$lim_{n to infty} dfrac{e^{c sqrt{ln n . ln ln n}}}{n^{epsilon}}=0$?
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The general rule is discussed here but that doesn't solve my problem. I want to prove that $$lim_{n to infty} dfrac{e^{c sqrt{ln n . ln ln n}}}{n^{epsilon}}=0$$ where $c>0$ a fixed constant and $epsilon>0$ is any small positive number. I used different methods like l'hopital, etc and I found by expanding $e^{f(n)}$ it is the easiest, but need to know :
$lim_{n to infty} dfrac{(ln n . ln ln n)^k}{n^{epsilon}}=0$ for any $k in mathbb{N}$. Does it imply $lim_{n to infty} dfrac{e^{c sqrt{ln n . ln ln n}}}{n^{epsilon}}=0$?
If not, any hint for solving $lim_{n to infty} dfrac{e^{c sqrt{ln n . ln ln n}}}{n^{epsilon}}=0$ would be appreciated.
limits analytic-number-theory
asked Dec 21 '18 at 10:02
72D72D
512116
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add a comment |
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The general rule is discussed here but that doesn't solve my problem. I want to prove that $$lim_{n to infty} dfrac{e^{c sqrt{ln n . ln ln n}}}{n^{epsilon}}=0$$ where $c>0$ a fixed constant and $epsilon>0$ is any small positive number. I used different methods like l'hopital, etc and I found by expanding $e^{f(n)}$ it is the easiest, but need to know :
$lim_{n to infty} dfrac{(ln n . ln ln n)^k}{n^{epsilon}}=0$ for any $k in mathbb{N}$. Does it imply $lim_{n to infty} dfrac{e^{c sqrt{ln n . ln ln n}}}{n^{epsilon}}=0$?
If not, any hint for solving $lim_{n to infty} dfrac{e^{c sqrt{ln n . ln ln n}}}{n^{epsilon}}=0$ would be appreciated.
limits analytic-number-theory
asked Dec 21 '18 at 10:02
72D72D
512116
$endgroup$
add a comment |
$begingroup$
The general rule is discussed here but that doesn't solve my problem. I want to prove that $$lim_{n to infty} dfrac{e^{c sqrt{ln n . ln ln n}}}{n^{epsilon}}=0$$ where $c>0$ a fixed constant and $epsilon>0$ is any small positive number. I used different methods like l'hopital, etc and I found by expanding $e^{f(n)}$ it is the easiest, but need to know :
$lim_{n to infty} dfrac{(ln n . ln ln n)^k}{n^{epsilon}}=0$ for any $k in mathbb{N}$. Does it imply $lim_{n to infty} dfrac{e^{c sqrt{ln n . ln ln n}}}{n^{epsilon}}=0$?
If not, any hint for solving $lim_{n to infty} dfrac{e^{c sqrt{ln n . ln ln n}}}{n^{epsilon}}=0$ would be appreciated.
limits analytic-number-theory
asked Dec 21 '18 at 10:02
72D72D
512116
$endgroup$
The general rule is discussed here but that doesn't solve my problem. I want to prove that $$lim_{n to infty} dfrac{e^{c sqrt{ln n . ln ln n}}}{n^{epsilon}}=0$$ where $c>0$ a fixed constant and $epsilon>0$ is any small positive number. I used different methods like l'hopital, etc and I found by expanding $e^{f(n)}$ it is the easiest, but need to know :
$lim_{n to infty} dfrac{(ln n . ln ln n)^k}{n^{epsilon}}=0$ for any $k in mathbb{N}$. Does it imply $lim_{n to infty} dfrac{e^{c sqrt{ln n . ln ln n}}}{n^{epsilon}}=0$?
If not, any hint for solving $lim_{n to infty} dfrac{e^{c sqrt{ln n . ln ln n}}}{n^{epsilon}}=0$ would be appreciated.
limits analytic-number-theory
limits analytic-number-theory
asked Dec 21 '18 at 10:02
72D72D
512116
asked Dec 21 '18 at 10:02
72D72D
512116
asked Dec 21 '18 at 10:02
72D72D
512116
asked Dec 21 '18 at 10:02
72D72D
512116
asked Dec 21 '18 at 10:02
72D72D
512116
512116
add a comment |
add a comment |
2 Answers
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$$displaystyle L=lim_{ntoinfty}frac{e^{csqrt{ln ncdotlnln n}}}{n^{epsilon}}$$
$displaystyleln L=lim_{ntoinfty}csqrt{ln ncdotlnln n}-epsilonln n$
$displaystyle=lim_{xtoinfty}csqrt{xcdotln x}-epsilon x$
$displaystyle=lim_{xtoinfty}frac{csqrt{frac{ln x}x}-epsilon}{frac1x}to-infty becauselim_{xtoinfty}frac{ln x}x=0 $
$therefore Lto e^{-infty}=0$
answered Dec 21 '18 at 10:16
Shubham JohriShubham Johri
5,097717
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The statement is equivalent to $csqrt {ln, n lnln , n}-epsilon ln, n to -infty$. To show this write this as $-ln, n [frac {csqrt {ln, n lnln , n}} {-ln , n} +epsilon]$. Can you show that the expression inside goes to $epsilon$? It amounts to showing that $frac {ln ln ,n} {ln, n} to 0$ for which you can use L'Hopital's Rule.
