Study convergence of the series $sumlimits_nleft(left(3+3^-n^3right)^n^2-3^k(n)right)$, where $k(n)=fracn^3 +n^63^n^21+n^43^n^2$

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The question is to study the convergence of the series $sum x(n)$, where
$$x(n)=left(3+3^-n^3right)^n^2-3^k(n)qquad k(n)=fracn^3 +n^63^n^21+n^43^n^2$$




I tried to put in evidence $3^k(n)$ or to transform every $3^x$ into $e^xlog(3)$, but with no success.




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    The question is to study the convergence of the series $sum x(n)$, where
    $$x(n)=left(3+3^-n^3right)^n^2-3^k(n)qquad k(n)=fracn^3 +n^63^n^21+n^43^n^2$$




    I tried to put in evidence $3^k(n)$ or to transform every $3^x$ into $e^xlog(3)$, but with no success.




    sequences-and-series




    share|cite|improve this question























      up vote
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      up vote
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      The question is to study the convergence of the series $sum x(n)$, where
      $$x(n)=left(3+3^-n^3right)^n^2-3^k(n)qquad k(n)=fracn^3 +n^63^n^21+n^43^n^2$$




      I tried to put in evidence $3^k(n)$ or to transform every $3^x$ into $e^xlog(3)$, but with no success.




      sequences-and-series




      share|cite|improve this question














      The question is to study the convergence of the series $sum x(n)$, where
      $$x(n)=left(3+3^-n^3right)^n^2-3^k(n)qquad k(n)=fracn^3 +n^63^n^21+n^43^n^2$$




      I tried to put in evidence $3^k(n)$ or to transform every $3^x$ into $e^xlog(3)$, but with no success.





      sequences-and-series





      share|cite|improve this question












      share|cite|improve this question




      share|cite|improve this question








      edited Aug 7 at 20:12









      Did

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      asked Aug 7 at 17:48









      Antony Magnavacchi

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          Tools:



          • If $a_nto0$ and $a_nb_nto0$, then $(1+a_n)^b_n-1sim a_nb_n$.

          • If $c>0$ and $d_nto0$, then $c^d_n-1sim d_nlog c$.



          Note that $$x(n)=3^n^2left(y(n)-3^ell(n)right)$$ with $$y(n)=left(1+3^-n^3-1right)^n^2$$ and $$ell(n)=k(n)-n^2=fracn^3-n^21+n^43^n^2sim n^-13^-n^2$$ Thus, $y(n)to1$ and $3^ell(n)to1$ with $$y(n)-1sim n^23^-n^3-1qquad3^ell(n)-1simell(n)log3sim n^-13^-n^2log3$$ This implies that $$y(n)-3^ell(n)sim-n^-13^-n^2log3$$ hence $$x(n)sim-n^-1log3$$ and the series $sum x(n)$ diverges.






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            up vote
            2
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            Tools:



            • If $a_nto0$ and $a_nb_nto0$, then $(1+a_n)^b_n-1sim a_nb_n$.

            • If $c>0$ and $d_nto0$, then $c^d_n-1sim d_nlog c$.



            Note that $$x(n)=3^n^2left(y(n)-3^ell(n)right)$$ with $$y(n)=left(1+3^-n^3-1right)^n^2$$ and $$ell(n)=k(n)-n^2=fracn^3-n^21+n^43^n^2sim n^-13^-n^2$$ Thus, $y(n)to1$ and $3^ell(n)to1$ with $$y(n)-1sim n^23^-n^3-1qquad3^ell(n)-1simell(n)log3sim n^-13^-n^2log3$$ This implies that $$y(n)-3^ell(n)sim-n^-13^-n^2log3$$ hence $$x(n)sim-n^-1log3$$ and the series $sum x(n)$ diverges.






            share|cite|improve this answer



























              up vote
              2
              down vote














              Tools:



              • If $a_nto0$ and $a_nb_nto0$, then $(1+a_n)^b_n-1sim a_nb_n$.

              • If $c>0$ and $d_nto0$, then $c^d_n-1sim d_nlog c$.



              Note that $$x(n)=3^n^2left(y(n)-3^ell(n)right)$$ with $$y(n)=left(1+3^-n^3-1right)^n^2$$ and $$ell(n)=k(n)-n^2=fracn^3-n^21+n^43^n^2sim n^-13^-n^2$$ Thus, $y(n)to1$ and $3^ell(n)to1$ with $$y(n)-1sim n^23^-n^3-1qquad3^ell(n)-1simell(n)log3sim n^-13^-n^2log3$$ This implies that $$y(n)-3^ell(n)sim-n^-13^-n^2log3$$ hence $$x(n)sim-n^-1log3$$ and the series $sum x(n)$ diverges.






              share|cite|improve this answer

























                up vote
                2
                down vote










                up vote
                2
                down vote










                Tools:



                • If $a_nto0$ and $a_nb_nto0$, then $(1+a_n)^b_n-1sim a_nb_n$.

                • If $c>0$ and $d_nto0$, then $c^d_n-1sim d_nlog c$.



                Note that $$x(n)=3^n^2left(y(n)-3^ell(n)right)$$ with $$y(n)=left(1+3^-n^3-1right)^n^2$$ and $$ell(n)=k(n)-n^2=fracn^3-n^21+n^43^n^2sim n^-13^-n^2$$ Thus, $y(n)to1$ and $3^ell(n)to1$ with $$y(n)-1sim n^23^-n^3-1qquad3^ell(n)-1simell(n)log3sim n^-13^-n^2log3$$ This implies that $$y(n)-3^ell(n)sim-n^-13^-n^2log3$$ hence $$x(n)sim-n^-1log3$$ and the series $sum x(n)$ diverges.






                share|cite|improve this answer
















                Tools:



                • If $a_nto0$ and $a_nb_nto0$, then $(1+a_n)^b_n-1sim a_nb_n$.

                • If $c>0$ and $d_nto0$, then $c^d_n-1sim d_nlog c$.



                Note that $$x(n)=3^n^2left(y(n)-3^ell(n)right)$$ with $$y(n)=left(1+3^-n^3-1right)^n^2$$ and $$ell(n)=k(n)-n^2=fracn^3-n^21+n^43^n^2sim n^-13^-n^2$$ Thus, $y(n)to1$ and $3^ell(n)to1$ with $$y(n)-1sim n^23^-n^3-1qquad3^ell(n)-1simell(n)log3sim n^-13^-n^2log3$$ This implies that $$y(n)-3^ell(n)sim-n^-13^-n^2log3$$ hence $$x(n)sim-n^-1log3$$ and the series $sum x(n)$ diverges.







                share|cite|improve this answer















                share|cite|improve this answer



                share|cite|improve this answer








                edited Aug 7 at 20:14


























                answered Aug 7 at 20:09









                Did

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