Limit of sequence: $lim_{ntoinfty}({2n+1over 3n})^n$











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0
down vote

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$lim_{ntoinfty}({2n+1over 3n})^n$



For sufficiently large values of $n$:



$0le ({2n+1over 3n})^nle({2over 3})^n$,



From Squeeze theorem:
$lim_{ntoinfty}({2n+1over 3n})^n=0$



Is it ok?



I also tried to do this with formula: $lim_{ntoinfty}(1+{aover n})^n=e^a$, but with no effect.










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  • You never have $$frac{2n+1}{3n} leq frac{2}{3}$$ no matter how big $n$ is. Try $$frac{2n+1}{3n} leq frac{2.5}{3}$$
    – Clement C.
    Nov 15 at 18:03












  • When I changed $ frac{2}{3}$ to $ frac{2.9999}{3}$ is it ok?
    – matematiccc
    Nov 15 at 18:11










  • Yes, now it is. That is not great to change the question though, as it makes a couple answers addressing this very point obsolete.
    – Clement C.
    Nov 15 at 18:12












  • Sorry. I fixed it.
    – matematiccc
    Nov 15 at 18:20















up vote
0
down vote

favorite












$lim_{ntoinfty}({2n+1over 3n})^n$



For sufficiently large values of $n$:



$0le ({2n+1over 3n})^nle({2over 3})^n$,



From Squeeze theorem:
$lim_{ntoinfty}({2n+1over 3n})^n=0$



Is it ok?



I also tried to do this with formula: $lim_{ntoinfty}(1+{aover n})^n=e^a$, but with no effect.










share|cite|improve this question
























  • You never have $$frac{2n+1}{3n} leq frac{2}{3}$$ no matter how big $n$ is. Try $$frac{2n+1}{3n} leq frac{2.5}{3}$$
    – Clement C.
    Nov 15 at 18:03












  • When I changed $ frac{2}{3}$ to $ frac{2.9999}{3}$ is it ok?
    – matematiccc
    Nov 15 at 18:11










  • Yes, now it is. That is not great to change the question though, as it makes a couple answers addressing this very point obsolete.
    – Clement C.
    Nov 15 at 18:12












  • Sorry. I fixed it.
    – matematiccc
    Nov 15 at 18:20













up vote
0
down vote

favorite









up vote
0
down vote

favorite











$lim_{ntoinfty}({2n+1over 3n})^n$



For sufficiently large values of $n$:



$0le ({2n+1over 3n})^nle({2over 3})^n$,



From Squeeze theorem:
$lim_{ntoinfty}({2n+1over 3n})^n=0$



Is it ok?



I also tried to do this with formula: $lim_{ntoinfty}(1+{aover n})^n=e^a$, but with no effect.










share|cite|improve this question















$lim_{ntoinfty}({2n+1over 3n})^n$



For sufficiently large values of $n$:



$0le ({2n+1over 3n})^nle({2over 3})^n$,



From Squeeze theorem:
$lim_{ntoinfty}({2n+1over 3n})^n=0$



Is it ok?



I also tried to do this with formula: $lim_{ntoinfty}(1+{aover n})^n=e^a$, but with no effect.







real-analysis sequences-and-series limits limits-without-lhopital






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edited Nov 15 at 18:19

























asked Nov 15 at 18:00









matematiccc

1145




1145












  • You never have $$frac{2n+1}{3n} leq frac{2}{3}$$ no matter how big $n$ is. Try $$frac{2n+1}{3n} leq frac{2.5}{3}$$
    – Clement C.
    Nov 15 at 18:03












  • When I changed $ frac{2}{3}$ to $ frac{2.9999}{3}$ is it ok?
    – matematiccc
    Nov 15 at 18:11










  • Yes, now it is. That is not great to change the question though, as it makes a couple answers addressing this very point obsolete.
    – Clement C.
    Nov 15 at 18:12












