$(a_n)_{n \in \mathbb{N}}$ 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!
Best Answer
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.