Limit of sequence $lim limits_{n to infty }(ln n)^{1+frac{(ln(ln n))^{2019}}{ln n}}-frac{sqrt[2018]n}{ln n}$
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$lim limits_{n to infty }(ln n)^{1+frac{(ln(ln n))^{2019}}{ln n}}-frac{sqrt[2018]n}{ln n}$
I can show that $lim limits_{n to infty } {frac{(ln(ln n))^{2019}}{ln n}}=0 $, so $lim limits_{n to infty }(ln n)^{1+frac{(ln(ln n))^{2019}}{ln n}}=+infty$
and $lim limits_{n to infty } frac{sqrt[2018]n}{ln n}=-infty$
But I don't know how take it together, and show that $lim limits_{n to infty }a_n=-infty$.
Could someone give me hints?
real-analysis sequences-and-series limits limits-without-lhopital
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add a comment |
$begingroup$
$lim limits_{n to infty }(ln n)^{1+frac{(ln(ln n))^{2019}}{ln n}}-frac{sqrt[2018]n}{ln n}$
I can show that $lim limits_{n to infty } {frac{(ln(ln n))^{2019}}{ln n}}=0 $, so $lim limits_{n to infty }(ln n)^{1+frac{(ln(ln n))^{2019}}{ln n}}=+infty$
and $lim limits_{n to infty } frac{sqrt[2018]n}{ln n}=-infty$
But I don't know how take it together, and show that $lim limits_{n to infty }a_n=-infty$.
Could someone give me hints?
real-analysis sequences-and-series limits limits-without-lhopital
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Are you familiar with Taylor series (as an available tool)?
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– Clement C.
Dec 10 '18 at 21:19
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No, I can't use it :/
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– matematiccc
Dec 10 '18 at 21:21
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Where does this problem come from? Could you write down the source?
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– user376343
Dec 10 '18 at 22:10
add a comment |
$begingroup$
$lim limits_{n to infty }(ln n)^{1+frac{(ln(ln n))^{2019}}{ln n}}-frac{sqrt[2018]n}{ln n}$
I can show that $lim limits_{n to infty } {frac{(ln(ln n))^{2019}}{ln n}}=0 $, so $lim limits_{n to infty }(ln n)^{1+frac{(ln(ln n))^{2019}}{ln n}}=+infty$
and $lim limits_{n to infty } frac{sqrt[2018]n}{ln n}=-infty$
But I don't know how take it together, and show that $lim limits_{n to infty }a_n=-infty$.
Could someone give me hints?
real-analysis sequences-and-series limits limits-without-lhopital
$endgroup$
$lim limits_{n to infty }(ln n)^{1+frac{(ln(ln n))^{2019}}{ln n}}-frac{sqrt[2018]n}{ln n}$
I can show that $lim limits_{n to infty } {frac{(ln(ln n))^{2019}}{ln n}}=0 $, so $lim limits_{n to infty }(ln n)^{1+frac{(ln(ln n))^{2019}}{ln n}}=+infty$
and $lim limits_{n to infty } frac{sqrt[2018]n}{ln n}=-infty$
But I don't know how take it together, and show that $lim limits_{n to infty }a_n=-infty$.
Could someone give me hints?
real-analysis sequences-and-series limits limits-without-lhopital
real-analysis sequences-and-series limits limits-without-lhopital
edited Dec 10 '18 at 21:11
user376343
3,7383827
3,7383827
asked Dec 10 '18 at 21:05
matematicccmatematiccc
1275
1275
$begingroup$
Are you familiar with Taylor series (as an available tool)?
$endgroup$
– Clement C.
Dec 10 '18 at 21:19
$begingroup$
No, I can't use it :/
$endgroup$
– matematiccc
Dec 10 '18 at 21:21
$begingroup$
Where does this problem come from? Could you write down the source?
$endgroup$
– user376343
Dec 10 '18 at 22:10
add a comment |
$begingroup$
Are you familiar with Taylor series (as an available tool)?
$endgroup$
– Clement C.
Dec 10 '18 at 21:19
$begingroup$
No, I can't use it :/
$endgroup$
– matematiccc
Dec 10 '18 at 21:21
$begingroup$
Where does this problem come from? Could you write down the source?
$endgroup$
– user376343
Dec 10 '18 at 22:10
$begingroup$
Are you familiar with Taylor series (as an available tool)?
$endgroup$
– Clement C.
Dec 10 '18 at 21:19
$begingroup$
Are you familiar with Taylor series (as an available tool)?
$endgroup$
– Clement C.
Dec 10 '18 at 21:19
$begingroup$
No, I can't use it :/
$endgroup$
– matematiccc
Dec 10 '18 at 21:21
$begingroup$
No, I can't use it :/
$endgroup$
– matematiccc
Dec 10 '18 at 21:21
$begingroup$
Where does this problem come from? Could you write down the source?
