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Determine whether the sequence converges or diverges. Exponential sequence and power sequence.



The Next CEO of Stack OverflowSeries diverges or convergesDetermine if the sequence converges or diverges.Proving a sequence diverges to infinityDetermine whether or not the series $sumlimits_n=2^infty (fracn+4n+8)^n$ converges or divergesHow to determine whether the series converges or diverges?Proving that a sequence diverges if another one divergesDetermine whether the series converges or diverges.How to determine whether a sequence converges or divergesDetermine whether the sequence converge or divergesDetermine whether the sequence converges or diverges.










0












$begingroup$


I have this sequence:



$$a_n = frac4^n1 + 9^n$$



How do I figure out if this diverges or converges? I tried using L'Hospital but that seems too complicated here. What else can I do?






sequences-and-series







share|cite|improve this question











$endgroup$











  • $begingroup$
    You shouldn't piggyback additional questions to your original question, especially since the original question has been answered. Instead, you should ask a second question.
    $endgroup$
    – MPW
    2 days ago
















0












$begingroup$


I have this sequence:



$$a_n = frac4^n1 + 9^n$$



How do I figure out if this diverges or converges? I tried using L'Hospital but that seems too complicated here. What else can I do?






sequences-and-series







share|cite|improve this question











$endgroup$











  • $begingroup$
    You shouldn't piggyback additional questions to your original question, especially since the original question has been answered. Instead, you should ask a second question.
    $endgroup$
    – MPW
    2 days ago














0












0








0





$begingroup$


I have this sequence:



$$a_n = frac4^n1 + 9^n$$



How do I figure out if this diverges or converges? I tried using L'Hospital but that seems too complicated here. What else can I do?






sequences-and-series







share|cite|improve this question











$endgroup$




I have this sequence:



$$a_n = frac4^n1 + 9^n$$



How do I figure out if this diverges or converges? I tried using L'Hospital but that seems too complicated here. What else can I do?






sequences-and-series




sequences-and-series






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 2 days ago







Kitty Capital

















asked 2 days ago









Kitty CapitalKitty Capital

1096




1096











  • $begingroup$
    You shouldn't piggyback additional questions to your original question, especially since the original question has been answered. Instead, you should ask a second question.
    $endgroup$
    – MPW
    2 days ago

















  • $begingroup$
    You shouldn't piggyback additional questions to your original question, especially since the original question has been answered. Instead, you should ask a second question.
    $endgroup$
    – MPW
    2 days ago
















$begingroup$
You shouldn't piggyback additional questions to your original question, especially since the original question has been answered. Instead, you should ask a second question.
$endgroup$
– MPW
2 days ago





$begingroup$
You shouldn't piggyback additional questions to your original question, especially since the original question has been answered. Instead, you should ask a second question.
$endgroup$
– MPW
2 days ago











4 Answers
4






active

oldest

votes


















1












$begingroup$

You can rewrite this (dividing by $4^n$ on top and bottom) as
$$frac1underbrace4^-n_to 0 + underbrace2.25^n_toinfty$$



As $ntoinfty$, the first term in the denominator shrinks to zero and the other term grows without bound; so their sum grows without bound.



Since the denominator grows without bound and the numerator is a nonzero constant, the whole fraction shrinks to zero.






share|cite|improve this answer









$endgroup$




















    1












    $begingroup$

    It converges to $0$:



    $$a_n=frac4^-n4^-na_n=frac1(1/4)^n+(9/4)^nto0$$



    because $(1/4)^nto 0$ and $(9/4)^nto+infty$ for $ntoinfty$.






    share|cite|improve this answer











    $endgroup$




















      0












      $begingroup$

      It converges and has limit $0$ because $a_n$ is non-negative and



      $$
      a_n lt frac4^n9^n=left(frac49right)^n
      $$
      and the latter is a basic null sequence i.e. $left(frac49right)^n to0$ as $n to infty$.






      share|cite|improve this answer









      $endgroup$




















        0












        $begingroup$

        For fun:



        Let $n ge 1$:



        $0< dfrac4^n1+9^n lt dfrac4^n9^n =left (dfrac49right )^n lt $



        $left (dfrac12 right )^n =dfrac12^n lt dfrac11+n lt dfrac 1n$.



