This was a question on an exam I recently took that I didn't do so great on, so I'm trying to understand this problem thoroughly. What flaws are there?

Suppose that $\{a_n\}$ and $\{b_n\}$ are two convergent sequences that both converge to the same $L \in \mathbb{R}$. Define a new sequence $c_n$ such that $c_n = a_n$ if $n$ is odd and $c_n = b_n$ if $n$ is even. Prove that $\{c_n\}$ converges to $L$.

Proof Since $a_n$ and $b_n$ both converge to $L$, we know that $$\lim \sup a_n = \lim \inf a_n = \lim \sup b_n = \lim \inf b_n = L$$ Define $i_n = \inf\{c_k : k \geq n\}$ and $s_n = \sup\{c_k : k \geq n\}$. Then $\lim i_n = \lim \inf c_n$ and $\lim s_n = \lim \sup c_n$.

$L$ is a subsequential limit of $c_n$ and so $\lim \inf c_n \leq L \leq \lim \sup c_n$.

Suppose for contradiction that $\lim \inf c_n < L$. Then there exists an $n_0$ such that $L \neq i_{n_0}$ and $i_{n_0} < a_{n_0}$ or $i_{n_0} < b_{n_0}$, but this contradicts the construction of $\{a_n\}$ and $\{b_n\}$ because one or both would not converge at all or would converge to $i_{n_0}$. Thus, $\lim \inf c_n = L$.

A similar argument can be used to show that $\lim \sup c_n = L$ by replacing $i_{n_0}$ with $s_{n_0}$, $\lim \inf c_n$ with $\lim \sup c_n$ and switching the inequality signs in the previous paragraph.

Since $\lim \inf c_n = L = \lim \sup c_n$, $c_n$ converges to $\lim a_n = \lim b_n = L$.

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    $\begingroup$ While it might be correct, this approach seems way to convoluted. Isnt it actually you picj an epsilon so you know that tgere are n0 and m0 such that all element a_n and b_m with n>n0 and m>m0 are closer than epsilon to L. Then pick max {n0,m0} and you are done. $\endgroup$ – lalala Oct 21 '18 at 20:35
  • $\begingroup$ @lalala. I noticed your comment after I was done typing up this sentiment. Full credit goes to you. $\endgroup$ – Mason Oct 21 '18 at 20:40
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    $\begingroup$ It would be nice if more than one of the answerers actually made an effort to address the asker's question of "what flaws are there" rather than just typing out the fastest proof of the claim that comes to mind. $\endgroup$ – T. Bongers Oct 21 '18 at 20:43
  • $\begingroup$ @T.Bongers I do appreciate all of the responses but I have to agree. ;) $\endgroup$ – Austin Oct 21 '18 at 20:44
  • $\begingroup$ @lalala That was also exacly my point, with the difference that I didn't give a full proof to let the asker prove by him/herself. Sometimes that is appreciated (do not give full answer!) sometime it is not, often depending upon who is answering! $\endgroup$ – gimusi Oct 21 '18 at 20:54

I am not sure whether your current proof is wrong but I am confident that it has some unneeded extra parts. This proof should be $3$ or $4$ sentences.

For a simpler approach try this without invoking $\sup$ or $\inf$. If you need help see below:

Let $\epsilon>0 $ be given we need to show that $\exists N_c$ such that $\forall n>N_c$ we have $|c_n-L|<\epsilon$ but because we know $\{a_n\}$ converges we can select $N_a$ such that $\forall n>N_a$ we have $|a_n-L|<\epsilon$ and $\{b_n\}$ converges we can select $N_b$ such that $\forall n>N_b$ we have $|b_n-L|<\epsilon$. So we can just take $N_c=\max\{N_a,N_b\}$ and then we are guaranteed to be within $\epsilon$ of our target.

