Prove that if $a,bin mathbb{R}$, $a>0$, then $lim frac{n^b}{(1+a)^n}=0$ as $ntoinfty$











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As the title says, i want to show that $$lim_{nto infty} frac{n^b}{(1+a)^n}=0$$ where $a,bin mathbb{R}$ with $a>0$.



I tried to bound the sequence $x_n = frac{n^b}{(1+a)^n}$ and use the sandwish theorem, but have no results. My problem is the exponent in the denominator, maybe it could be bounded with Bernoulli inequality ($(1+a)^ngeq 1+na$) but then the problem is de numerator $n^b$.



I'm grateful for any hints for doing this problem, or some steps to clarify.






calculus sequences-and-series limits







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    up vote
    3
    down vote

    favorite
    1












    As the title says, i want to show that $$lim_{nto infty} frac{n^b}{(1+a)^n}=0$$ where $a,bin mathbb{R}$ with $a>0$.



    I tried to bound the sequence $x_n = frac{n^b}{(1+a)^n}$ and use the sandwish theorem, but have no results. My problem is the exponent in the denominator, maybe it could be bounded with Bernoulli inequality ($(1+a)^ngeq 1+na$) but then the problem is de numerator $n^b$.



    I'm grateful for any hints for doing this problem, or some steps to clarify.






    calculus sequences-and-series limits







    share|cite|improve this question







    New contributor




    Raul_MFer is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.






















      up vote
      3
      down vote

      favorite
      1









      up vote
      3
      down vote

      favorite
      1






      1





      As the title says, i want to show that $$lim_{nto infty} frac{n^b}{(1+a)^n}=0$$ where $a,bin mathbb{R}$ with $a>0$.



      I tried to bound the sequence $x_n = frac{n^b}{(1+a)^n}$ and use the sandwish theorem, but have no results. My problem is the exponent in the denominator, maybe it could be bounded with Bernoulli inequality ($(1+a)^ngeq 1+na$) but then the problem is de numerator $n^b$.



      I'm grateful for any hints for doing this problem, or some steps to clarify.






      calculus sequences-and-series limits







      share|cite|improve this question







      New contributor




      Raul_MFer is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      As the title says, i want to show that $$lim_{nto infty} frac{n^b}{(1+a)^n}=0$$ where $a,bin mathbb{R}$ with $a>0$.



      I tried to bound the sequence $x_n = frac{n^b}{(1+a)^n}$ and use the sandwish theorem, but have no results. My problem is the exponent in the denominator, maybe it could be bounded with Bernoulli inequality ($(1+a)^ngeq 1+na$) but then the problem is de numerator $n^b$.



      I'm grateful for any hints for doing this problem, or some steps to clarify.






      calculus sequences-and-series limits




      calculus sequences-and-series limits






      share|cite|improve this question







      New contributor




      Raul_MFer is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      share|cite|improve this question







      New contributor




      Raul_MFer is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.









      share|cite|improve this question




      share|cite|improve this question






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      New contributor





      Raul_MFer is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.






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      Check out our Code of Conduct.






















          3 Answers
          3






          active

          oldest

          votes

















          up vote
          3
          down vote













          HINT:



          Bernoulli's inequality won't suffice when $bge1$, but the binomial theorem will suffice.



          From the binomial theorem, with $n>b$, we see that



          $$(1+a)^nge binom{n}{lfloor brfloor +1}a^{lfloor b rfloor +1}$$



          where $lfloor cdot rfloor$ is the floor function.



