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So I'm doing some exercise on sequences, and one of the question is giving a hard time. It requires me to give an example of 2 bounded non-negative sequences $x_n$ and $y_n$ such that

lim sup ($x_ny_n$) < lim sup($x_n$) . lim sup($y_n$)

All the examples I thought of ended up in an equality instead of the strictly lesser than inequality. Does anyone know any simple examples for this ?

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Consider the sequences $(a_n), (b_n)$, where $a_{2k} = 1, \; a_{2k+1} = 2 , \; b_{2k} = 2$ and $ b_{2k+1} = 1$. then we have $$\limsup a_n = 2$$ $$\limsup b_n = 2$$ But $c_n = a_n \cdot b_n$ = 2. Therefore $$ \limsup a_n b_n = \limsup c_n = 2 < \limsup a_n \cdot \limsup b_n = 2 \cdot 2 = 4 $$

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  • $\begingroup$ damn, I knew there was a simple answer. cracked my head over this, thank you for the reply. $\endgroup$ – some1fromhell Mar 16 '16 at 13:13
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    $\begingroup$ Bear in mind that in such problems the most efficient way to find a solution is to "play" with the subsequences! That is also the main idea behind various other problems in real (or complex) analysis, such as finding an oscilating sequence whose set of limits of its converging subsequences is the entire $\Bbb N$. $\endgroup$ – christina_g Mar 16 '16 at 13:27
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HINT: Choose $x_n$ and $y_n$ so that $x_ny_n=0$ for each $n$, but $\limsup_nx_n=\limsup_ny_n>0$.

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