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, 2018 at 16:26
  • $\begingroup$ $k = 2n$ - you proved what you started from. $\endgroup$
    – NickD
    May 4, 2018 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 $\endgroup$ May 4, 2018 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, 2018 at 16:30
  • $\begingroup$ You did not justify your writing of "$\Rightarrow$" $\endgroup$ May 4, 2018 at 16:31

3 Answers 3


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. $\endgroup$ May 4, 2018 at 16:51
  • $\begingroup$ Are you saying my definition of the limit is incorrect? $\endgroup$
    – Arnaut B
    May 4, 2018 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. $\endgroup$ May 4, 2018 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, 2018 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 ;) $\endgroup$ May 4, 2018 at 18:52

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, 2018 at 7:12
  • $\begingroup$ @ArnautB: forget my last deleted comment. I am glad you understood the issue here. $\endgroup$
    – Paramanand Singh
    May 6, 2018 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, 2018 at 8:06

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, 2018 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, 2018 at 22:55

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