        The limit $n rightarrow infty$ is ?



        Used: $2^n=(1+1)^n gt 1+ n$ (Binomial expansion).






        share|cite|improve this answer











        $endgroup$













          Your Answer





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          4 Answers
          4






          active

          oldest

          votes








          4 Answers
          4






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          1












          $begingroup$

          You can rewrite this (dividing by $4^n$ on top and bottom) as
          $$frac1underbrace4^-n_to 0 + underbrace2.25^n_toinfty$$



          As $ntoinfty$, the first term in the denominator shrinks to zero and the other term grows without bound; so their sum grows without bound.



          Since the denominator grows without bound and the numerator is a nonzero constant, the whole fraction shrinks to zero.






          share|cite|improve this answer









          $endgroup$

















            1












            $begingroup$

            You can rewrite this (dividing by $4^n$ on top and bottom) as
            $$frac1underbrace4^-n_to 0 + underbrace2.25^n_toinfty$$



            As $ntoinfty$, the first term in the denominator shrinks to zero and the other term grows without bound; so their sum grows without bound.



            Since the denominator grows without bound and the numerator is a nonzero constant, the whole fraction shrinks to zero.






            share|cite|improve this answer









            $endgroup$















              1












              1








              1





              $begingroup$

              You can rewrite this (dividing by $4^n$ on top and bottom) as
              $$frac1underbrace4^-n_to 0 + underbrace2.25^n_toinfty$$



              As $ntoinfty$, the first term in the denominator shrinks to zero and the other term grows without bound; so their sum grows without bound.



              Since the denominator grows without bound and the numerator is a nonzero constant, the whole fraction shrinks to zero.






              share|cite|improve this answer









              $endgroup$



              You can rewrite this (dividing by $4^n$ on top and bottom) as
              $$frac1underbrace4^-n_to 0 + underbrace2.25^n_toinfty$$



              As $ntoinfty$, the first term in the denominator shrinks to zero and the other term grows without bound; so their sum grows without bound.



              Since the denominator grows without bound and the numerator is a nonzero constant, the whole fraction shrinks to zero.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered 2 days ago









              MPWMPW

              31k12157




              31k12157





















                  1












                  $begingroup$

                  It converges to $0$:



                  $$a_n=frac4^-n4^-na_n=frac1(1/4)^n+(9/4)^nto0$$



                  because $(1/4)^nto 0$ and $(9/4)^nto+infty$ for $ntoinfty$.






                  share|cite|improve this answer











                  $endgroup$

















                    1












                    $begingroup$

                    It converges to $0$:



                    $$a_n=frac4^-n4^-na_n=frac1(1/4)^n+(9/4)^nto0$$



                    because $(1/4)^nto 0$ and $(9/4)^nto+infty$ for $ntoinfty$.






                    share|cite|improve this answer











                    $endgroup$















                      1












                      1








                      1





                      $begingroup$

                      It converges to $0$:



                      $$a_n=frac4^-n4^-na_n=frac1(1/4)^n+(9/4)^nto0$$



                      because $(1/4)^nto 0$ and $(9/4)^nto+infty$ for $ntoinfty$.






                      share|cite|improve this answer











                      $endgroup$



                      It converges to $0$:



                      $$a_n=frac4^-n4^-na_n=frac1(1/4)^n+(9/4)^nto0$$



                      because $(1/4)^nto 0$ and $(9/4)^nto+infty$ for $ntoinfty$.







                      share|cite|improve this answer














                      share|cite|improve this answer



                      share|cite|improve this answer








                      edited 2 days ago

























                      answered 2 days ago









                      st.mathst.math

                      1,05615




                      1,05615





















                          0












                          $begingroup$

                          It converges and has limit $0$ because $a_n$ is non-negative and



                          $$
                          a_n lt frac4^n9^n=left(frac49right)^n
                          $$
                          and the latter is a basic null sequence i.e. $left(frac49right)^n to0$ as $n to infty$.






