Prove if sequence $s_{2n}$ converges, then also $s_n$ converges 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!
 A: 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.
A: 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
