Convergence or divergence of a series given divergent series












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If $displaystylesumlimits_{n=0}^{infty}a_n$ is divergent and $a_n > 0$ for all $n$, then does $displaystylesumlimits_{n=0}^{infty} dfrac{a_n}{1+n^2 a_n}$ converge or diverge?



The only progress I have is that if you consider the harmonic series, then we get the series with terms $dfrac{1}{n(n+1)}$, which converges.






sequences-and-series convergence







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    1












    $begingroup$


    If $displaystylesumlimits_{n=0}^{infty}a_n$ is divergent and $a_n > 0$ for all $n$, then does $displaystylesumlimits_{n=0}^{infty} dfrac{a_n}{1+n^2 a_n}$ converge or diverge?



    The only progress I have is that if you consider the harmonic series, then we get the series with terms $dfrac{1}{n(n+1)}$, which converges.






    sequences-and-series convergence







    share|cite|improve this question









    $endgroup$















      1












      1








      1


      0



      $begingroup$


      If $displaystylesumlimits_{n=0}^{infty}a_n$ is divergent and $a_n > 0$ for all $n$, then does $displaystylesumlimits_{n=0}^{infty} dfrac{a_n}{1+n^2 a_n}$ converge or diverge?



      The only progress I have is that if you consider the harmonic series, then we get the series with terms $dfrac{1}{n(n+1)}$, which converges.






      sequences-and-series convergence







      share|cite|improve this question









      $endgroup$




      If $displaystylesumlimits_{n=0}^{infty}a_n$ is divergent and $a_n > 0$ for all $n$, then does $displaystylesumlimits_{n=0}^{infty} dfrac{a_n}{1+n^2 a_n}$ converge or diverge?



      The only progress I have is that if you consider the harmonic series, then we get the series with terms $dfrac{1}{n(n+1)}$, which converges.






      sequences-and-series convergence




      sequences-and-series convergence






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked 1 hour ago









      A. SmithA. Smith

      435




      435






















          1 Answer
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          $begingroup$

          HINT
          You have
          $$
          0<sum frac{a_n}{1+n^2a_n} < sum frac{a_n}{n^2a_n} = sum frac{1}{n^2}
          $$






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            (+1) Looks like the assumption that $sum a_n$ diverges is unnecessary.
            $endgroup$
            – Yanko
            1 hour ago










          • $begingroup$
            @Yanko likely professor tried to trick them into thinking this may diverge as well, not sure. Definitely not needed...
            $endgroup$
            – gt6989b
            1 hour ago











          Your Answer





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          1 Answer
          1






          active

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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          7












          $begingroup$

          HINT
          You have
          $$
          0<sum frac{a_n}{1+n^2a_n} < sum frac{a_n}{n^2a_n} = sum frac{1}{n^2}
          $$






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            (+1) Looks like the assumption that $sum a_n$ diverges is unnecessary.
            $endgroup$
            – Yanko
            1 hour ago










          • $begingroup$
            @Yanko likely professor tried to trick them into thinking this may diverge as well, not sure. Definitely not needed...
            $endgroup$
            – gt6989b
            1 hour ago
















          7












          $begingroup$

          HINT
          You have
          $$
          0<sum frac{a_n}{1+n^2a_n} < sum frac{a_n}{n^2a_n} = sum frac{1}{n^2}
          $$






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            (+1) Looks like the assumption that $sum a_n$ diverges is unnecessary.
            $endgroup$
            – Yanko
            1 hour ago










          • $begingroup$
            @Yanko likely professor tried to trick them into thinking this may diverge as well, not sure. Definitely not needed...
            $endgroup$
            – gt6989b
            1 hour ago














          7












          7








          7





          $begingroup$

          HINT
          You have
          $$
          0<sum frac{a_n}{1+n^2a_n} < sum frac{a_n}{n^2a_n} = sum frac{1}{n^2}
          $$






          share|cite|improve this answer









          $endgroup$



          HINT
          You have
          $$
          0<sum frac{a_n}{1+n^2a_n} < sum frac{a_n}{n^2a_n} = sum frac{1}{n^2}
          $$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 1 hour ago









          gt6989bgt6989b

          33.6k22453




          33.6k22453








          • 1




            $begingroup$
            (+1) Looks like the assumption that $sum a_n$ diverges is unnecessary.
            $endgroup$
            – Yanko
            1 hour ago










          • $begingroup$
            @Yanko likely professor tried to trick them into thinking this may diverge as well, not sure. Definitely not needed...
            $endgroup$
            – gt6989b
            1 hour ago














          • 1




            $begingroup$
            (+1) Looks like the assumption that $sum a_n$ diverges is unnecessary.
            $endgroup$
            – Yanko
            1 hour ago










          • $begingroup$
            @Yanko likely professor tried to trick them into thinking this may diverge as well, not sure. Definitely not needed...
            $endgroup$
            – gt6989b
            1 hour ago








          1




          1




          $begingroup$
          (+1) Looks like the assumption that $sum a_n$ diverges is unnecessary.
          $endgroup$
          – Yanko
          1 hour ago




          $begingroup$
          (+1) Looks like the assumption that $sum a_n$ diverges is unnecessary.
          $endgroup$
          – Yanko
          1 hour ago












          $begingroup$
          @Yanko likely professor tried to trick them into thinking this may diverge as well, not sure. Definitely not needed...
          $endgroup$
          – gt6989b
          1 hour ago




          $begingroup$
          @Yanko likely professor tried to trick them into thinking this may diverge as well, not sure. Definitely not needed...
          $endgroup$
          – gt6989b
          1 hour ago


















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