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How to construct such a sequence? please help. thank you!

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    $a_n = (-1)^{n+1}$ – 2017-02-08
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    @user284275: In your title, didn't you mean "$\ge$", not "$\le$"? Also, don't you want some kind of boundedness (e.g., nonnegative terms)? – 2017-02-08
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    @quasi Replacing $\leq$ with $\geq$ changes pretty much nothing. – 2017-02-08
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    @Wojowu: With the current statement of the problem, any increasing unbounded sequence would be a trivial example. – 2017-02-08
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    @quasi No, since we want $a_{2n+3}\leq a_{2n+1}$, the sequence can't be increasing. – 2017-02-08
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    @Wojowu: Ah! Leave it to me to not fully read the question. – 2017-02-08
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    @user284275: Ignore my comment about your title -- I misread the problem. – 2017-02-08

2 Answers 2

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If you want a sequence with strict inequalities, you can go for a variation on @Dark's comment above, something like $$ (-1)^{n+1}\left(1+\frac1{n+1}\right) $$

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Idea: if $n$ is even, choose a nondecreasing sequence (for example, constant). If $n$ is odd, choose a descreasing sequence that is bounded from below by the other one (and has a different limit). For instance, $a_n=1$ if $n$ is even, $a_n=2+\frac{1}{n}$ if $n$ is odd.