Series of reciprocal of integers












4












$begingroup$


This is a question I asked myself today...



$ $



Do you know if it is possible to build a strictly-increasing sequence $(u_n)_{ninmathbb{N}^star}$ of positive integers such that $displaystylesum_{n=1}^{+infty}frac{1}{u_n}<+infty$ and such that for any given $0<varepsilon<1$, one has $displaystylesum_{n=1}^{+infty}frac{1}{u_n^varepsilon}=+infty$ ?



$ $



This is looking for a Dirichlet series (with reciprocal of $u_n$ coefficients) with abscissa of convergence $sigma=1$ and that has a finite limit at $s=1$, does that exist ? It would mean that the corresponding L-function (if the series was extendable around $s=1$) would have no pole at $s=1$. This is tough because the L-function would not be in the Selberg class, and those are hard to study...










share|cite|improve this question











$endgroup$

















    4












    $begingroup$


    This is a question I asked myself today...



    $ $



    Do you know if it is possible to build a strictly-increasing sequence $(u_n)_{ninmathbb{N}^star}$ of positive integers such that $displaystylesum_{n=1}^{+infty}frac{1}{u_n}<+infty$ and such that for any given $0<varepsilon<1$, one has $displaystylesum_{n=1}^{+infty}frac{1}{u_n^varepsilon}=+infty$ ?



    $ $



    This is looking for a Dirichlet series (with reciprocal of $u_n$ coefficients) with abscissa of convergence $sigma=1$ and that has a finite limit at $s=1$, does that exist ? It would mean that the corresponding L-function (if the series was extendable around $s=1$) would have no pole at $s=1$. This is tough because the L-function would not be in the Selberg class, and those are hard to study...










    share|cite|improve this question











    $endgroup$















      4












      4








      4


      1



      $begingroup$


      This is a question I asked myself today...



      $ $



      Do you know if it is possible to build a strictly-increasing sequence $(u_n)_{ninmathbb{N}^star}$ of positive integers such that $displaystylesum_{n=1}^{+infty}frac{1}{u_n}<+infty$ and such that for any given $0<varepsilon<1$, one has $displaystylesum_{n=1}^{+infty}frac{1}{u_n^varepsilon}=+infty$ ?



      $ $



      This is looking for a Dirichlet series (with reciprocal of $u_n$ coefficients) with abscissa of convergence $sigma=1$ and that has a finite limit at $s=1$, does that exist ? It would mean that the corresponding L-function (if the series was extendable around $s=1$) would have no pole at $s=1$. This is tough because the L-function would not be in the Selberg class, and those are hard to study...










      share|cite|improve this question











      $endgroup$




      This is a question I asked myself today...



      $ $



      Do you know if it is possible to build a strictly-increasing sequence $(u_n)_{ninmathbb{N}^star}$ of positive integers such that $displaystylesum_{n=1}^{+infty}frac{1}{u_n}<+infty$ and such that for any given $0<varepsilon<1$, one has $displaystylesum_{n=1}^{+infty}frac{1}{u_n^varepsilon}=+infty$ ?



      $ $



      This is looking for a Dirichlet series (with reciprocal of $u_n$ coefficients) with abscissa of convergence $sigma=1$ and that has a finite limit at $s=1$, does that exist ? It would mean that the corresponding L-function (if the series was extendable around $s=1$) would have no pole at $s=1$. This is tough because the L-function would not be in the Selberg class, and those are hard to study...







      sequences-and-series l-functions






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      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 7 '18 at 1:46







      Anthony

















      asked Dec 7 '18 at 1:26









      AnthonyAnthony

      628




      628






















          1 Answer
          1






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          3












          $begingroup$

          You can take $u_n = lceil n log^2 nrceil$:
          $$
          sum_{n=1}^infty frac{1}{n log^2 n} < infty
          $$

          but
          $$
          sum_{n=1}^infty frac{1}{n^varepsilon log^{2varepsilon} n} = infty
          $$

          for every $varepsilon < 1$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Oh that's totally true ! I tried $u_n=nlog(n)$ (the PNT makes sure $p_n$ works), but I didn't try that ! Thanks !
            $endgroup$
            – Anthony
            Dec 7 '18 at 2:01










