If $\{x_n\}$ is a sequence s.t. $x_{2k} \rightarrow x$ and $x_{2k-1} \rightarrow x$, then $x_k \rightarrow x$

Just looking for feedback on my proof attemtpt

Proof Attempt

given that the odd and even terms of the sequence both individually converge that means: $$1) \ \exists \ N_1 \ s.t \ \forall \ k \leq N_1 \ |x_{2k} - x| < \frac{\epsilon}{2} \\ 2) \ \exists \ N_2 \ s.t \ \forall \ k \leq N_2 \ |x_{2k-1} - x| < \frac{\epsilon}{2} $$

If we choose $N = max\{N_1,N_2\}$

Then: $$|x_{2k} - x_{2k-1}| \leq |x_{2k} - x| + |x_{2k-1} - x| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon$$

I don't know if it is a big leap, but I'm assuming that $|x_{2k} - x_{2k-1}|$ represent consecutive sequential terms so in essence I am bringing together the odd sequential terms and the even sequential terms. Is this the right idea?

  • $\begingroup$ What you've shown is that the sequence is Cauchy. For sequences in $\Bbb R$ this is equivalent to convergence but it is not immediately obvious that that's the case. $\endgroup$ – Lukas Kofler Oct 21 '18 at 22:40
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    $\begingroup$ I think you meant $k\geq N_1$ and $l\geq N_2$, not the other way around. $\endgroup$ – Mee Seong Im Oct 21 '18 at 22:40
  • $\begingroup$ @LukasKofler that is true.....and this is supposed to be in $\mathbb{R}$, but since I wasn't thinking about it in those terms, what could I do to prove it? $\endgroup$ – dc3rd Oct 21 '18 at 22:43
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    $\begingroup$ If you are in $\mathbb R$, you are done, since $\mathbb R$ is complete $\endgroup$ – Don Thousand Oct 21 '18 at 22:45
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    $\begingroup$ You can just say that for all $n>N$ that $|x_n-x|<\epsilon$ because $n$ is either of the form $2k$ or $2k-1$ for some $k>N$. $\endgroup$ – kingW3 Oct 21 '18 at 22:47

so in essence I am bringing together the odd sequential terms and the even sequential terms. Is this the right idea?

Ooooh. Ouch. No. That is not the right idea. I'd doesn't matter how close sequential terms get. The classic counter example is the harmonic series in which $a_n = \sum_{k = 1}^n \frac 1n$. $|x_n - x_{n-1}|=\frac 1n \to 0$ but $\{a_n\}$ does not converge.

In this case you are lucky in that you actually have and $x$ which $x_{2k}$ and $x_{2k - 1}$ converge to.

So for any $\epsilon$ you have an $N_1$ so that $n > N_1 \implies |x_{2n} - x| < \epsilon$ and you have an $N_2$ so that $n > N_2 \implies |x_{2n-1} - x| < \epsilon$.

So if you have $m > 2n > 2n-1$ where $n \ge \max (N_1, N_2)$ then if $m$ is odd then $m = 2k - 1$ for $k > n> N_2$ so $|x_m - x|= |x_{2k -1} - x| < \epsilon$. But if $m$ is even then $m = 2k$ for $k > n > N_1$ so $|x_m - x| = |x_{2k} - x| < \epsilon$.

In other words, let $M = 2\max (N_1, N_2)$. Then if $m > M$ then if $m = 2k$we have $k > N_1$ and if $m = 2k -1$ then we have $k > N_2$. ANd either way $|x_m - x| < \epsilon$.


Now if you hadn't been given that $x_{2k}, x_{2k-1}\to x$ and where given that $x_{2k}$ and $x_{2k-1}$ were Chauchy and needed to prove $x_m$ was cauchy you would have had the right idea only you don't prove it only for the subsequent terms you must prove it for any TWO terms $m_1, m_2 > N$.

And we'd do this by taken cases.

Case 1: if $m_1, m_2$ are both even then $m_1, m_2 \ge N_1$ (um, why did you write $k \le N_1$? That was a typo I assume) so $|x_{m_1} - x_{m_2}| < \frac {\epsilon}2 < \epsilon.$

Case 2: if $m_1, m_2$ are both odd... some thing but with $N_2$.

Case 3: If $m_1$ is even , $m_2$ is odd are opposite parity then $|x_{m_1} - x_{m_2} \le |x_{m_1} - x_{2k}| + |x_{2k} - x_{2k -1}| + |x_{2k-1} - x_{m_2}| < \frac \epsilon 2 + \frac \epsilon 2 + \frac \epsilon 2 = \frac 32 \epsilon$.

So you'd have to modify for $\frac \epsilon 3$ instead.

  • $\begingroup$ Your solution is very interesting. I haven't been able to remove the notion from my head, but I'm always under the impression that since these are "exercises" or questions from an assignment they will always have an all encompassing solution that captures everything at once. But your solution had to break it down into cases. If we were to expand on this, say that there were 100 or 1000 cases, I guess we would have to just explicitly show them all and possibly over time refine it. Thanks @fleablood for the thorough explanation. $\endgroup$ – dc3rd Oct 22 '18 at 1:34
  • $\begingroup$ If there were a thousand case we just need to consider the maximum of the resulting $N_i$s If $n > \max N_i$ then $n$ will qualify no matter what. $\endgroup$ – fleablood Oct 22 '18 at 1:36

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