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let $g : \mathbb N \rightarrow \mathbb R$ be a sequence defined by $$g(n) = \frac{(c\log(1+\log(1+n)))^{n-1} + 1}{n}$$ where $1\ge c>0$ It seems like this sequence is strictly decreasing, I guess one could take the same function on the reals and compute its derivative, but I was wondering if there is a particular inequality that would give a shorter/more elegant proof, that would help showing $g(n+1)<g(n)$. Thank you in advance :).

Edit: It seems I am quite wrong. Now I wonder if there are any c for which $g(n)$ is decreasing.

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    $\begingroup$ If $c>1$ this sequence cannot be decreasing: it tends to $\infty$. $\endgroup$
    – Anne Bauval
    Commented May 17 at 5:52
  • $\begingroup$ Thank you very much, it seems I overlooked that. Could you make a comment about how one could show it for $c\in (0,1]$ ? $\endgroup$
    – PD_Sathya
    Commented May 17 at 6:06
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    $\begingroup$ Even for $c=0.7$ this is wrong: wolframalpha.com/… $\endgroup$
    – Anne Bauval
    Commented May 17 at 6:11
  • $\begingroup$ desmos Thank you for the reply, either something is wrong with Desmos (here I've put c =1) or I'm just not seeing it carefully. But wolfram shows it right wolfram. @AnneBauval $\endgroup$
    – PD_Sathya
    Commented May 17 at 6:23
  • $\begingroup$ Yes I know, I did the same query before (c=1). What I am reporting is even worse (c=0.7) $\endgroup$
    – Anne Bauval
    Commented May 17 at 6:30

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