

Let $(a_{2n})$ converge to $l_1$, $(a_{2n+1})$ to $l_2$ and $(a_{3n})$ to $l_3$.

$(a_{6n})$ is a subsequence both of $(a_{2n})$ and of $(a_{3n})$, so it converges to both $l_1$ and $l_3$, so $l_1=l_3$.

$(a_{6n+3})$ is a subsequence both of $(a_{2n+1})$ and of $(a_{3n})$, so it converges to both $l_2$ and $l_3$, so $l_2=l_3$.

So $l_1=l_2=l_3=l$, say. Moreover, $(a_{2n})$ and $(a_{2n+1})$ comprise the whole of $(a_n)$. So $(a_n)$ is Cauchy, and thus convergent, with limit $l$.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .