What step in this "proof" that $\lim_{n \to\infty} a_{2n} = L$ implies $\lim_{n \to\infty} a_n = L$ is wrong:

1) $\lim_{n \to\infty} a_{2n} = L$

2) $\Rightarrow \forall \epsilon > 0$ $\exists N \in\mathbb{N} \ni$ if $n \ge N$ then $|a_{2n} - L| < \epsilon$

3) let $k = 2n$ $\wedge$ $n \ge N$ $\iff$ $n = k/2 \ge N$ $\iff$ $k \ge 2N$

4) $\Rightarrow$ $\forall \epsilon > 0$ $\exists N' \in\mathbb{N} \ni$ if $k \ge N'$ then $|a_k -L| < \epsilon$ where $N' = 2N$

5) $\Rightarrow$ $\lim_{k \to\infty} a_k = L$

The theorem is obviously wrong, just take $a_n = (-1)^n$ as a counterexample.


To be precise, I'm only interested in what step in the above "proof" is wrong, not why the above "theorem" is wrong in general.


Some people have had issues with my definition of the limit, specifically the "if $n \ge N$" clause. I got this straight from "Foundations of Mathematical Analysis" by Johnsonbaugh and Pfaffenberger (a Stanford University textbook):

$\lim_{n \to\infty} a_n = L$ if $\forall \epsilon > 0$ $\exists N \in \mathbb{N} \ni$ if $n \ge N$ then $|a_n - L| < \epsilon$

Note, some have taken issue with the fact that I've substituted quantifiers for some wording from text in the above definition, so here is the exact definition from the text:

$\{a_n\}_{n=1}^{\infty}$ has limit $L \in \mathbb{R}$ if for every $\epsilon > 0$, there exists a positive integer $N$, such that if $n \ge N$, then $|a_n -L| < \epsilon$

  • $\begingroup$ The sequence $a_k$ is not the original sequence $a_n$ if $k = 2n$. $\endgroup$ – dannum May 4 '18 at 16:26
  • $\begingroup$ $k = 2n$ - you proved what you started from. $\endgroup$ – NickD May 4 '18 at 16:27
  • $\begingroup$ All the following terms after $a_{k}$ needs also to be within this interval. Or take $a_{k+1}$, you can't divide it by two since your $k$ is even, so you don't have a proof, since a following term in your sequence doesn't follow the rule of the limit. Your limit definition is wrong, it is \forall{n} \get $\endgroup$ – Jean Rostan May 4 '18 at 16:27
  • $\begingroup$ the fact is that $\lim_{n \to\infty} a_{2n} = L$ can't give any information on the behavour of $a_k$ when $k$ is odd $\endgroup$ – user May 4 '18 at 16:30
  • $\begingroup$ You did not justify your writing of "$\Rightarrow$" $\endgroup$ – Hagen von Eitzen May 4 '18 at 16:31

You can note that $n$ is a positive integer and hence $k=2n$ implies that $k$ is an even positive integer. Your final statement $(4)$ thus deals with all even $k>N'$ and not all $k$ as required by definition of limit.

  • $\begingroup$ Thanks, that's the crux of the problem. To symbolizes it, in step 4) I implicitly say $...\forall k[k \in \mathbb{2N} \wedge k \ge N' => |a_k -L| < \epsilon]...$ but the definition of the limit requires $...\forall k[k \in \mathbb{N} \wedge k \ge N' => |a_k -L| < \epsilon]...$ and you can't get the latter from the former. $\endgroup$ – Arnaut B May 6 '18 at 7:12
  • $\begingroup$ @ArnautB: forget my last deleted comment. I am glad you understood the issue here. $\endgroup$ – Paramanand Singh May 6 '18 at 7:39
  • $\begingroup$ @ArnautB: just to confirm your symbolic interpretation is exactly same as in my answer and what you have wrote in your comment is the correct argument. $\endgroup$ – Paramanand Singh May 6 '18 at 8:06

The problem lies in your definition of a limit:

$$ \forall \epsilon >0, \exists N \in \mathbb{N}, \forall n\geq N, |a_n-L|<\epsilon $$.

The second $\forall$ is key.

See your mistake?

  • $\begingroup$ Bill.Exactly, +1. $\endgroup$ – Peter Szilas May 4 '18 at 16:51
  • $\begingroup$ Are you saying my definition of the limit is incorrect? $\endgroup$ – Arnaut B May 4 '18 at 16:59
  • $\begingroup$ I think your definition is quite ambiguous and it is that ambiguity that lead you to this mistake: "if $n \geq N$" does not denote a set but rather a property where $n$ would have been introduced beforehand. $\endgroup$ – Bill O'Haran May 4 '18 at 17:05
  • $\begingroup$ As I noted elsewhere this definition is right out of Foundations of Mathematical Analysis by Johnsonbaugh and Pfaffenberger, a textbook used in Stanford. Sorry, I fail to see the ambiguity, can you elaborate? $n$ is implictly a natural number since $a_n$ is a sequence. $\endgroup$ – Arnaut B May 4 '18 at 17:15
  • 1
    $\begingroup$ Again, a book being used somewhere is no argument. I never said the definition given in the book was incorrect, having looked at it, it is actually perfectly equivalent to the one I gave. I'll be honest with you: I do not have any reference to provide. However, I would like you to understand that the whole problem you have comes down to translating that if correctly. Since you seem a little protective of what you think, remember that you (rightfully) accepted that your proof is wrong, so accept that there is something wrong with your proof ;) $\endgroup$ – Bill O'Haran May 4 '18 at 18:52

Your highly abbreviated language hides the misstatement.

It is incorrect to say that "$\lim\limits_{n\to\infty} a_{2n}=L$" implies that "if $n \geq N$ then $|a_{2n}-L|<\epsilon$".

Rather, the correct conclusion is that "if $n$ is an integer with $n \geq N$ then $|a_{2n}-L|<\epsilon$".

This is implicit when using sequences, but it must be considered nonetheless.

  • $\begingroup$ The textbook Foundations of Mathematical Analysis by Johnsonbaugh and Pfaffenberger states it like that. This is a textbook used in Stanford. As you say it is implicit $n$ is an integer since we're using sequences, therefore I don't think the wording is incorrect. $\endgroup$ – Arnaut B May 4 '18 at 16:50
  • $\begingroup$ @ArnautB : I don’t mean to say your wording is incorrect—it is perfectly acceptable. I just mean you have to be careful to only use expressions which represent integers for subscripts. I think that’s what got glossed over in the proof. $\endgroup$ – MPW May 4 '18 at 22:55

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