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If an arbitrary sequence contains just a finite number of floor terms, and the last one is $a_F$, form a strictly decreasing subsequence with $a_{F+ 1}$ as its first term.

We call a term of a sequence a ‘floor term’ when none of its successors is strictly less. Given that definition I am not sure how to construct such a sequence. One (non)-example is $1, 2, 3, 4, 5, 1000000, 1000000 - 1,\ldots, 1000000 - n, \ldots$ For our purposes, no term after $5$ should be less than $5$, but in this sequence that's not guaranteed. The only way I see this working is if the strictly decreasing subsequence is finite, like so $1, 2, 3, 4 , 5, 100, 99, 98, 97, 96, 95, 94, 94, 94, 94, 94 \ldots$ where all the terms after $94$ are $94.$

Is it possible to have a sequence with a finite number of floor terms and one of whose subsequences is strictly decreasing and infinite? I can't seem to think of one. Thanks.

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  • $\begingroup$ Not entirely sure I follow. Would it suffice to take $a_{F+i}=a_F+\frac 1i$? $\endgroup$
    – lulu
    Mar 10 '16 at 23:27
  • $\begingroup$ I am trying to make sense of this answer to the given question. I am not sure if I understand it correctly. $\endgroup$
    – user321384
    Mar 10 '16 at 23:31
  • $\begingroup$ I don't see that as a constructive answer, just as an (accurate) observation. The writer is saying "suppose you have a sequence such that $a_F$ is the last floor term. Then $a_{F+1}$ is not a floor term so we can find some $a_{F+k}<a_{F+1}$ But then $a_{F+k}$ is also not a floor term so we can find some $a_{F+k+m}<a_{F+k}$ and so on, thereby obtaining a strictly decreasing subsequence. $\endgroup$
    – lulu
    Mar 10 '16 at 23:35
  • $\begingroup$ Thanks. I see. But I am still not sure how to construct at least a specific example of such a sequence. Can you, please, give an example. $\endgroup$
    – user321384
    Mar 10 '16 at 23:51
  • $\begingroup$ My example works. Suppose, for clarity that $a_1=0$ is the only floor term. Then for $i>1$ we define $a_i=\frac 1i$ That is a strictly decreasing, and all of the terms past the first are strictly greater than $a_1$. $\endgroup$
    – lulu
    Mar 10 '16 at 23:54
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Obviously, only the last decreasing subsequence can be infinite.

Since $(n+1)^2 =n^2+2n+1 $, for each $n$ do $(n+1)^2, (n+1)^2-1, (n+1)^2-2, ... (n+1)^2-2n $ as many times as you want.

To finish with an infinite decreasing subsequence, do $n^2-\sum_{k=0}^m \frac1{2^k} $ for each successive $m$.

Also obviously, any infinite decreasing sequence of integers has to go to $-\infty$, so non-integers have to be used for the last sequence.

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