How to show divergence of $sum_{n = 1}^{infty} 2^{-o(1)}frac{1}{n}$?












0












$begingroup$


I am trying to show the divergence of the following sum:



$$sum_{n = 1}^{infty} 2^{-o(1)}frac{1}{n}$$



where the exponent of the 2 is negative little-o of 1. I have tried some ideas such as the following:



since $o(1)$ means that $lim_{n rightarrow infty} f(n)/1 rightarrow 0$, then we can potentially look at $max_n f(n)$ and $min_n f(n)$ and use those to bound the term $2^{-o(1)}$ in the sum, but I'm not sure how to handle functions $f(n)$ that oscillate. Any thoughts are appreciated!






sequences-and-series analysis computational-complexity







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$endgroup$












  • $begingroup$
    An $o(1)$ function cannot oscillate wildly.
    $endgroup$
    – Yves Daoust
    Dec 10 '18 at 21:56












  • $begingroup$
    apologies I have fixed that in the question
    $endgroup$
    – hologram
    Dec 10 '18 at 21:59
















0












$begingroup$


I am trying to show the divergence of the following sum:



$$sum_{n = 1}^{infty} 2^{-o(1)}frac{1}{n}$$



where the exponent of the 2 is negative little-o of 1. I have tried some ideas such as the following:



since $o(1)$ means that $lim_{n rightarrow infty} f(n)/1 rightarrow 0$, then we can potentially look at $max_n f(n)$ and $min_n f(n)$ and use those to bound the term $2^{-o(1)}$ in the sum, but I'm not sure how to handle functions $f(n)$ that oscillate. Any thoughts are appreciated!






sequences-and-series analysis computational-complexity







share|cite|improve this question











$endgroup$












  • $begingroup$
    An $o(1)$ function cannot oscillate wildly.
    $endgroup$
    – Yves Daoust
    Dec 10 '18 at 21:56












  • $begingroup$
    apologies I have fixed that in the question
    $endgroup$
    – hologram
    Dec 10 '18 at 21:59














0












0








0





$begingroup$


I am trying to show the divergence of the following sum:



$$sum_{n = 1}^{infty} 2^{-o(1)}frac{1}{n}$$



where the exponent of the 2 is negative little-o of 1. I have tried some ideas such as the following:



since $o(1)$ means that $lim_{n rightarrow infty} f(n)/1 rightarrow 0$, then we can potentially look at $max_n f(n)$ and $min_n f(n)$ and use those to bound the term $2^{-o(1)}$ in the sum, but I'm not sure how to handle functions $f(n)$ that oscillate. Any thoughts are appreciated!






sequences-and-series analysis computational-complexity







share|cite|improve this question











$endgroup$




I am trying to show the divergence of the following sum:



$$sum_{n = 1}^{infty} 2^{-o(1)}frac{1}{n}$$



where the exponent of the 2 is negative little-o of 1. I have tried some ideas such as the following:



since $o(1)$ means that $lim_{n rightarrow infty} f(n)/1 rightarrow 0$, then we can potentially look at $max_n f(n)$ and $min_n f(n)$ and use those to bound the term $2^{-o(1)}$ in the sum, but I'm not sure how to handle functions $f(n)$ that oscillate. Any thoughts are appreciated!






sequences-and-series analysis computational-complexity




sequences-and-series analysis computational-complexity






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 10 '18 at 21:58







hologram

















asked Dec 10 '18 at 21:43









hologramhologram

14312




14312












  • $begingroup$
    An $o(1)$ function cannot oscillate wildly.
    $endgroup$
    – Yves Daoust
    Dec 10 '18 at 21:56












  • $begingroup$
    apologies I have fixed that in the question
    $endgroup$
    – hologram
    Dec 10 '18 at 21:59


















  • $begingroup$
    An $o(1)$ function cannot oscillate wildly.
    $endgroup$
    – Yves Daoust
    Dec 10 '18 at 21:56












  • $begingroup$
    apologies I have fixed that in the question
    $endgroup$
    – hologram
    Dec 10 '18 at 21:59
















$begingroup$
An $o(1)$ function cannot oscillate wildly.
$endgroup$
– Yves Daoust
Dec 10 '18 at 21:56






$begingroup$
An $o(1)$ function cannot oscillate wildly.
$endgroup$
– Yves Daoust
Dec 10 '18 at 21:56














$begingroup$
apologies I have fixed that in the question
$endgroup$
– hologram
Dec 10 '18 at 21:59




$begingroup$
apologies I have fixed that in the question
$endgroup$
– hologram
Dec 10 '18 at 21:59










1 Answer
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$begingroup$

If $f(n)=o(1)$, by definition, for some $epsilon>0$, you will find an $N$ such that



$$n>Nimplies|f(n)|<epsilonimplies 2^{-f(n)}>2^{-epsilon}.$$



Then the tail of the sum diverges as it is bounded below by an harmonic series.






share|cite|improve this answer











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    1 Answer
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    active

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    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    3












    $begingroup$

    If $f(n)=o(1)$, by definition, for some $epsilon>0$, you will find an $N$ such that



    $$n>Nimplies|f(n)|<epsilonimplies 2^{-f(n)}>2^{-epsilon}.$$



    Then the tail of the sum diverges as it is bounded below by an harmonic series.






    share|cite|improve this answer











    $endgroup$


















      3












      $begingroup$

      If $f(n)=o(1)$, by definition, for some $epsilon>0$, you will find an $N$ such that



      $$n>Nimplies|f(n)|<epsilonimplies 2^{-f(n)}>2^{-epsilon}.$$



      Then the tail of the sum diverges as it is bounded below by an harmonic series.






      share|cite|improve this answer











      $endgroup$
















        3












        3








        3





        $begingroup$

        If $f(n)=o(1)$, by definition, for some $epsilon>0$, you will find an $N$ such that



        $$n>Nimplies|f(n)|<epsilonimplies 2^{-f(n)}>2^{-epsilon}.$$



        Then the tail of the sum diverges as it is bounded below by an harmonic series.






        share|cite|improve this answer











        $endgroup$



        If $f(n)=o(1)$, by definition, for some $epsilon>0$, you will find an $N$ such that



        $$n>Nimplies|f(n)|<epsilonimplies 2^{-f(n)}>2^{-epsilon}.$$



        Then the tail of the sum diverges as it is bounded below by an harmonic series.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 10 '18 at 22:08

























        answered Dec 10 '18 at 21:55









        Yves DaoustYves Daoust

        126k672226




        126k672226






























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