Calculate the limit of $lim_krightarrowinfty(k!)^frac1k$

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$$lim_krightarrowinfty(k!)^frac1k$$
I want to determine the limit as formal as possible, that is, to prove, for example
$$lim_krightarrowinftyfracln(k!)k$$
I had the intention of using the L'Hopital rule and I do not know that it does exist, despite only representing a number. But it is not very justified.



Can you use sanduche theorem? Can I limit?




calculus real-analysis complex-analysis




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  • Possible duplicate of $limlimits_n to+inftysqrt[n]n!$ is infinite
    – Hans Lundmark
    8 hours ago














up vote
1
down vote

favorite
1












$$lim_krightarrowinfty(k!)^frac1k$$
I want to determine the limit as formal as possible, that is, to prove, for example
$$lim_krightarrowinftyfracln(k!)k$$
I had the intention of using the L'Hopital rule and I do not know that it does exist, despite only representing a number. But it is not very justified.



Can you use sanduche theorem? Can I limit?




calculus real-analysis complex-analysis




share|cite|improve this question





















  • Possible duplicate of $limlimits_n to+inftysqrt[n]n!$ is infinite
    – Hans Lundmark
    8 hours ago












up vote
1
down vote

favorite
1









up vote
1
down vote

favorite
1






1





$$lim_krightarrowinfty(k!)^frac1k$$
I want to determine the limit as formal as possible, that is, to prove, for example
$$lim_krightarrowinftyfracln(k!)k$$
I had the intention of using the L'Hopital rule and I do not know that it does exist, despite only representing a number. But it is not very justified.



Can you use sanduche theorem? Can I limit?




calculus real-analysis complex-analysis




share|cite|improve this question













$$lim_krightarrowinfty(k!)^frac1k$$
I want to determine the limit as formal as possible, that is, to prove, for example
$$lim_krightarrowinftyfracln(k!)k$$
I had the intention of using the L'Hopital rule and I do not know that it does exist, despite only representing a number. But it is not very justified.



Can you use sanduche theorem? Can I limit?





calculus real-analysis complex-analysis





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited 4 hours ago









Asaf Karagila

291k31400731




291k31400731









asked 9 hours ago









Santiago Seeker

577




577











  • Possible duplicate of $limlimits_n to+inftysqrt[n]n!$ is infinite
    – Hans Lundmark
    8 hours ago
















  • Possible duplicate of $limlimits_n to+inftysqrt[n]n!$ is infinite
    – Hans Lundmark
    8 hours ago















Possible duplicate of $limlimits_n to+inftysqrt[n]n!$ is infinite
– Hans Lundmark
8 hours ago




Possible duplicate of $limlimits_n to+inftysqrt[n]n!$ is infinite
– Hans Lundmark
8 hours ago










3 Answers
3






active

oldest

votes

















up vote
2
down vote



accepted










As an alternative by Stolz-Cesaro



$$lim_krightarrowinftyfracln(k!)k=lim_krightarrowinftyfracln((k+1)!)-ln(k!)k+1-k=lim_krightarrowinfty log (k+1) =infty$$






share|cite|improve this answer




























    up vote
    2
    down vote













    Hint: for a sequence of positive numbers $a_n$, if $lim_nrightarrowinfty(fraca_n+1a_n)$ exists and equals $L$ ($L$ can be a real number of infinity, it doesn't matter) then $lim_nrightarrowinfty(sqrt[n]a_n)=L$.






    share|cite|improve this answer





















    • Thank you very much, I had not considered it.
      – Santiago Seeker
      8 hours ago

















    up vote
    2
    down vote













    HINT:



    Note that



    $$beginalign
    fraclog(k!)k&=frac1ksum_n=1^k log(n)\\
    &=log(k)+frac1ksum_n=1^k log(n/k)tag1
    endalign$$



    The sum on the right-hand side of $(1)$ is the Riemann sum for the improper integral $lim_varepsilonto0int_varepsilon^1 log(x),dx$.



    Can you finish this?






    share|cite|improve this answer





















    • Oh, wow! that great ... is the criterion of the integral and with that I show that it is divergent :) thank you very much !!
      – Santiago Seeker
      8 hours ago










    • You're welcome. My pleasure.
      – Mark Viola
      8 hours ago










    • Avoiding the problem of why those Riemann sums converge to the improper integral, we could estimate $$sum_j=1^kln j ge (k/2)ln (k/2),$$ from which the result follows.
      – zhw.
      5 hours ago










    • No concavity needed, just need $ln j$ increasing. And I'm doing fine thank you, and I hope the same is true your way.
      – zhw.
      5 hours ago










    • I see. You're using $n!ge (n/2)^n/2$ and $log(x)$ is increasing. I missed that upon first glance.
      – Mark Viola
      4 hours ago










