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This question already has an answer here:

I am trying to show:

If $(x_n)$ converges to $a>0$ and $x_n>0, y_n \geq 0$ then $\lim \sup (x_n y_n) = a\lim \sup y_n$. But have not been able to make much progress.

I can show that if $p$ is a limit point of $a(y_n)$ then it is a limit point of $(x_n y_n)$ but I cannot show the converse of this, which would complete the proof. Any hints?

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marked as duplicate by 6005, user223391, Greg Martin, Chill2Macht, Daniel W. Farlow Aug 11 '16 at 0:48

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  • $\begingroup$ Are you also assuming $\limsup y_n$ exists? $\endgroup$ – user223391 Aug 10 '16 at 21:38
  • $\begingroup$ No I am not, $y_n$ may not be bounded here. $\endgroup$ – fosho Aug 10 '16 at 21:40
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    $\begingroup$ See this: math.stackexchange.com/questions/113121/… $\endgroup$ – Brenton Aug 10 '16 at 21:40
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If $(x_n)$ converges to $a>0$, for any $\varepsilon >0$, there's an $n_0$ such that $\; a-\varepsilon <x_n < a+\varepsilon\;$ for each $n\ge n_0$. We choose $\varepsilon$ so that $a-\varepsilon>0$. To prove the assertion, supposing $n\ge n_0$ does not change the $ \limsup$. We'll compute the limits in $\overline{\mathbf R\!}\,$.

So let $n\ge n_0$, and $k\ge n$. Multiplying by $y_k$, we obtain $$(a-\varepsilon)y_k \le x_k y_k \le a+\varepsilon)y_k,$$ whence $$(a-\varepsilon)\sup_{k\ge n}y_k \le \sup_{k\ge n}x_ky_k\le (a+\varepsilon)\sup_{k\ge n}y_k$$ and finally \begin{alignat*}{2}(a-\varepsilon)\lim_n\Bigl(\sup_{k\ge n}y_k\Bigr) &\le \lim_n\Bigl(\sup_{k\ge n}x_ky_k\Bigr)&&\le (a+\varepsilon)\lim_n\Bigl(\sup_{k\ge n}y_k\Bigr), \\\text{i.e.}\qquad (a-\varepsilon)\limsup_{n}y_n&\le \limsup_{n}(x_ny_n)&&\le(a+\varepsilon)\limsup_{n}y_n. \end{alignat*} As these inequalities are true for any $\varepsilon>0$, this means $$a\limsup_{n}y_n = \limsup_{n}(x_ny_n)=a\limsup_{n}y_n.$$

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