let $f,g : \mathbb N \rightarrow \mathbb R^+$. Does there exists such $f$ where $f$ is increasing (however slowly) and unbounded and $$g(n) = \frac{(f(n) - 1)^{n-1} +1}{n}$$ is strictly decreasing? Please feel free to assume extra conditions/restrictions (for example maybe $f(n) > 1$ for each $n$). It would be great to get a few constructive examples of such $f$. I was thinking that maybe something like $1 +\log(1+\log(1+...\log(1+n))...)$ could be a candidate. I would also very much apricate any intuition as to why/why not this would be possible. I understand that the question is quite vague, and if this question is not appropriate here, do let me know. Thank you in advance.
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1$\begingroup$ Since $f$ is unbounded and increasing, there is a $k$ such that $f(k)\geq3.$ For any $n\geq0$ we have $f(n+k)\geq3.$ Hence, $g(n+k)\geq \frac{(f(k)-1)^{n+k-1}+1}{n+k}\geq \frac{2^{n+k-1}+1}{n+k}.$ Consequently, $g(n+k)\to\infty$ as $n\to\infty.$ $\endgroup$– Steen82Commented May 17 at 12:30
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$\begingroup$ Wow! that is clean, thank you so much! This should be an answer :). $\endgroup$– PD_SathyaCommented May 17 at 12:53
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$\begingroup$ My pleasure. Been taken to task by a super-mod for just answering questions without investigating if they are similar to previously answered questions, so made it a comment. $\endgroup$– Steen82Commented May 17 at 12:58
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$\begingroup$ @Steen82 All good. Thank you. $\endgroup$– PD_SathyaCommented May 18 at 0:22
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