          Can you proceed?






          share|cite|improve this answer























          • Could you explain me more of your idea? I tried to use it to bound the sequence but i still don't see it. Thank you :)
            – Raul_MFer
            2 days ago










          • Write out the term $binom{n}{lfloor b rfloor +1}$. Do you see that it is $O(n^{lfloor b rfloor +1})$
            – Mark Viola
            2 days ago




















          up vote
          0
          down vote













          We need to show that



          $$logleft(frac{n^b}{(1+a)^n}right)=blog n-nlog(1+a)=nleft(bfrac{log n}n-log(1+a)right) to -infty$$



          that is



          $$frac{log n}n to 0$$






          share|cite|improve this answer























          • God dammit I was seconds away from posting.
            – Rushabh Mehta
            2 days ago










          • @RushabhMehta Sorry! :)
            – gimusi
            2 days ago










          • @RushabhMehta Anyway in my opinion if your answer was near to be completed you could post it. I think it is not agains any rule and maybe your way to present things could be more clear to the asker or to other readers than mine.
            – gimusi
            2 days ago


















          up vote
          -1
          down vote













          By the ratio test, the sequence converges if
          $$frac{(n+1)^b}{(1+a)n^b}<1.$$



          This is achieved when



          $$n>frac1{sqrt[b]{1+a}-1},$$ a finite number.




          E.g., with $dfrac{n^3}{2^n}$, for $n>3$,



          $$frac{(n+1)^3}{2n^3}<frac{125}{2cdot64}<1.$$






          share|cite|improve this answer























          • Sorry, I don't see how to use it to prove that the sequence converges to $0$. Could you show a little more?
            – Raul_MFer
            2 days ago












          • @Raul_MFer: did you downvote ?
            – Yves Daoust
            2 days ago











          Your Answer





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          3 Answers
          3






          active

          oldest

          votes








          3 Answers
          3






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes








          up vote
          3
          down vote













          HINT:



          Bernoulli's inequality won't suffice when $bge1$, but the binomial theorem will suffice.



          From the binomial theorem, with $n>b$, we see that



          $$(1+a)^nge binom{n}{lfloor brfloor +1}a^{lfloor b rfloor +1}$$



          where $lfloor cdot rfloor$ is the floor function.



          Can you proceed?






          share|cite|improve this answer























          • Could you explain me more of your idea? I tried to use it to bound the sequence but i still don't see it. Thank you :)
            – Raul_MFer
            2 days ago










          • Write out the term $binom{n}{lfloor b rfloor +1}$. Do you see that it is $O(n^{lfloor b rfloor +1})$
            – Mark Viola
            2 days ago

















          up vote
          3
          down vote













          HINT:



          Bernoulli's inequality won't suffice when $bge1$, but the binomial theorem will suffice.



          From the binomial theorem, with $n>b$, we see that



          $$(1+a)^nge binom{n}{lfloor brfloor +1}a^{lfloor b rfloor +1}$$



          where $lfloor cdot rfloor$ is the floor function.



          Can you proceed?






          share|cite|improve this answer























          • Could you explain me more of your idea? I tried to use it to bound the sequence but i still don't see it. Thank you :)
            – Raul_MFer
            2 days ago










          • Write out the term $binom{n}{lfloor b rfloor +1}$. Do you see that it is $O(n^{lfloor b rfloor +1})$
            – Mark Viola
            2 days ago















          up vote
          3
          down vote










          up vote
          3
          down vote









          HINT:



          Bernoulli's inequality won't suffice when $bge1$, but the binomial theorem will suffice.



          From the binomial theorem, with $n>b$, we see that



          $$(1+a)^nge binom{n}{lfloor brfloor +1}a^{lfloor b rfloor +1}$$



          where $lfloor cdot rfloor$ is the floor function.



          Can you proceed?






          share|cite|improve this answer














          HINT:



          Bernoulli's inequality won't suffice when $bge1$, but the binomial theorem will suffice.



          From the binomial theorem, with $n>b$, we see that



          $$(1+a)^nge binom{n}{lfloor brfloor +1}a^{lfloor b rfloor +1}$$



          where $lfloor cdot rfloor$ is the floor function.



          Can you proceed?