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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    2
    down vote



    accepted










    As an alternative by Stolz-Cesaro



    $$lim_krightarrowinftyfracln(k!)k=lim_krightarrowinftyfracln((k+1)!)-ln(k!)k+1-k=lim_krightarrowinfty log (k+1) =infty$$






    share|cite|improve this answer

























      up vote
      2
      down vote



      accepted










      As an alternative by Stolz-Cesaro



      $$lim_krightarrowinftyfracln(k!)k=lim_krightarrowinftyfracln((k+1)!)-ln(k!)k+1-k=lim_krightarrowinfty log (k+1) =infty$$






      share|cite|improve this answer























        up vote
        2
        down vote



        accepted







        up vote
        2
        down vote



        accepted






        As an alternative by Stolz-Cesaro



        $$lim_krightarrowinftyfracln(k!)k=lim_krightarrowinftyfracln((k+1)!)-ln(k!)k+1-k=lim_krightarrowinfty log (k+1) =infty$$






        share|cite|improve this answer













        As an alternative by Stolz-Cesaro



        $$lim_krightarrowinftyfracln(k!)k=lim_krightarrowinftyfracln((k+1)!)-ln(k!)k+1-k=lim_krightarrowinfty log (k+1) =infty$$







        share|cite|improve this answer













        share|cite|improve this answer



        share|cite|improve this answer











        answered 8 hours ago









        gimusi

        63.4k73379




        63.4k73379




















            up vote
            2
            down vote













            Hint: for a sequence of positive numbers $a_n$, if $lim_nrightarrowinfty(fraca_n+1a_n)$ exists and equals $L$ ($L$ can be a real number of infinity, it doesn't matter) then $lim_nrightarrowinfty(sqrt[n]a_n)=L$.






            share|cite|improve this answer





















            • Thank you very much, I had not considered it.
              – Santiago Seeker
              8 hours ago














            up vote
            2
            down vote













            Hint: for a sequence of positive numbers $a_n$, if $lim_nrightarrowinfty(fraca_n+1a_n)$ exists and equals $L$ ($L$ can be a real number of infinity, it doesn't matter) then $lim_nrightarrowinfty(sqrt[n]a_n)=L$.






            share|cite|improve this answer





















            • Thank you very much, I had not considered it.
              – Santiago Seeker
              8 hours ago












            up vote
            2
            down vote










            up vote
            2
            down vote









            Hint: for a sequence of positive numbers $a_n$, if $lim_nrightarrowinfty(fraca_n+1a_n)$ exists and equals $L$ ($L$ can be a real number of infinity, it doesn't matter) then $lim_nrightarrowinfty(sqrt[n]a_n)=L$.






            share|cite|improve this answer













            Hint: for a sequence of positive numbers $a_n$, if $lim_nrightarrowinfty(fraca_n+1a_n)$ exists and equals $L$ ($L$ can be a real number of infinity, it doesn't matter) then $lim_nrightarrowinfty(sqrt[n]a_n)=L$.







            share|cite|improve this answer













            share|cite|improve this answer



            share|cite|improve this answer











            answered 9 hours ago









            Mark

            5849




            5849











            • Thank you very much, I had not considered it.
              – Santiago Seeker
              8 hours ago
















            • Thank you very much, I had not considered it.
              – Santiago Seeker
              8 hours ago















            Thank you very much, I had not considered it.
            – Santiago Seeker
            8 hours ago




            Thank you very much, I had not considered it.
            – Santiago Seeker
            8 hours ago










            up vote
            2
            down vote













            HINT:



            Note that



            $$beginalign
            fraclog(k!)k&=frac1ksum_n=1^k log(n)\\
            &=log(k)+frac1ksum_n=1^k log(n/k)tag1
            endalign$$



            The sum on the right-hand side of $(1)$ is the Riemann sum for the improper integral $lim_varepsilonto0int_varepsilon^1 log(x),dx$.



            Can you finish this?






            share|cite|improve this answer





















            • Oh, wow! that great ... is the criterion of the integral and with that I show that it is divergent :) thank you very much !!
              – Santiago Seeker
              8 hours ago










            • You're welcome. My pleasure.
              – Mark Viola
              8 hours ago










            • Avoiding the problem of why those Riemann sums converge to the improper integral, we could estimate $$sum_j=1^kln j ge (k/2)ln (k/2),$$ from which the result follows.
              – zhw.
              5 hours ago










            • No concavity needed, just need $ln j$ increasing. And I'm doing fine thank you, and I hope the same is true your way.
              – zhw.
              5 hours ago










            • I see. You're using $n!ge (n/2)^n/2$ and $log(x)$ is increasing. I missed that upon first glance.
              – Mark Viola
              4 hours ago














            up vote
            2
            down vote













            HINT:



            Note that



            $$beginalign
            fraclog(k!)k&=frac1ksum_n=1^k log(n)\\
            &=log(k)+frac1ksum_n=1^k log(n/k)tag1
            endalign$$



            The sum on the right-hand side of $(1)$ is the Riemann sum for the improper integral $lim_varepsilonto0int_varepsilon^1 log(x),dx$.