  • $\begingroup$ That was also my point! It can be proved directly from the definition without such kind of complicated approach! $\endgroup$ – gimusi Oct 21 '18 at 20:55
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    $\begingroup$ Maybe we can interpret T Bongers comment like this. The OP didn't really ask us to prove anything so the ideal thing to do is give the feedback that we had intended: The argument is unnecessarily convoluted and not 'spoil' the fun by showing too much. $\endgroup$ – Mason Oct 21 '18 at 21:02
  • $\begingroup$ @Mason Exactly: I read this as a feedback question, rather than a "let's show how quickly I can recall the proof" question. So (+1) for your answer. $\endgroup$ – T. Bongers Oct 21 '18 at 21:04
  • $\begingroup$ @Mason Indeed it was exactly my point! I admit that maybe my answer at first was not clear on that point but my goal was exactly your one. I'v eonly chosen to do not give a full solution to let the pleasure to the asker to find it by himself. You answer also is very good in my opinion. Bye $\endgroup$ – gimusi Oct 21 '18 at 21:06

The other answers give quick proofs, but if you want to use sups and infs, that's ok too. But the flaw in your proof, in my opinion, starts here: "Then there exists an $n_0$....". Let's pick it up at "Suppose for contradiction that $\liminf c_n<L$", which is fine. Here is a sketch of what comes next. I will leave the details to you: if $\liminf c_n$ is strictly less than $L$, then there is an infinite number of terms of $c_n$ less than $L$. And this means that-----






$$\lim_{n\to+\infty} c_{2n}=\lim_{n\to+\infty} c_{2n+1}=L$$

and $$\lim_{n\to+\infty} c_n=L$$

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    $\begingroup$ This assumes a number of facts about subsequential limits that the asker may or may not know, and sort of trivializes the problem. $\endgroup$ – T. Bongers Oct 21 '18 at 20:44


As noticed, I am also not sure whether your current proof is wrong but it seems too much complicated indeed we don't need to use $\liminf$ and $\limsup$ concept since it can be carried out directly by the definition of limit.

Indeed, since $c_n \to L$ for $n$ even and $c_n \to L$ for $n$ odd, since we are considering two subsequences wich cover all the possible values for $n$, we can easily shown that $c_n \to L$.

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    $\begingroup$ This isn't a "hint," nor does it really address the question of "what flaws are there." Several of the other answers don't really address the question either, though. $\endgroup$ – T. Bongers Oct 21 '18 at 20:41
  • $\begingroup$ @T.Bongers Yes maybe I need to clarify that. My point is as that raised up by Mason. We don't need a so complicated proof but we can proceed in a simpler way directly by the definition of limit. I'm indeed answering to the main question the asker is trying to solve. Use a not effective method is the problem in my opinion. Note that the asker is not requiring to use a particular kind of proof and moreover he/she doesn't seem to be aware about that simple way. $\endgroup$ – gimusi Oct 21 '18 at 20:44
  • $\begingroup$ Then perhaps you could expand your answer to explain how you have addressed the only question contained in the body: "What flaws are there?" $\endgroup$ – T. Bongers Oct 21 '18 at 20:46

Here's a simple proof

$\textbf{Proof:}$ Assume $\epsilon>0$, and by convergence choose $N_1,N_2\in\mathbb{N}$ such that $a_{n_1},b_{n_2}\in\{x\in\mathbb{C}|\text{dist}(x,L)<\epsilon\}$ for all $n_1\geq N_1$ and $n_2\geq N_2$. Then choosing $N=\max(N_1,N_2)$ we find $c_n\in\{x\in\mathbb{C}|\text{dist}(x,L)<\epsilon\}$ for all $n\geq N$, therefore $c_n\to L$, for $\epsilon$ was arbitrary.$\square$

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    $\begingroup$ The first word of your proof is "assume" but don't we know that there exist such values $N_1$ and $N_2$ from the premise? I think you may be rearranging the quantifiers... Usually we let $\epsilon>0$ be given and then select $N$s that do the job we want. $\endgroup$ – Mason Oct 21 '18 at 21:16
  • $\begingroup$ Woops. We gotta choose $\epsilon$ before we choose $N_1,N_2$. Corrected that. Thank you. After we choose $\epsilon$, then we can choose $N_1,N_2$ from the $\epsilon,N$ characterization of convergence. $\endgroup$ – Melody Oct 21 '18 at 21:25

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