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited 2 days ago

























          answered 2 days ago









          Mark Viola

          129k1273170




          129k1273170












          • Could you explain me more of your idea? I tried to use it to bound the sequence but i still don't see it. Thank you :)
            – Raul_MFer
            2 days ago










          • Write out the term $binom{n}{lfloor b rfloor +1}$. Do you see that it is $O(n^{lfloor b rfloor +1})$
            – Mark Viola
            2 days ago




















          • Could you explain me more of your idea? I tried to use it to bound the sequence but i still don't see it. Thank you :)
            – Raul_MFer
            2 days ago










          • Write out the term $binom{n}{lfloor b rfloor +1}$. Do you see that it is $O(n^{lfloor b rfloor +1})$
            – Mark Viola
            2 days ago


















          Could you explain me more of your idea? I tried to use it to bound the sequence but i still don't see it. Thank you :)
          – Raul_MFer
          2 days ago




          Could you explain me more of your idea? I tried to use it to bound the sequence but i still don't see it. Thank you :)
          – Raul_MFer
          2 days ago












          Write out the term $binom{n}{lfloor b rfloor +1}$. Do you see that it is $O(n^{lfloor b rfloor +1})$
          – Mark Viola
          2 days ago






          Write out the term $binom{n}{lfloor b rfloor +1}$. Do you see that it is $O(n^{lfloor b rfloor +1})$
          – Mark Viola
          2 days ago












          up vote
          0
          down vote













          We need to show that



          $$logleft(frac{n^b}{(1+a)^n}right)=blog n-nlog(1+a)=nleft(bfrac{log n}n-log(1+a)right) to -infty$$



          that is



          $$frac{log n}n to 0$$






          share|cite|improve this answer























          • God dammit I was seconds away from posting.
            – Rushabh Mehta
            2 days ago










          • @RushabhMehta Sorry! :)
            – gimusi
            2 days ago










          • @RushabhMehta Anyway in my opinion if your answer was near to be completed you could post it. I think it is not agains any rule and maybe your way to present things could be more clear to the asker or to other readers than mine.
            – gimusi
            2 days ago















          up vote
          0
          down vote













          We need to show that



          $$logleft(frac{n^b}{(1+a)^n}right)=blog n-nlog(1+a)=nleft(bfrac{log n}n-log(1+a)right) to -infty$$



          that is



          $$frac{log n}n to 0$$






          share|cite|improve this answer























          • God dammit I was seconds away from posting.
            – Rushabh Mehta
            2 days ago










          • @RushabhMehta Sorry! :)
            – gimusi
            2 days ago










          • @RushabhMehta Anyway in my opinion if your answer was near to be completed you could post it. I think it is not agains any rule and maybe your way to present things could be more clear to the asker or to other readers than mine.
            – gimusi
            2 days ago













          up vote
          0
          down vote










          up vote
          0
          down vote









          We need to show that



          $$logleft(frac{n^b}{(1+a)^n}right)=blog n-nlog(1+a)=nleft(bfrac{log n}n-log(1+a)right) to -infty$$



          that is



          $$frac{log n}n to 0$$






          share|cite|improve this answer














          We need to show that



          $$logleft(frac{n^b}{(1+a)^n}right)=blog n-nlog(1+a)=nleft(bfrac{log n}n-log(1+a)right) to -infty$$



          that is



          $$frac{log n}n to 0$$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited 2 days ago

























          answered 2 days ago









          gimusi

          86.1k74392




          86.1k74392












          • God dammit I was seconds away from posting.
            – Rushabh Mehta
            2 days ago










          • @RushabhMehta Sorry! :)
            – gimusi
            2 days ago










          • @RushabhMehta Anyway in my opinion if your answer was near to be completed you could post it. I think it is not agains any rule and maybe your way to present things could be more clear to the asker or to other readers than mine.
            – gimusi
            2 days ago


















          • God dammit I was seconds away from posting.
            – Rushabh Mehta
            2 days ago










          • @RushabhMehta Sorry! :)
            – gimusi
            2 days ago










          • @RushabhMehta Anyway in my opinion if your answer was near to be completed you could post it. I think it is not agains any rule and maybe your way to present things could be more clear to the asker or to other readers than mine.
            – gimusi
            2 days ago
















          God dammit I was seconds away from posting.
          – Rushabh Mehta
          2 days ago




          God dammit I was seconds away from posting.
          – Rushabh Mehta
          2 days ago