            Can you finish this?






            share|cite|improve this answer





















            • Oh, wow! that great ... is the criterion of the integral and with that I show that it is divergent :) thank you very much !!
              – Santiago Seeker
              8 hours ago










            • You're welcome. My pleasure.
              – Mark Viola
              8 hours ago










            • Avoiding the problem of why those Riemann sums converge to the improper integral, we could estimate $$sum_j=1^kln j ge (k/2)ln (k/2),$$ from which the result follows.
              – zhw.
              5 hours ago










            • No concavity needed, just need $ln j$ increasing. And I'm doing fine thank you, and I hope the same is true your way.
              – zhw.
              5 hours ago










            • I see. You're using $n!ge (n/2)^n/2$ and $log(x)$ is increasing. I missed that upon first glance.
              – Mark Viola
              4 hours ago












            up vote
            2
            down vote










            up vote
            2
            down vote









            HINT:



            Note that



            $$beginalign
            fraclog(k!)k&=frac1ksum_n=1^k log(n)\\
            &=log(k)+frac1ksum_n=1^k log(n/k)tag1
            endalign$$



            The sum on the right-hand side of $(1)$ is the Riemann sum for the improper integral $lim_varepsilonto0int_varepsilon^1 log(x),dx$.



            Can you finish this?






            share|cite|improve this answer













            HINT:



            Note that



            $$beginalign
            fraclog(k!)k&=frac1ksum_n=1^k log(n)\\
            &=log(k)+frac1ksum_n=1^k log(n/k)tag1
            endalign$$



            The sum on the right-hand side of $(1)$ is the Riemann sum for the improper integral $lim_varepsilonto0int_varepsilon^1 log(x),dx$.



            Can you finish this?







            share|cite|improve this answer













            share|cite|improve this answer



            share|cite|improve this answer











            answered 9 hours ago









            Mark Viola

            126k1170165




            126k1170165











            • Oh, wow! that great ... is the criterion of the integral and with that I show that it is divergent :) thank you very much !!
              – Santiago Seeker
              8 hours ago










            • You're welcome. My pleasure.
              – Mark Viola
              8 hours ago










            • Avoiding the problem of why those Riemann sums converge to the improper integral, we could estimate $$sum_j=1^kln j ge (k/2)ln (k/2),$$ from which the result follows.
              – zhw.
              5 hours ago










            • No concavity needed, just need $ln j$ increasing. And I'm doing fine thank you, and I hope the same is true your way.
              – zhw.
              5 hours ago










            • I see. You're using $n!ge (n/2)^n/2$ and $log(x)$ is increasing. I missed that upon first glance.
              – Mark Viola
              4 hours ago
















            • Oh, wow! that great ... is the criterion of the integral and with that I show that it is divergent :) thank you very much !!
              – Santiago Seeker
              8 hours ago










            • You're welcome. My pleasure.
              – Mark Viola
              8 hours ago










            • Avoiding the problem of why those Riemann sums converge to the improper integral, we could estimate $$sum_j=1^kln j ge (k/2)ln (k/2),$$ from which the result follows.
              – zhw.
              5 hours ago










            • No concavity needed, just need $ln j$ increasing. And I'm doing fine thank you, and I hope the same is true your way.
              – zhw.
              5 hours ago










            • I see. You're using $n!ge (n/2)^n/2$ and $log(x)$ is increasing. I missed that upon first glance.
              – Mark Viola
              4 hours ago















            Oh, wow! that great ... is the criterion of the integral and with that I show that it is divergent :) thank you very much !!
            – Santiago Seeker
            8 hours ago




            Oh, wow! that great ... is the criterion of the integral and with that I show that it is divergent :) thank you very much !!
            – Santiago Seeker
            8 hours ago












            You're welcome. My pleasure.
            – Mark Viola
            8 hours ago




            You're welcome. My pleasure.
            – Mark Viola
            8 hours ago












            Avoiding the problem of why those Riemann sums converge to the improper integral, we could estimate $$sum_j=1^kln j ge (k/2)ln (k/2),$$ from which the result follows.
            – zhw.
            5 hours ago




            Avoiding the problem of why those Riemann sums converge to the improper integral, we could estimate $$sum_j=1^kln j ge (k/2)ln (k/2),$$ from which the result follows.
            – zhw.
            5 hours ago












            No concavity needed, just need $ln j$ increasing. And I'm doing fine thank you, and I hope the same is true your way.
            – zhw.
            5 hours ago




            No concavity needed, just need $ln j$ increasing. And I'm doing fine thank you, and I hope the same is true your way.
            – zhw.
            5 hours ago












            I see. You're using $n!ge (n/2)^n/2$ and $log(x)$ is increasing. I missed that upon first glance.
            – Mark Viola
            4 hours ago




            I see. You're using $n!ge (n/2)^n/2$ and $log(x)$ is increasing. I missed that upon first glance.
            – Mark Viola
            4 hours ago












             

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