HACK Corel DRAW Graphics Suite X7l [REPACK]

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HACK Corel DRAW Graphics Suite X7l

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CorelDRAW Graphics Suite X7 – Free Download Corel Draw Free Trial – PC. free version of CorelDRAW Graphics Suite 2018 in CorelDRAW Graphics Suite is.Q:

Dividing a sequence into subsequences of equal lengths

In trying to proof sufficiency of subsequences for proving Bolzano’s weak form of the Bolzano-Weierstrass property (i.e. that every infinite sequence has a convergent subsequence), I came across a trivial application of the theorem which I have not been able to reduce to an elementary calculation yet.
Let $\{a_n\}$ be an infinite sequence of real numbers. Define $a_1,a_2,\ldots,a_{2n},\ldots$ such that $a_{2n}=a_n$ for $n=1,2,\ldots$ and $a_{2n+1}=a_{2n}$.
I am trying to prove the existence of a subsequence of $\{a_n\}$ which consists only of $a_n$s with $n$ odd.
My current approach uses the sequence $(-1)^n$. It is easy to show that $a_n=-1$ if and only if $n$ is even. Therefore $\{a_n\}$ consists of subsequences of equal lengths, namely of $0$s and $2$s (the subsequence of $n$s which is even).
Is my reasoning correct? Is there a better/simpler way to approach this problem?

A:

It is not true that $a_{2n+1}=a_{2n}$. You can take any $\delta$-sequence and take $a_{2n+1}$ in the next $2n+1$ terms in the sequence that have not been selected before.

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