  • Sorry. I fixed it.
    – matematiccc
    Nov 15 at 18:20


















  • You never have $$frac{2n+1}{3n} leq frac{2}{3}$$ no matter how big $n$ is. Try $$frac{2n+1}{3n} leq frac{2.5}{3}$$
    – Clement C.
    Nov 15 at 18:03












  • When I changed $ frac{2}{3}$ to $ frac{2.9999}{3}$ is it ok?
    – matematiccc
    Nov 15 at 18:11










  • Yes, now it is. That is not great to change the question though, as it makes a couple answers addressing this very point obsolete.
    – Clement C.
    Nov 15 at 18:12












  • Sorry. I fixed it.
    – matematiccc
    Nov 15 at 18:20
















You never have $$frac{2n+1}{3n} leq frac{2}{3}$$ no matter how big $n$ is. Try $$frac{2n+1}{3n} leq frac{2.5}{3}$$
– Clement C.
Nov 15 at 18:03






You never have $$frac{2n+1}{3n} leq frac{2}{3}$$ no matter how big $n$ is. Try $$frac{2n+1}{3n} leq frac{2.5}{3}$$
– Clement C.
Nov 15 at 18:03














When I changed $ frac{2}{3}$ to $ frac{2.9999}{3}$ is it ok?
– matematiccc
Nov 15 at 18:11




When I changed $ frac{2}{3}$ to $ frac{2.9999}{3}$ is it ok?
– matematiccc
Nov 15 at 18:11












Yes, now it is. That is not great to change the question though, as it makes a couple answers addressing this very point obsolete.
– Clement C.
Nov 15 at 18:12






Yes, now it is. That is not great to change the question though, as it makes a couple answers addressing this very point obsolete.
– Clement C.
Nov 15 at 18:12














Sorry. I fixed it.
– matematiccc
Nov 15 at 18:20




Sorry. I fixed it.
– matematiccc
Nov 15 at 18:20










4 Answers
4






active

oldest

votes

















up vote
3
down vote



accepted










No, it is not correct, because the inequality $left(frac{2n+1}{3n}right)^nleqslantleft(frac23right)^n$ never holds. But you can use the fact that$$lim_{ntoinfty}left(frac{2n+1}{3n}right)^n=lim_{ntoinfty}left(frac23right)^ntimeslim_{ntoinfty}left(1+frac1{2n}right)^n=0.$$






share|cite|improve this answer

















  • 2




    I might be wrong but I don't think that it is wise to split limits this way. Doesn't always work (works this time though).
    – Makina
    Nov 15 at 18:12












  • If it didn't work this time, I would not have written it.
    – José Carlos Santos
    Nov 15 at 18:13










  • True, but for educational purposes I'd say one shouldn't write things this way.
    – Makina
    Nov 15 at 18:14


















up vote
0
down vote













Hint: in the original expression take out $frac{2}{3}$ from your expression inside the brackets and then split.






share|cite|improve this answer




























    up vote
    0
    down vote













    Note that



    $${2n+1over 3n}le{2over 3}$$



    is not true, use that eventually



    $${2n+1over 3n}le frac45<1 iff 10n+5<12n iff 2n>5$$



    As a simpler alternative by root test



    $$sqrt[n]{a_n}=sqrt[n]{left({2n+1over 3n}right)^n}={2n+1over 3n}tofrac23<1implies a_n to 0$$






    share|cite|improve this answer























    • Which is not enough... since $1-1/n<1$ "eventually" (actually, always) as well. You need $< c$ for some constant $0leq c<1$.
      – Clement C.
      Nov 15 at 18:04












    • After the edit, fixing the solution, downvote removed. But you should really make sure your solution is correct before posting it (it's not the first time...).
      – Clement C.
      Nov 15 at 18:07