answered Dec 21 '18 at 10:14
Kavi Rama MurthyKavi Rama Murthy
57.3k42160
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add a comment |
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2 Answers
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2 Answers
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$begingroup$
$$displaystyle L=lim_{ntoinfty}frac{e^{csqrt{ln ncdotlnln n}}}{n^{epsilon}}$$
$displaystyleln L=lim_{ntoinfty}csqrt{ln ncdotlnln n}-epsilonln n$
$displaystyle=lim_{xtoinfty}csqrt{xcdotln x}-epsilon x$
$displaystyle=lim_{xtoinfty}frac{csqrt{frac{ln x}x}-epsilon}{frac1x}to-infty becauselim_{xtoinfty}frac{ln x}x=0 $
$therefore Lto e^{-infty}=0$
answered Dec 21 '18 at 10:16
Shubham JohriShubham Johri
5,097717
$endgroup$
add a comment |
$begingroup$
$$displaystyle L=lim_{ntoinfty}frac{e^{csqrt{ln ncdotlnln n}}}{n^{epsilon}}$$
$displaystyleln L=lim_{ntoinfty}csqrt{ln ncdotlnln n}-epsilonln n$
$displaystyle=lim_{xtoinfty}csqrt{xcdotln x}-epsilon x$
$displaystyle=lim_{xtoinfty}frac{csqrt{frac{ln x}x}-epsilon}{frac1x}to-infty becauselim_{xtoinfty}frac{ln x}x=0 $
$therefore Lto e^{-infty}=0$
answered Dec 21 '18 at 10:16
Shubham JohriShubham Johri
5,097717
$endgroup$
add a comment |
$begingroup$
$$displaystyle L=lim_{ntoinfty}frac{e^{csqrt{ln ncdotlnln n}}}{n^{epsilon}}$$
$displaystyleln L=lim_{ntoinfty}csqrt{ln ncdotlnln n}-epsilonln n$
$displaystyle=lim_{xtoinfty}csqrt{xcdotln x}-epsilon x$
$displaystyle=lim_{xtoinfty}frac{csqrt{frac{ln x}x}-epsilon}{frac1x}to-infty becauselim_{xtoinfty}frac{ln x}x=0 $
$therefore Lto e^{-infty}=0$
answered Dec 21 '18 at 10:16
Shubham JohriShubham Johri
5,097717
$endgroup$
$$displaystyle L=lim_{ntoinfty}frac{e^{csqrt{ln ncdotlnln n}}}{n^{epsilon}}$$
$displaystyleln L=lim_{ntoinfty}csqrt{ln ncdotlnln n}-epsilonln n$
$displaystyle=lim_{xtoinfty}csqrt{xcdotln x}-epsilon x$
$displaystyle=lim_{xtoinfty}frac{csqrt{frac{ln x}x}-epsilon}{frac1x}to-infty becauselim_{xtoinfty}frac{ln x}x=0 $
$therefore Lto e^{-infty}=0$
answered Dec 21 '18 at 10:16
Shubham JohriShubham Johri
5,097717
answered Dec 21 '18 at 10:16
Shubham JohriShubham Johri
5,097717
answered Dec 21 '18 at 10:16
Shubham JohriShubham Johri
5,097717
answered Dec 21 '18 at 10:16
Shubham JohriShubham Johri
5,097717
5,097717
add a comment |
add a comment |
$begingroup$
The statement is equivalent to $csqrt {ln, n lnln , n}-epsilon ln, n to -infty$. To show this write this as $-ln, n [frac {csqrt {ln, n lnln , n}} {-ln , n} +epsilon]$. Can you show that the expression inside goes to $epsilon$? It amounts to showing that $frac {ln ln ,n} {ln, n} to 0$ for which you can use L'Hopital's Rule.
answered Dec 21 '18 at 10:14
Kavi Rama MurthyKavi Rama Murthy
57.3k42160
$endgroup$
add a comment |
$begingroup$
The statement is equivalent to $csqrt {ln, n lnln , n}-epsilon ln, n to -infty$. To show this write this as $-ln, n [frac {csqrt {ln, n lnln , n}} {-ln , n} +epsilon]$. Can you show that the expression inside goes to $epsilon$? It amounts to showing that $frac {ln ln ,n} {ln, n} to 0$ for which you can use L'Hopital's Rule.
answered Dec 21 '18 at 10:14
Kavi Rama MurthyKavi Rama Murthy
57.3k42160
$endgroup$
add a comment |
$begingroup$
The statement is equivalent to $csqrt {ln, n lnln , n}-epsilon ln, n to -infty$. To show this write this as $-ln, n [frac {csqrt {ln, n lnln , n}} {-ln , n} +epsilon]$. Can you show that the expression inside goes to $epsilon$? It amounts to showing that $frac {ln ln ,n} {ln, n} to 0$ for which you can use L'Hopital's Rule.
answered Dec 21 '18 at 10:14
Kavi Rama MurthyKavi Rama Murthy
57.3k42160
$endgroup$
The statement is equivalent to $csqrt {ln, n lnln , n}-epsilon ln, n to -infty$. To show this write this as $-ln, n [frac {csqrt {ln, n lnln , n}} {-ln , n} +epsilon]$. Can you show that the expression inside goes to $epsilon$? It amounts to showing that $frac {ln ln ,n} {ln, n} to 0$ for which you can use L'Hopital's Rule.
answered Dec 21 '18 at 10:14
Kavi Rama MurthyKavi Rama Murthy
57.3k42160
answered Dec 21 '18 at 10:14
Kavi Rama MurthyKavi Rama Murthy
57.3k42160
answered Dec 21 '18 at 10:14
Kavi Rama MurthyKavi Rama Murthy
57.3k42160
answered Dec 21 '18 at 10:14
Kavi Rama MurthyKavi Rama Murthy
57.3k42160
57.3k42160
add a comment |
add a comment |
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