Limit of a sequence where $U_n = frac{(log n )^p}{n}$












0














What is the limit of this sequence where , $U_n = frac{(log n )^p}{n}$ where $p ge 0$.



I have done this problem when $p$ is an integer.
Sorry but I am not too much familiar with writing questions in stack exchange.






sequences-and-series limits







share|cite|improve this question
























  • See if you can exploit propositions 2.1 and 2.2 from here
    – rtybase
    Nov 25 '18 at 10:07






  • 1




    If you know the results for $p in mathbb N$, note that for general $p$, there is a $kin mathbb N$ s.t. $k leqslant p<k+1$, and to get the limit, try the squeezing theorem.
    – xbh
    Nov 25 '18 at 13:19
















0














What is the limit of this sequence where , $U_n = frac{(log n )^p}{n}$ where $p ge 0$.



I have done this problem when $p$ is an integer.
Sorry but I am not too much familiar with writing questions in stack exchange.






sequences-and-series limits







share|cite|improve this question
























  • See if you can exploit propositions 2.1 and 2.2 from here
    – rtybase
    Nov 25 '18 at 10:07






  • 1




    If you know the results for $p in mathbb N$, note that for general $p$, there is a $kin mathbb N$ s.t. $k leqslant p<k+1$, and to get the limit, try the squeezing theorem.
    – xbh
    Nov 25 '18 at 13:19














0












0








0







What is the limit of this sequence where , $U_n = frac{(log n )^p}{n}$ where $p ge 0$.



I have done this problem when $p$ is an integer.
Sorry but I am not too much familiar with writing questions in stack exchange.






sequences-and-series limits







share|cite|improve this question















What is the limit of this sequence where , $U_n = frac{(log n )^p}{n}$ where $p ge 0$.



I have done this problem when $p$ is an integer.
Sorry but I am not too much familiar with writing questions in stack exchange.






sequences-and-series limits




sequences-and-series limits






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 25 '18 at 9:55







user616975

















asked Nov 25 '18 at 9:46









Supriyo Banerjee

595




595












  • See if you can exploit propositions 2.1 and 2.2 from here
    – rtybase
    Nov 25 '18 at 10:07






  • 1




    If you know the results for $p in mathbb N$, note that for general $p$, there is a $kin mathbb N$ s.t. $k leqslant p<k+1$, and to get the limit, try the squeezing theorem.
    – xbh
    Nov 25 '18 at 13:19


















  • See if you can exploit propositions 2.1 and 2.2 from here
    – rtybase
    Nov 25 '18 at 10:07






  • 1




    If you know the results for $p in mathbb N$, note that for general $p$, there is a $kin mathbb N$ s.t. $k leqslant p<k+1$, and to get the limit, try the squeezing theorem.
    – xbh
    Nov 25 '18 at 13:19
















See if you can exploit propositions 2.1 and 2.2 from here
– rtybase
Nov 25 '18 at 10:07




See if you can exploit propositions 2.1 and 2.2 from here
– rtybase
Nov 25 '18 at 10:07




1




1




If you know the results for $p in mathbb N$, note that for general $p$, there is a $kin mathbb N$ s.t. $k leqslant p<k+1$, and to get the limit, try the squeezing theorem.
– xbh
Nov 25 '18 at 13:19




If you know the results for $p in mathbb N$, note that for general $p$, there is a $kin mathbb N$ s.t. $k leqslant p<k+1$, and to get the limit, try the squeezing theorem.
– xbh
Nov 25 '18 at 13:19










1 Answer
1






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0














We have that



$$frac{(log n )^p}{n}=e^{plog(log n)-log n} to 0$$



indeed



$$plog(log n)-log n=log nleft(pfrac{log(log n)}{log n}-1right)to -infty$$



since



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



which can be easily proved by $frac{log x}x to 0$ as $x to infty$ by $x=log y$ and $y to infty$.






share|cite|improve this answer





















  • Why the downvote, something wrong?
    – gimusi
    Nov 25 '18 at 18:53










  • Can we write $lim$ $x tends to infinity$ $f(n)g(n)$ = $lim$ $x tends to infinity$ $f(n)$ × $lim$ $x tends to infinity$ $g(n)$. When one of the sequence diverges? I don't know...
    – Supriyo Banerjee
    Nov 26 '18 at 3:08






  • 1




    Note that here we are using something different that is $$lim_{nto infty} f(n)=lim_{nto infty} g(n)cdot h(n)$$ and using the fact that $g(n)to infty$ and $h(n)to -1$ and that’s leads to a not indeterminate form.
    – gimusi
    Nov 26 '18 at 6:28











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1 Answer
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1 Answer
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0














We have that



$$frac{(log n )^p}{n}=e^{plog(log n)-log n} to 0$$



indeed



$$plog(log n)-log n=log nleft(pfrac{log(log n)}{log n}-1right)to -infty$$



since



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



which can be easily proved by $frac{log x}x to 0$ as $x to infty$ by $x=log y$ and $y to infty$.