    • @ClementC. Yes of course! the inequalities do not hold in the limit!
      – gimusi
      Nov 15 at 18:10


















    up vote
    0
    down vote













    You can use the root test to get a result quite easily. Since $$dfrac{2n+1}{3n}todfrac{2}{3}<1$$
    by the ratio test
    $$bigg(dfrac{2n+1}{3n}bigg)^nto0.$$






    share|cite|improve this answer





















    • I did not have it yet.
      – matematiccc
      Nov 15 at 18:18











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    4 Answers
    4






    active

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    4 Answers
    4






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    up vote
    3
    down vote



    accepted










    No, it is not correct, because the inequality $left(frac{2n+1}{3n}right)^nleqslantleft(frac23right)^n$ never holds. But you can use the fact that$$lim_{ntoinfty}left(frac{2n+1}{3n}right)^n=lim_{ntoinfty}left(frac23right)^ntimeslim_{ntoinfty}left(1+frac1{2n}right)^n=0.$$






    share|cite|improve this answer

















    • 2




      I might be wrong but I don't think that it is wise to split limits this way. Doesn't always work (works this time though).
      – Makina
      Nov 15 at 18:12












    • If it didn't work this time, I would not have written it.
      – José Carlos Santos
      Nov 15 at 18:13










    • True, but for educational purposes I'd say one shouldn't write things this way.
      – Makina
      Nov 15 at 18:14















    up vote
    3
    down vote



    accepted










    No, it is not correct, because the inequality $left(frac{2n+1}{3n}right)^nleqslantleft(frac23right)^n$ never holds. But you can use the fact that$$lim_{ntoinfty}left(frac{2n+1}{3n}right)^n=lim_{ntoinfty}left(frac23right)^ntimeslim_{ntoinfty}left(1+frac1{2n}right)^n=0.$$






    share|cite|improve this answer

















    • 2




      I might be wrong but I don't think that it is wise to split limits this way. Doesn't always work (works this time though).
      – Makina
      Nov 15 at 18:12












    • If it didn't work this time, I would not have written it.
      – José Carlos Santos
      Nov 15 at 18:13










    • True, but for educational purposes I'd say one shouldn't write things this way.
      – Makina
      Nov 15 at 18:14













    up vote
    3
    down vote



    accepted







    up vote
    3
    down vote



    accepted






    No, it is not correct, because the inequality $left(frac{2n+1}{3n}right)^nleqslantleft(frac23right)^n$ never holds. But you can use the fact that$$lim_{ntoinfty}left(frac{2n+1}{3n}right)^n=lim_{ntoinfty}left(frac23right)^ntimeslim_{ntoinfty}left(1+frac1{2n}right)^n=0.$$






    share|cite|improve this answer












    No, it is not correct, because the inequality $left(frac{2n+1}{3n}right)^nleqslantleft(frac23right)^n$ never holds. But you can use the fact that$$lim_{ntoinfty}left(frac{2n+1}{3n}right)^n=lim_{ntoinfty}left(frac23right)^ntimeslim_{ntoinfty}left(1+frac1{2n}right)^n=0.$$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Nov 15 at 18:08









    José Carlos Santos

    142k20112208




    142k20112208








    • 2




      I might be wrong but I don't think that it is wise to split limits this way. Doesn't always work (works this time though).
      – Makina
      Nov 15 at 18:12












    • If it didn't work this time, I would not have written it.
      – José Carlos Santos
      Nov 15 at 18:13










    • True, but for educational purposes I'd say one shouldn't write things this way.
      – Makina
      Nov 15 at 18:14














    • 2




      I might be wrong but I don't think that it is wise to split limits this way. Doesn't always work (works this time though).
      – Makina
      Nov 15 at 18:12












    • If it didn't work this time, I would not have written it.
      – José Carlos Santos
      Nov 15 at 18:13










    • True, but for educational purposes I'd say one shouldn't write things this way.
      – Makina
      Nov 15 at 18:14








    2




    2




    I might be wrong but I don't think that it is wise to split limits this way. Doesn't always work (works this time though).
    – Makina
    Nov 15 at 18:12






