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$\mathbb {P}(\limsup\limits_{n\to\infty}Z_n/\ln n=1)=1$ for $Z_n$ iid $Exp(1)$

I guess I have to use Borel-Cantelli but I am struggling with especially one part. Shouldn't we have

$\sum_{n=0}^\infty \{Z_n/\ln n=1\} = 0 < \infty$ as $Z_n$ are continues Random Variables?

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    Indeed, use Borel-Cantelli with the events $$A_n^x=\{Z_n>x\ln n\}$$ for various values of $x$. – 2017-02-08
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    @Did So $\sum_{n=0}^\infty \mathbb {P}(A_n^x)\begin{cases} = \infty, & \text{if $x \le$1} \\ < \infty, & \text{if $x >1$} \end{cases}$ and I can use Borel Cantelli but how do I get the final result from there? – 2017-02-08
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    You first: which conclusion(s) do you get using BC? – 2017-02-08
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    @Did For $x \le 1$: $\mathbb{P}((\limsup\limits_{n\to\infty}Z_n/\ln n>x)=1$ For $x > 1$: $\mathbb{P}((\limsup\limits_{n\to\infty}Z_n/\ln n>x)=0$ From the first I can conclude that $\mathbb{P}((\limsup\limits_{n\to\infty}Z_n/\ln n \le x)=1$ for $x\le 1$ and thus we have $\mathbb {P}(\limsup\limits_{n\to\infty}Z_n/\ln n \le 1)=1$ but how can I get to other inequality? – 2017-02-08
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    Well, you know that $$\limsup Z_n/\ln n>x$$ almost surely for every $x<1$ and that $$\limsup Z_n/\ln n1$, hence $$\bigcap_k\{\limsup Z_n/\ln n>1-\tfrac1k\}\cap\{\limsup Z_n/\ln n<1+\tfrac1k\}$$ has probability $1$. Now, this event is $____$ hence you are done. – 2017-02-08
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    Finally understood it, thank you! – 2017-02-08

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