Limit of a sequence where $U_n = frac{(log n )^p}{n}$
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
asked Nov 25 '18 at 9:46
Supriyo Banerjee
595
add a comment |
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
asked Nov 25 '18 at 9:46
Supriyo Banerjee
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
add a comment |
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
asked Nov 25 '18 at 9:46
Supriyo Banerjee
595
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
asked Nov 25 '18 at 9:46
Supriyo Banerjee
595
asked Nov 25 '18 at 9:46
Supriyo Banerjee
595
edited Nov 25 '18 at 9:55
user616975
asked Nov 25 '18 at 9:46
Supriyo Banerjee
595
asked Nov 25 '18 at 9:46
Supriyo Banerjee
595
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
add a comment |
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
add a comment |
1 Answer
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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 $y to infty$.
answered Nov 25 '18 at 18:32
gimusi
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
add a comment |
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1 Answer
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1 Answer
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active
oldest
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votes
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 $y to infty$.
answered Nov 25 '18 at 18:32
gimusi
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
add a comment |
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 $y to infty$.
answered Nov 25 '18 at 18:32
gimusi
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
add a comment |
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 $y to infty$.
answered Nov 25 '18 at 18:32
gimusi
1
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 $y to infty$.
answered Nov 25 '18 at 18:32
gimusi
1
answered Nov 25 '18 at 18:32
gimusi
1
answered Nov 25 '18 at 18:32
gimusi
1
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
add a comment |
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
add a comment |
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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