    I might be wrong but I don't think that it is wise to split limits this way. Doesn't always work (works this time though).
    – Makina
    Nov 15 at 18:12














    If it didn't work this time, I would not have written it.
    – José Carlos Santos
    Nov 15 at 18:13




    If it didn't work this time, I would not have written it.
    – José Carlos Santos
    Nov 15 at 18:13












    True, but for educational purposes I'd say one shouldn't write things this way.
    – Makina
    Nov 15 at 18:14




    True, but for educational purposes I'd say one shouldn't write things this way.
    – Makina
    Nov 15 at 18:14










    up vote
    0
    down vote













    Hint: in the original expression take out $frac{2}{3}$ from your expression inside the brackets and then split.






    share|cite|improve this answer

























      up vote
      0
      down vote













      Hint: in the original expression take out $frac{2}{3}$ from your expression inside the brackets and then split.






      share|cite|improve this answer























        up vote
        0
        down vote










        up vote
        0
        down vote









        Hint: in the original expression take out $frac{2}{3}$ from your expression inside the brackets and then split.






        share|cite|improve this answer












        Hint: in the original expression take out $frac{2}{3}$ from your expression inside the brackets and then split.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 15 at 18:05









        Makina

        1,006113




        1,006113






















            up vote
            0
            down vote













            Note that



            $${2n+1over 3n}le{2over 3}$$



            is not true, use that eventually



            $${2n+1over 3n}le frac45<1 iff 10n+5<12n iff 2n>5$$



            As a simpler alternative by root test



            $$sqrt[n]{a_n}=sqrt[n]{left({2n+1over 3n}right)^n}={2n+1over 3n}tofrac23<1implies a_n to 0$$






            share|cite|improve this answer























            • Which is not enough... since $1-1/n<1$ "eventually" (actually, always) as well. You need $< c$ for some constant $0leq c<1$.
              – Clement C.
              Nov 15 at 18:04












            • After the edit, fixing the solution, downvote removed. But you should really make sure your solution is correct before posting it (it's not the first time...).
              – Clement C.
              Nov 15 at 18:07












            • @ClementC. Yes of course! the inequalities do not hold in the limit!
              – gimusi
              Nov 15 at 18:10















            up vote
            0
            down vote













            Note that



            $${2n+1over 3n}le{2over 3}$$



            is not true, use that eventually



            $${2n+1over 3n}le frac45<1 iff 10n+5<12n iff 2n>5$$



            As a simpler alternative by root test



            $$sqrt[n]{a_n}=sqrt[n]{left({2n+1over 3n}right)^n}={2n+1over 3n}tofrac23<1implies a_n to 0$$






            share|cite|improve this answer























            • Which is not enough... since $1-1/n<1$ "eventually" (actually, always) as well. You need $< c$ for some constant $0leq c<1$.
              – Clement C.
              Nov 15 at 18:04












            • After the edit, fixing the solution, downvote removed. But you should really make sure your solution is correct before posting it (it's not the first time...).
              – Clement C.
              Nov 15 at 18:07












            • @ClementC. Yes of course! the inequalities do not hold in the limit!
              – gimusi
              Nov 15 at 18:10













            up vote
            0
            down vote










            up vote
            0
            down vote









            Note that



            $${2n+1over 3n}le{2over 3}$$



            is not true, use that eventually



            $${2n+1over 3n}le frac45<1 iff 10n+5<12n iff 2n>5$$



            As a simpler alternative by root test



            $$sqrt[n]{a_n}=sqrt[n]{left({2n+1over 3n}right)^n}={2n+1over 3n}tofrac23<1implies a_n to 0$$






            share|cite|improve this answer














            Note that



            $${2n+1over 3n}le{2over 3}$$



            is not true, use that eventually



            $${2n+1over 3n}le frac45<1 iff 10n+5<12n iff 2n>5$$



            As a simpler alternative by root test



            $$sqrt[n]{a_n}=sqrt[n]{left({2n+1over 3n}right)^n}={2n+1over 3n}tofrac23<1implies a_n to 0$$







