If $(a_{2n+1})$ and $(a_{2n})$ converge to $a$ then $(a_n)$ converges to $a$.

Question

If $(a_{2n+1})$ and $(a_{2n})$ converge to $a$ then $(a_n)$ converges to $a$.

So far I realize that if $(a_{2n+1})$ and $(a_{2n})$ converge then for each $\epsilon>0$, there exists $N$ such that for all $n>N$, $|a_{2n+1}|, |a_{2n}| < \epsilon$. As eventually even and odd values of $n$ converge (which is to say that eventually for all natural numbers $n>N$) $a_n$ converges.

I just don't know how to prove it. Perhaps I could show that the set of all odd and even numbers is the set of Natural numbers but I'm struggling to prove it properly.

P.S. Sorry for the simplicity of the question, I'm not used to proving such things.

Consider $|a_n-l|\leq\max(|a_{2E(n/2)}-l|,|a_{2E(n/2)+1}-l|)$ were l is the shared limit of $a_{2n}$ and $a_{2n+1}$ — Bertrand R, Oct 20 '13 at 22:13
The statement in your first sentence is only true for $a=0$. — Carsten S, Oct 21 '13 at 15:21

score 5 · Accepted Answer · answered Oct 20 '13 at 22:14

By assumption, for $\epsilon>0$ there exists $N_o$ such that $n>N_o$ implies $|a_{2n+1}-a|<\epsilon$ and there exists $N_e$ such that $n>N_e$ implies $|a_{2n}-a|<\epsilon$. I claim that for $N:=\max\{2N_o+1,2N_e\}$ we have $|a_n-a|<\epsilon$ for all $n>N$.

Indeed, assume $n>N$. Then either $n$ is even or $n$ is odd. In the first case $n=2m$ for some $m$ and $m=\frac n2 > \frac N2\ge \frac{2N_e}2=N_e$, hence $|a_n-a|=|a_{2m}-a|<\epsilon$. In the other case $n=2m+1$ for some $m$ and $m=\frac{n-1}2>\frac{N-1}2\ge\frac{(2N_o+1)-1}2=N_o$, hence $|a_n-a|=|a_{2m+1}-a|<\epsilon$.

Something makes me think there should be another way, without division and such. — dfeuer, Oct 20 '13 at 22:24

score 2 · Answer 2 · answered Oct 20 '13 at 22:59

Here's a simpler way to do it. Note that $$ \limsup a_n \leq \max \{\limsup a_{2n},\limsup a_{2n+1}\} = a, $$ $$ \liminf a_n \geq \min \{\liminf a_{2n}, \liminf a_{2n+1}\} = a $$ since any subsequence must take infinitely many odds or infinitely many evens and so it follows that $\limsup a_n = \liminf a_n = \lim a_n = a$.

score 0 · Answer 3 · answered Oct 20 '13 at 22:43

If $a_n$ does not converge to $a$ then there is a $a_1$ and a subsequence of $a_{k_n}$ such that $a_{k_n} \rightarrow a_1 \ne a$. Possible $a_1 =\pm \infty$ Let \begin{align} A_0 =\{ n | k_n \equiv 0 \mod 2\}\\ A_1 =\{ n | k_n \equiv 1 \mod 2\} \end{align} Pigeonhole theorem suggests $A_0$ or $A_1$ is infinite. But this leads to a contradiction because of the fact that in a convergent sequence all of its subsequences converge to the same number.

score 0 · Answer 4 · answered Oct 21 '13 at 15:15

Let $(p_j)$ and $(q_k)$ be strictly increasing sequences of integers such that $\{p_j:j\in \Bbb N\}\cup\{q_k:k\in\Bbb N\}=\Bbb N$.

Suppose that $(a_{p_j})$ and $(a_{q_k})$ converge to $a$.

Let $N$ be a neighborhood of $a$.

Then for some $J,K\in\Bbb N$, $j\ge J\implies a_{p_j}\in N$ and $k\ge K\implies a_{q_k}\in N$.

Let $m=\max\{p_J,q_K\}$.

If $n\ge m$, then either $n=p_j$ for some $j$ or $n=q_k$ for some $k$.

Assume without loss of generality that $n=p_j$ for some $j$.

Since $n\ge m$, $p_j=n\ge p_J$ by the definition of $m$.

Since $(p_j)$ is strictly increasing, $j\ge J$, so $a_n=a_{p_j}\in N$.

So we see that $n\ge m\implies n\in N$.

This argument can easily be extended to any finite number of sequences, and I believe also to transfinite sequences. The "strictly increasing" criterion can almost certainly be weakened to an "eventually succeeeds each natural" one. I don't see a really natural extension to nets, but that is likely a failure of my imagination.

If $(a_{2n+1})$ and $(a_{2n})$ converge to $a$ then $(a_n)$ converges to $a$.

4 Answers4