share|cite|improve this answer





















  • Why the downvote, something wrong?
    – gimusi
    Nov 25 '18 at 18:53










  • Can we write $lim$ $x tends to infinity$ $f(n)g(n)$ = $lim$ $x tends to infinity$ $f(n)$ × $lim$ $x tends to infinity$ $g(n)$. When one of the sequence diverges? I don't know...
    – Supriyo Banerjee
    Nov 26 '18 at 3:08






  • 1




    Note that here we are using something different that is $$lim_{nto infty} f(n)=lim_{nto infty} g(n)cdot h(n)$$ and using the fact that $g(n)to infty$ and $h(n)to -1$ and that’s leads to a not indeterminate form.
    – gimusi
    Nov 26 '18 at 6:28
















0














We have that



$$frac{(log n )^p}{n}=e^{plog(log n)-log n} to 0$$



indeed



$$plog(log n)-log n=log nleft(pfrac{log(log n)}{log n}-1right)to -infty$$



since



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



which can be easily proved by $frac{log x}x to 0$ as $x to infty$ by $x=log y$ and $y to infty$.






share|cite|improve this answer





















  • Why the downvote, something wrong?
    – gimusi
    Nov 25 '18 at 18:53










  • Can we write $lim$ $x tends to infinity$ $f(n)g(n)$ = $lim$ $x tends to infinity$ $f(n)$ × $lim$ $x tends to infinity$ $g(n)$. When one of the sequence diverges? I don't know...
    – Supriyo Banerjee
    Nov 26 '18 at 3:08






  • 1




    Note that here we are using something different that is $$lim_{nto infty} f(n)=lim_{nto infty} g(n)cdot h(n)$$ and using the fact that $g(n)to infty$ and $h(n)to -1$ and that’s leads to a not indeterminate form.
    – gimusi
    Nov 26 '18 at 6:28














0












0








0






We have that



$$frac{(log n )^p}{n}=e^{plog(log n)-log n} to 0$$



indeed



$$plog(log n)-log n=log nleft(pfrac{log(log n)}{log n}-1right)to -infty$$



since



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



which can be easily proved by $frac{log x}x to 0$ as $x to infty$ by $x=log y$ and $y to infty$.






share|cite|improve this answer












We have that



$$frac{(log n )^p}{n}=e^{plog(log n)-log n} to 0$$



indeed



$$plog(log n)-log n=log nleft(pfrac{log(log n)}{log n}-1right)to -infty$$



since



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



which can be easily proved by $frac{log x}x to 0$ as $x to infty$ by $x=log y$ and $y to infty$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 25 '18 at 18:32









gimusi

1




1












  • Why the downvote, something wrong?
    – gimusi
    Nov 25 '18 at 18:53










  • Can we write $lim$ $x tends to infinity$ $f(n)g(n)$ = $lim$ $x tends to infinity$ $f(n)$ × $lim$ $x tends to infinity$ $g(n)$. When one of the sequence diverges? I don't know...
    – Supriyo Banerjee
    Nov 26 '18 at 3:08






  • 1




    Note that here we are using something different that is $$lim_{nto infty} f(n)=lim_{nto infty} g(n)cdot h(n)$$ and using the fact that $g(n)to infty$ and $h(n)to -1$ and that’s leads to a not indeterminate form.
    – gimusi
    Nov 26 '18 at 6:28


















  • Why the downvote, something wrong?
    – gimusi
    Nov 25 '18 at 18:53










  • Can we write $lim$ $x tends to infinity$ $f(n)g(n)$ = $lim$ $x tends to infinity$ $f(n)$ × $lim$ $x tends to infinity$ $g(n)$. When one of the sequence diverges? I don't know...
    – Supriyo Banerjee
    Nov 26 '18 at 3:08






  • 1




    Note that here we are using something different that is $$lim_{nto infty} f(n)=lim_{nto infty} g(n)cdot h(n)$$ and using the fact that $g(n)to infty$ and $h(n)to -1$ and that’s leads to a not indeterminate form.
    – gimusi
    Nov 26 '18 at 6:28
















Why the downvote, something wrong?
– gimusi
Nov 25 '18 at 18:53




Why the downvote, something wrong?
– gimusi
Nov 25 '18 at 18:53












Can we write $lim$ $x tends to infinity$ $f(n)g(n)$ = $lim$ $x tends to infinity$ $f(n)$ × $lim$ $x tends to infinity$ $g(n)$. When one of the sequence diverges? I don't know...
– Supriyo Banerjee
Nov 26 '18 at 3:08




Can we write $lim$ $x tends to infinity$ $f(n)g(n)$ = $lim$ $x tends to infinity$ $f(n)$ × $lim$ $x tends to infinity$ $g(n)$. When one of the sequence diverges? I don't know...
– Supriyo Banerjee
Nov 26 '18 at 3:08




1




1




Note that here we are using something different that is $$lim_{nto infty} f(n)=lim_{nto infty} g(n)cdot h(n)$$ and using the fact that $g(n)to infty$ and $h(n)to -1$ and that’s leads to a not indeterminate form.
– gimusi
Nov 26 '18 at 6:28




Note that here we are using something different that is $$lim_{nto infty} f(n)=lim_{nto infty} g(n)cdot h(n)$$ and using the fact that $g(n)to infty$ and $h(n)to -1$ and that’s leads to a not indeterminate form.
– gimusi
Nov 26 '18 at 6:28


















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