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Bonjour. Show that $$\sum{\frac{1}{n^{2+\cos{n}}}}$$ is a divergent serie. $$\\$$

My main problem is: If $\epsilon$ is “infinitely small positive real number” define $A_{\epsilon}$ as the set of all $n, |2+\cos n|\leq 1+\epsilon$ $(n \in A_{\epsilon}\iff-1\leq \cos n \leq -1+\epsilon)$. The divergence should come from the sum over $A_{\epsilon}$ but I have no idea to how to handle this. $$\\$$

DINEDINE
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    Bonjour. What have you tried? – Wojowu Feb 01 '19 at 11:42
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    Bonjour. I don’t even know where to start. The problem is $2+\cos n$ can be arbitrarily closer to $1$ for infinitely many n. Summing those value we can end with everything. – DINEDINE Feb 01 '19 at 11:46
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    If the downvoters can explain why they downvoted, that will be great! – mathcounterexamples.net Feb 01 '19 at 12:30
  • The downvote button is a toy for some of our friends. – DINEDINE Feb 01 '19 at 12:50
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    Physicists are currently working on a universal law that unifies general relativity with quantum mechanics. Once they are finished they will tackle the more difficult question of why people cast downvote and yet offer no constructive criticism or alternatives. – Oscar Lanzi Feb 01 '19 at 14:12
  • Corrected @peterag. – DINEDINE Feb 01 '19 at 14:17
  • Do you know that $\sup_{\epsilon > 0} \sum_{n=1}^\infty \frac{1}{n^{1+\epsilon}} = \infty$ ? – H1ghfiv3 Feb 01 '19 at 14:24
  • @BerniWaterman Yes but not $\sup_{\epsilon>0} \sum_{n=1}^\infty\frac{1}{n^{2+\epsilon}}$. – DINEDINE Feb 01 '19 at 14:30
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    Divergence of this series has already been addressed here. You should either delete your question, ask for someone to close it as a duplicate, or ask Andrés E. Caicedo to copy part of his answer here. – Gabriel Romon Feb 01 '19 at 14:37
  • Since you're a French speaker, you can also directly look for the issue 119-1 of the Revue de la filière mathématique. There's an entry by Brighi and Chevallier dedicated to the divergence of the series. They give a nice elementary exposition. Here's a screenshot. – Gabriel Romon Feb 01 '19 at 14:44
  • @GabrielRomon I’ve seen that paper using a weak version Denjoy-Koksma inequality, I was just looking for an elementary proof without so sophisticated stuff. – DINEDINE Feb 01 '19 at 14:49

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