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I'm having some trouble understanding why it is justified to introduce substitution when computing the limits of sequences.

The definition of sequence conversion is a usual, where $m$ are positive integers:

$\tag{1} \lim_{m \to +\infty} s_m = L \iff [\forall \epsilon > 0. \exists M > 0. m > M \implies |s_m - L| < \epsilon]$

Now, because $2n > n$, I know that $\lim_{n \to +\infty} s_n = L \implies \lim_{n \to +\infty} s_{2n} = L$, but I'm not sure how to prove it the other way around:

$\lim_{n \to +\infty} s_{2n} = L \implies \lim_{n \to +\infty} s_n = L$

It's just a substitution (if we replace $m = 2n$ in $(1)$), we get that, but could someone help me prove it as well, by using the definition?

Edit:

I will give an example, because it looks like it is not clear what I meant with $s_{2n}$. For example:

$\tag{2} s_n = \frac{1}{n}\sum_{k=1}^{n} \sin(\frac{k \pi}{n})$

I can prove that $\tag{3} t_{n} = \frac{1}{2n}\sum_{k=1}^{2n} \sin(\frac{k \pi}{2n})$ converges, but I don't know if from that I can conclude that $(2)$ converges as well.

Thanks!

S11n
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    It is false. E.g. $s_n=(-1)^n$. – Anne Bauval Oct 04 '22 at 10:53
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    Might be worth noting that it is true once we assume that the sequence is Cauchy (see here https://math.stackexchange.com/questions/354965/cauchy-sequence-is-convergent-iff-it-has-a-convergent-subsequence) – Giulio R Oct 04 '22 at 14:12

2 Answers2

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We can not Prove it because it is not true.

You have Proved that
$\lim_{n \to +\infty} s_n = L \implies \lim_{n \to +\infty} s_{2n} = L$
which is Correct & true.

You want to Prove that
$\lim_{n \to +\infty} s_{2n} = L \implies \lim_{n \to +\infty} s_n = L$
which is not true.

Here is the general outline of making a Chosen Counter-Example :

Let the Even terms [ that is $2n$ ] of the Sequence $S$ be a Converging Sequence.
Let the ODD terms [ that is $2n+1$ ] of the Sequence $S$ be $[1+L]$ or $[(-1)^{2n+1}]$ or $[2^{2n+1}]$ [ Choose some Positive Linear Sequence or Oscillating Sequence or Diverging Sequence XXXXX ]
Then we can see that this Sequence $S$ will NOT Converge.

UPDATE to handle the new specific Example given by OP :

Comment 1 : Sequences & Series are not the same & better to separate them at the moment.

Comment 2 : The Series $\Sigma W(n)$ might converge , but the Series $\Sigma W(2n)$ might not converge.
This is because, the Even terms might be cancelling with the ODD terms & when considered in Isolation, it might not converge.

Comment 3 : Likewise, the Series $\Sigma W(2n)$ might converge , but the Series $\Sigma W(n)$ might not converge.
It might occur that every Even term cancels the next Even term, & the whole will converge, but the ODD terms do not get cancelled.

Comment 4 : Coming back to Sequences :
When the given Sequence $s_n$ converges, then the given Sequence $t_n$ (taking every other element , where we are throwing out known elements) might also converge.
When we given only the $t_n$ (& the original $s_n$ values are unknown to us) & we can insert values between 2 elements to generate the new $s_n$ , then it may or may not converge , because we do not know the original $s_n$ values & we can insert the Examples given near XXXXX, which have a guarantee to not converge.

Prem
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  • Thank you for the answer, but there might have been a confusion about what I meant with $s_{2n}$. Do you think your answer is applicable even for the example I gave? By $2n$ I didn't mean even only terms, but actually the $2n$th member of the sequence. – S11n Oct 04 '22 at 14:09
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    @S11n That is the even-indexed terms. The issue is that you delete an entire subsequence (the odd-indexed terms), and if that subsequence didn't converge, or if it converged to something different from the even-indexed terms, then the whole sequence can't converge. – Ian Oct 04 '22 at 14:12
  • @Ian I don't see how I'm deleting the odd-indexed terms if I just take the first $n$ members of the sequence which is long $2n$ members (where $2n \to +\infty$)? Have you seen the example I added in the question, after seeing Prem's answer? – S11n Oct 04 '22 at 14:15
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    @S11n I think you are getting a little bit confused between sequences and series; the language around "deleting terms" is especially prone to confusion in this regard. I suggest that you restrict attention to sequences in your mind for a little bit until you're really comfortable with the concepts, and then you can revisit series by interpreting them as sequences of partial sums. – Ian Oct 04 '22 at 14:31
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    In that example, each term of the sequence happens to be an expression that involves a sum, but note that the $s_i$ are not partial sums. The terms in the sum for $s_i$ differ from the terms in the sum for $s_{i+1}$. The value of $s_n$ is basically the average y-coordinate of the vertices in the top half of a regular $2n$-gon. If you restrict yourself to $s_{2n}$, you are only looking at $4n$-gons, skipping over the hexagon, decagon, etc. Regardless, even if both $s_n$ and $s_{2n}$ converge in that example, it only needs one counterexample to show the proposed principle fails in general. – Jaap Scherphuis Oct 04 '22 at 14:48
  • Thank you all, now it's clear! – S11n Oct 05 '22 at 08:28
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This is true if you assume that every subsequence of $s_n$ converge to the limit of $s_n$.

Suppose $s_n \to p’$, and the subsequences of $s_n$ converge to $p$, $p’\neq p$. Since $s_n$ converges, it is Cauchy, so the $d(s_n,s_m)< \epsilon$ for $n,m > N$. Then for $n_k>N$, $d(p,p’)< d(s_{n_k},p’)+d(s_{n_k},s_m)+d(s_m,p)<\epsilon$. Contradiction

Alan Cai
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  • Thanks! I still didn't get to what it means for a sequence to be Cauchy. In your proof, I don't understand the last inequality, namely the part between $d(p, p')$ and $\epsilon$. – S11n Oct 05 '22 at 08:38