$endgroup$
– user376343
Dec 10 '18 at 22:10
$begingroup$
Where does this problem come from? Could you write down the source?
$endgroup$
– user376343
Dec 10 '18 at 22:10
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
We have that
$$(ln n)^{1+frac{(ln(ln n))^{2019}}{ln n}}=ln n cdot e^{frac{(ln(ln n))^{2020}}{ln n}}$$
and then
$$(ln n)^{1+frac{(ln(ln n))^{2019}}{ln n}}-frac{sqrt[2018]n}{ln n}= ln n cdot e^{frac{(ln(ln n))^{2020}}{ln n}}-frac{sqrt[2018]n}{ln n}=ln n left(e^{frac{(ln(ln n))^{2020}}{ln n}}-frac{sqrt[2018]n}{ln^2 n}right)to-infty$$
indeed $forall a>0$
- $frac{n^a}{log n} to infty$
- $frac{(ln(ln n))^{a}}{ln n} to 0$
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add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
We have that
$$(ln n)^{1+frac{(ln(ln n))^{2019}}{ln n}}=ln n cdot e^{frac{(ln(ln n))^{2020}}{ln n}}$$
and then
$$(ln n)^{1+frac{(ln(ln n))^{2019}}{ln n}}-frac{sqrt[2018]n}{ln n}= ln n cdot e^{frac{(ln(ln n))^{2020}}{ln n}}-frac{sqrt[2018]n}{ln n}=ln n left(e^{frac{(ln(ln n))^{2020}}{ln n}}-frac{sqrt[2018]n}{ln^2 n}right)to-infty$$
indeed $forall a>0$
- $frac{n^a}{log n} to infty$
- $frac{(ln(ln n))^{a}}{ln n} to 0$
$endgroup$
add a comment |
$begingroup$
We have that
$$(ln n)^{1+frac{(ln(ln n))^{2019}}{ln n}}=ln n cdot e^{frac{(ln(ln n))^{2020}}{ln n}}$$
and then
$$(ln n)^{1+frac{(ln(ln n))^{2019}}{ln n}}-frac{sqrt[2018]n}{ln n}= ln n cdot e^{frac{(ln(ln n))^{2020}}{ln n}}-frac{sqrt[2018]n}{ln n}=ln n left(e^{frac{(ln(ln n))^{2020}}{ln n}}-frac{sqrt[2018]n}{ln^2 n}right)to-infty$$
indeed $forall a>0$
- $frac{n^a}{log n} to infty$
- $frac{(ln(ln n))^{a}}{ln n} to 0$
$endgroup$
add a comment |
$begingroup$
We have that
$$(ln n)^{1+frac{(ln(ln n))^{2019}}{ln n}}=ln n cdot e^{frac{(ln(ln n))^{2020}}{ln n}}$$
and then
$$(ln n)^{1+frac{(ln(ln n))^{2019}}{ln n}}-frac{sqrt[2018]n}{ln n}= ln n cdot e^{frac{(ln(ln n))^{2020}}{ln n}}-frac{sqrt[2018]n}{ln n}=ln n left(e^{frac{(ln(ln n))^{2020}}{ln n}}-frac{sqrt[2018]n}{ln^2 n}right)to-infty$$
indeed $forall a>0$
- $frac{n^a}{log n} to infty$
- $frac{(ln(ln n))^{a}}{ln n} to 0$
$endgroup$
We have that
$$(ln n)^{1+frac{(ln(ln n))^{2019}}{ln n}}=ln n cdot e^{frac{(ln(ln n))^{2020}}{ln n}}$$
and then
$$(ln n)^{1+frac{(ln(ln n))^{2019}}{ln n}}-frac{sqrt[2018]n}{ln n}= ln n cdot e^{frac{(ln(ln n))^{2020}}{ln n}}-frac{sqrt[2018]n}{ln n}=ln n left(e^{frac{(ln(ln n))^{2020}}{ln n}}-frac{sqrt[2018]n}{ln^2 n}right)to-infty$$
indeed $forall a>0$
- $frac{n^a}{log n} to infty$
- $frac{(ln(ln n))^{a}}{ln n} to 0$
answered Dec 10 '18 at 22:27
gimusigimusi
92.8k84494
92.8k84494
add a comment |
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$begingroup$
Are you familiar with Taylor series (as an available tool)?
$endgroup$
– Clement C.
Dec 10 '18 at 21:19
$begingroup$
No, I can't use it :/
$endgroup$
– matematiccc
Dec 10 '18 at 21:21
$begingroup$
Where does this problem come from? Could you write down the source?
$endgroup$
– user376343
Dec 10 '18 at 22:10