                          share|cite|improve this answer









                          $endgroup$

















                            0












                            $begingroup$

                            It converges and has limit $0$ because $a_n$ is non-negative and



                            $$
                            a_n lt frac4^n9^n=left(frac49right)^n
                            $$
                            and the latter is a basic null sequence i.e. $left(frac49right)^n to0$ as $n to infty$.






                            share|cite|improve this answer









                            $endgroup$















                              0












                              0








                              0





                              $begingroup$

                              It converges and has limit $0$ because $a_n$ is non-negative and



                              $$
                              a_n lt frac4^n9^n=left(frac49right)^n
                              $$
                              and the latter is a basic null sequence i.e. $left(frac49right)^n to0$ as $n to infty$.






                              share|cite|improve this answer









                              $endgroup$



                              It converges and has limit $0$ because $a_n$ is non-negative and



                              $$
                              a_n lt frac4^n9^n=left(frac49right)^n
                              $$
                              and the latter is a basic null sequence i.e. $left(frac49right)^n to0$ as $n to infty$.







                              share|cite|improve this answer












                              share|cite|improve this answer



                              share|cite|improve this answer










                              answered 2 days ago









                              PM.PM.

                              3,4432925




                              3,4432925





















                                  0












                                  $begingroup$

                                  For fun:



                                  Let $n ge 1$:



                                  $0< dfrac4^n1+9^n lt dfrac4^n9^n =left (dfrac49right )^n lt $



                                  $left (dfrac12 right )^n =dfrac12^n lt dfrac11+n lt dfrac 1n$.



                                  The limit $n rightarrow infty$ is ?



                                  Used: $2^n=(1+1)^n gt 1+ n$ (Binomial expansion).






                                  share|cite|improve this answer











                                  $endgroup$

















                                    0












                                    $begingroup$

                                    For fun:



                                    Let $n ge 1$:



                                    $0< dfrac4^n1+9^n lt dfrac4^n9^n =left (dfrac49right )^n lt $



                                    $left (dfrac12 right )^n =dfrac12^n lt dfrac11+n lt dfrac 1n$.



                                    The limit $n rightarrow infty$ is ?



                                    Used: $2^n=(1+1)^n gt 1+ n$ (Binomial expansion).






                                    share|cite|improve this answer











                                    $endgroup$















                                      0












                                      0








                                      0





                                      $begingroup$

                                      For fun:



                                      Let $n ge 1$:



                                      $0< dfrac4^n1+9^n lt dfrac4^n9^n =left (dfrac49right )^n lt $



                                      $left (dfrac12 right )^n =dfrac12^n lt dfrac11+n lt dfrac 1n$.



                                      The limit $n rightarrow infty$ is ?



                                      Used: $2^n=(1+1)^n gt 1+ n$ (Binomial expansion).






                                      share|cite|improve this answer











                                      $endgroup$



                                      For fun:



                                      Let $n ge 1$:



                                      $0< dfrac4^n1+9^n lt dfrac4^n9^n =left (dfrac49right )^n lt $



                                      $left (dfrac12 right )^n =dfrac12^n lt dfrac11+n lt dfrac 1n$.



                                      The limit $n rightarrow infty$ is ?



                                      Used: $2^n=(1+1)^n gt 1+ n$ (Binomial expansion).







                                      share|cite|improve this answer














                                      share|cite|improve this answer



                                      share|cite|improve this answer








                                      edited 2 days ago

























                                      answered 2 days ago









                                      Peter SzilasPeter Szilas

                                      11.6k2822




                                      11.6k2822



























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