          @RushabhMehta Sorry! :)
          – gimusi
          2 days ago




          @RushabhMehta Sorry! :)
          – gimusi
          2 days ago












          @RushabhMehta Anyway in my opinion if your answer was near to be completed you could post it. I think it is not agains any rule and maybe your way to present things could be more clear to the asker or to other readers than mine.
          – gimusi
          2 days ago




          @RushabhMehta Anyway in my opinion if your answer was near to be completed you could post it. I think it is not agains any rule and maybe your way to present things could be more clear to the asker or to other readers than mine.
          – gimusi
          2 days ago










          up vote
          -1
          down vote













          By the ratio test, the sequence converges if
          $$frac{(n+1)^b}{(1+a)n^b}<1.$$



          This is achieved when



          $$n>frac1{sqrt[b]{1+a}-1},$$ a finite number.




          E.g., with $dfrac{n^3}{2^n}$, for $n>3$,



          $$frac{(n+1)^3}{2n^3}<frac{125}{2cdot64}<1.$$






          share|cite|improve this answer























          • Sorry, I don't see how to use it to prove that the sequence converges to $0$. Could you show a little more?
            – Raul_MFer
            2 days ago












          • @Raul_MFer: did you downvote ?
            – Yves Daoust
            2 days ago















          up vote
          -1
          down vote













          By the ratio test, the sequence converges if
          $$frac{(n+1)^b}{(1+a)n^b}<1.$$



          This is achieved when



          $$n>frac1{sqrt[b]{1+a}-1},$$ a finite number.




          E.g., with $dfrac{n^3}{2^n}$, for $n>3$,



          $$frac{(n+1)^3}{2n^3}<frac{125}{2cdot64}<1.$$






          share|cite|improve this answer























          • Sorry, I don't see how to use it to prove that the sequence converges to $0$. Could you show a little more?
            – Raul_MFer
            2 days ago












          • @Raul_MFer: did you downvote ?
            – Yves Daoust
            2 days ago













          up vote
          -1
          down vote










          up vote
          -1
          down vote









          By the ratio test, the sequence converges if
          $$frac{(n+1)^b}{(1+a)n^b}<1.$$



          This is achieved when



          $$n>frac1{sqrt[b]{1+a}-1},$$ a finite number.




          E.g., with $dfrac{n^3}{2^n}$, for $n>3$,



          $$frac{(n+1)^3}{2n^3}<frac{125}{2cdot64}<1.$$






          share|cite|improve this answer














          By the ratio test, the sequence converges if
          $$frac{(n+1)^b}{(1+a)n^b}<1.$$



          This is achieved when



          $$n>frac1{sqrt[b]{1+a}-1},$$ a finite number.




          E.g., with $dfrac{n^3}{2^n}$, for $n>3$,



          $$frac{(n+1)^3}{2n^3}<frac{125}{2cdot64}<1.$$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited 2 days ago

























          answered 2 days ago









          Yves Daoust

          121k668216




          121k668216












          • Sorry, I don't see how to use it to prove that the sequence converges to $0$. Could you show a little more?
            – Raul_MFer
            2 days ago












          • @Raul_MFer: did you downvote ?
            – Yves Daoust
            2 days ago


















          • Sorry, I don't see how to use it to prove that the sequence converges to $0$. Could you show a little more?
            – Raul_MFer
            2 days ago












          • @Raul_MFer: did you downvote ?
            – Yves Daoust
            2 days ago
















          Sorry, I don't see how to use it to prove that the sequence converges to $0$. Could you show a little more?
          – Raul_MFer
          2 days ago






          Sorry, I don't see how to use it to prove that the sequence converges to $0$. Could you show a little more?
          – Raul_MFer
          2 days ago














          @Raul_MFer: did you downvote ?
          – Yves Daoust
          2 days ago




          @Raul_MFer: did you downvote ?
          – Yves Daoust
          2 days ago










          Raul_MFer is a new contributor. Be nice, and check out our Code of Conduct.










           

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