I'm reading the proof that $\inf \left\{\exp((-1)^nn)\right\} = 0$ where the solution I'm reading says that we can choose $n=2 \max\left\{ \lceil \log(1/\varepsilon)\rceil,1 \right\}+1$ then $\exp((-1)^nn)< \varepsilon $ to show that any $\varepsilon > 0$ is not a lower bound for the sequence. I'm confused by where the $1$ in $\max\left\{ \lceil \log(1/\varepsilon)\rceil,1 \right\}$ comes from. In my mind, we could choose, say, $n=2 \lceil \log(1/ \varepsilon)\rceil+1$. So clearly my understanding is not right here.

So where does it come from? Thanks in advance.

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    $\begingroup$ The 1 is there in case $\varepsilon$ is large. Another way to do it would be to use $\varepsilon'=\min \{ 1,\varepsilon\}$ and then your idea would be fine with $\varepsilon$ replaced by $\varepsilon'$. $\endgroup$ – Ian Mar 10 '18 at 4:54

For $\epsilon > 1$, $\log(1/\epsilon)$ is negative, which could produce a nonsensical result. But with the max in place, then for large $\epsilon$, we choose $n=3$ and are fine since $e^{-3} < \epsilon$.

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  • $\begingroup$ Thank you. Instead of picking out a specific $n$, if I say we pick odd $n$ such that $n > \log(1/\varepsilon)$ that would work because when $\varepsilon > 1$ this says we pick $n=1$ or greater, right? In other words, we wouldn't need the max{..} bit unless we pick $n$ with equality? $\endgroup$ – Ryan Mar 10 '18 at 6:23
  • $\begingroup$ Yes, I believe that would work as well. $\endgroup$ – philbo_baggins Mar 10 '18 at 20:32

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