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Nov 15 at 18:12

























            answered Nov 15 at 18:03









            gimusi

            88.5k74394




            88.5k74394












            • Which is not enough... since $1-1/n<1$ "eventually" (actually, always) as well. You need $< c$ for some constant $0leq c<1$.
              – Clement C.
              Nov 15 at 18:04












            • After the edit, fixing the solution, downvote removed. But you should really make sure your solution is correct before posting it (it's not the first time...).
              – Clement C.
              Nov 15 at 18:07












            • @ClementC. Yes of course! the inequalities do not hold in the limit!
              – gimusi
              Nov 15 at 18:10


















            • Which is not enough... since $1-1/n<1$ "eventually" (actually, always) as well. You need $< c$ for some constant $0leq c<1$.
              – Clement C.
              Nov 15 at 18:04












            • After the edit, fixing the solution, downvote removed. But you should really make sure your solution is correct before posting it (it's not the first time...).
              – Clement C.
              Nov 15 at 18:07












            • @ClementC. Yes of course! the inequalities do not hold in the limit!
              – gimusi
              Nov 15 at 18:10
















            Which is not enough... since $1-1/n<1$ "eventually" (actually, always) as well. You need $< c$ for some constant $0leq c<1$.
            – Clement C.
            Nov 15 at 18:04






            Which is not enough... since $1-1/n<1$ "eventually" (actually, always) as well. You need $< c$ for some constant $0leq c<1$.
            – Clement C.
            Nov 15 at 18:04














            After the edit, fixing the solution, downvote removed. But you should really make sure your solution is correct before posting it (it's not the first time...).
            – Clement C.
            Nov 15 at 18:07






            After the edit, fixing the solution, downvote removed. But you should really make sure your solution is correct before posting it (it's not the first time...).
            – Clement C.
            Nov 15 at 18:07














            @ClementC. Yes of course! the inequalities do not hold in the limit!
            – gimusi
            Nov 15 at 18:10




            @ClementC. Yes of course! the inequalities do not hold in the limit!
            – gimusi
            Nov 15 at 18:10










            up vote
            0
            down vote













            You can use the root test to get a result quite easily. Since $$dfrac{2n+1}{3n}todfrac{2}{3}<1$$
            by the ratio test
            $$bigg(dfrac{2n+1}{3n}bigg)^nto0.$$






            share|cite|improve this answer





















            • I did not have it yet.
              – matematiccc
              Nov 15 at 18:18















            up vote
            0
            down vote













            You can use the root test to get a result quite easily. Since $$dfrac{2n+1}{3n}todfrac{2}{3}<1$$
            by the ratio test
            $$bigg(dfrac{2n+1}{3n}bigg)^nto0.$$






            share|cite|improve this answer





















            • I did not have it yet.
              – matematiccc
              Nov 15 at 18:18













            up vote
            0
            down vote










            up vote
            0
            down vote









            You can use the root test to get a result quite easily. Since $$dfrac{2n+1}{3n}todfrac{2}{3}<1$$
            by the ratio test
            $$bigg(dfrac{2n+1}{3n}bigg)^nto0.$$






            share|cite|improve this answer












            You can use the root test to get a result quite easily. Since $$dfrac{2n+1}{3n}todfrac{2}{3}<1$$
            by the ratio test
            $$bigg(dfrac{2n+1}{3n}bigg)^nto0.$$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Nov 15 at 18:16









            Melody

            45211




            45211












            • I did not have it yet.
              – matematiccc
              Nov 15 at 18:18


















            • I did not have it yet.
              – matematiccc
              Nov 15 at 18:18
















            I did not have it yet.
            – matematiccc
            Nov 15 at 18:18




            I did not have it yet.
            – matematiccc
            Nov 15 at 18:18


















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