          • $begingroup$
            @Anthony Glad this helped!
            $endgroup$
            – Clement C.
            Dec 7 '18 at 2:02











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          1 Answer
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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          3












          $begingroup$

          You can take $u_n = lceil n log^2 nrceil$:
          $$
          sum_{n=1}^infty frac{1}{n log^2 n} < infty
          $$

          but
          $$
          sum_{n=1}^infty frac{1}{n^varepsilon log^{2varepsilon} n} = infty
          $$

          for every $varepsilon < 1$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Oh that's totally true ! I tried $u_n=nlog(n)$ (the PNT makes sure $p_n$ works), but I didn't try that ! Thanks !
            $endgroup$
            – Anthony
            Dec 7 '18 at 2:01










          • $begingroup$
            @Anthony Glad this helped!
            $endgroup$
            – Clement C.
            Dec 7 '18 at 2:02
















          3












          $begingroup$

          You can take $u_n = lceil n log^2 nrceil$:
          $$
          sum_{n=1}^infty frac{1}{n log^2 n} < infty
          $$

          but
          $$
          sum_{n=1}^infty frac{1}{n^varepsilon log^{2varepsilon} n} = infty
          $$

          for every $varepsilon < 1$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Oh that's totally true ! I tried $u_n=nlog(n)$ (the PNT makes sure $p_n$ works), but I didn't try that ! Thanks !
            $endgroup$
            – Anthony
            Dec 7 '18 at 2:01










          • $begingroup$
            @Anthony Glad this helped!
            $endgroup$
            – Clement C.
            Dec 7 '18 at 2:02














          3












          3








          3





          $begingroup$

          You can take $u_n = lceil n log^2 nrceil$:
          $$
          sum_{n=1}^infty frac{1}{n log^2 n} < infty
          $$

          but
          $$
          sum_{n=1}^infty frac{1}{n^varepsilon log^{2varepsilon} n} = infty
          $$

          for every $varepsilon < 1$.






          share|cite|improve this answer









          $endgroup$



          You can take $u_n = lceil n log^2 nrceil$:
          $$
          sum_{n=1}^infty frac{1}{n log^2 n} < infty
          $$

          but
          $$
          sum_{n=1}^infty frac{1}{n^varepsilon log^{2varepsilon} n} = infty
          $$

          for every $varepsilon < 1$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 7 '18 at 2:00









          Clement C.Clement C.

          49.9k33887




          49.9k33887












          • $begingroup$
            Oh that's totally true ! I tried $u_n=nlog(n)$ (the PNT makes sure $p_n$ works), but I didn't try that ! Thanks !
            $endgroup$
            – Anthony
            Dec 7 '18 at 2:01










          • $begingroup$
            @Anthony Glad this helped!
            $endgroup$
            – Clement C.
            Dec 7 '18 at 2:02


















          • $begingroup$
            Oh that's totally true ! I tried $u_n=nlog(n)$ (the PNT makes sure $p_n$ works), but I didn't try that ! Thanks !
            $endgroup$
            – Anthony
            Dec 7 '18 at 2:01










          • $begingroup$
            @Anthony Glad this helped!
            $endgroup$
            – Clement C.
            Dec 7 '18 at 2:02
















          $begingroup$
          Oh that's totally true ! I tried $u_n=nlog(n)$ (the PNT makes sure $p_n$ works), but I didn't try that ! Thanks !
          $endgroup$
          – Anthony
          Dec 7 '18 at 2:01




          $begingroup$
          Oh that's totally true ! I tried $u_n=nlog(n)$ (the PNT makes sure $p_n$ works), but I didn't try that ! Thanks !
          $endgroup$
          – Anthony
          Dec 7 '18 at 2:01












          $begingroup$
          @Anthony Glad this helped!
          $endgroup$
          – Clement C.
          Dec 7 '18 at 2:02




          $begingroup$
          @Anthony Glad this helped!
          $endgroup$
          – Clement C.
          Dec 7 '18 at 2:02


















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