Mixing
Evaluting $lim_{n to infty } sqrt[n]{25n+n^3}$
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$lim limits_{n to infty } sqrt[n]{25n+n^3}$
$1leftarrowsqrt[n]{25n}lesqrt[n]{25n+n^3}lesqrt[n]{25n^3+n^3}le sqrt[n]{26n^3}=sqrt[n]{26}cdotsqrt[n]{n^3}to1cdot1=1$
But is it obvious that $ forall _{kin Bbb N} (sqrt[n]n)^k to1 ?$
Since I know $(sqrt[n]n) to1$.
Is it enought to say $(sqrt[n]n)^kto1^k? $
real-analysis limits limits-without-lhopital
edited yesterday
amWhy
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asked yesterday
matematiccc
1035
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up vote
2
down vote
favorite
$lim limits_{n to infty } sqrt[n]{25n+n^3}$
$1leftarrowsqrt[n]{25n}lesqrt[n]{25n+n^3}lesqrt[n]{25n^3+n^3}le sqrt[n]{26n^3}=sqrt[n]{26}cdotsqrt[n]{n^3}to1cdot1=1$
But is it obvious that $ forall _{kin Bbb N} (sqrt[n]n)^k to1 ?$
Since I know $(sqrt[n]n) to1$.
Is it enought to say $(sqrt[n]n)^kto1^k? $
real-analysis limits limits-without-lhopital
edited yesterday
amWhy
191k27223437
asked yesterday
matematiccc
1035
$n^{3/n}=(n/3)^{3/n}cdot 3^{3/n}to 1$.
– user254433
yesterday
1
You can use the basic result: if $a_nto a$ and $b_nto b$ then $a_nb_n to ab$ to get $sqrt[n]{n}^k to 1^k = 1$. This works for integer $k$ (induction). In general you can use continuity of the power-function.
– Winther
yesterday
These answers actually show that if $p(n)$ is a polynomial in $n$, then $(p(n))^{1/n} to 1$.
– JavaMan
yesterday
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
$lim limits_{n to infty } sqrt[n]{25n+n^3}$
$1leftarrowsqrt[n]{25n}lesqrt[n]{25n+n^3}lesqrt[n]{25n^3+n^3}le sqrt[n]{26n^3}=sqrt[n]{26}cdotsqrt[n]{n^3}to1cdot1=1$
But is it obvious that $ forall _{kin Bbb N} (sqrt[n]n)^k to1 ?$
Since I know $(sqrt[n]n) to1$.
Is it enought to say $(sqrt[n]n)^kto1^k? $
real-analysis limits limits-without-lhopital
edited yesterday
amWhy
191k27223437
asked yesterday
matematiccc
1035
$lim limits_{n to infty } sqrt[n]{25n+n^3}$
$1leftarrowsqrt[n]{25n}lesqrt[n]{25n+n^3}lesqrt[n]{25n^3+n^3}le sqrt[n]{26n^3}=sqrt[n]{26}cdotsqrt[n]{n^3}to1cdot1=1$
But is it obvious that $ forall _{kin Bbb N} (sqrt[n]n)^k to1 ?$
Since I know $(sqrt[n]n) to1$.
Is it enought to say $(sqrt[n]n)^kto1^k? $
real-analysis limits limits-without-lhopital
real-analysis limits limits-without-lhopital
edited yesterday
amWhy
191k27223437
asked yesterday
matematiccc
1035
edited yesterday
amWhy
191k27223437
asked yesterday
matematiccc
1035
edited yesterday
amWhy
191k27223437
edited yesterday
amWhy
191k27223437
edited yesterday
amWhy
191k27223437
191k27223437
asked yesterday
matematiccc
1035
asked yesterday
matematiccc
1035
asked yesterday
matematiccc
1035
1035
$n^{3/n}=(n/3)^{3/n}cdot 3^{3/n}to 1$.
– user254433
yesterday
1
You can use the basic result: if $a_nto a$ and $b_nto b$ then $a_nb_n to ab$ to get $sqrt[n]{n}^k to 1^k = 1$. This works for integer $k$ (induction). In general you can use continuity of the power-function.
– Winther
yesterday
These answers actually show that if $p(n)$ is a polynomial in $n$, then $(p(n))^{1/n} to 1$.
– JavaMan
yesterday
add a comment |
$n^{3/n}=(n/3)^{3/n}cdot 3^{3/n}to 1$.
– user254433
yesterday
1
You can use the basic result: if $a_nto a$ and $b_nto b$ then $a_nb_n to ab$ to get $sqrt[n]{n}^k to 1^k = 1$. This works for integer $k$ (induction). In general you can use continuity of the power-function.
– Winther
yesterday
These answers actually show that if $p(n)$ is a polynomial in $n$, then $(p(n))^{1/n} to 1$.
– JavaMan
yesterday
$n^{3/n}=(n/3)^{3/n}cdot 3^{3/n}to 1$.
– user254433
yesterday
$n^{3/n}=(n/3)^{3/n}cdot 3^{3/n}to 1$.
– user254433
yesterday
1
1
You can use the basic result: if $a_nto a$ and $b_nto b$ then $a_nb_n to ab$ to get $sqrt[n]{n}^k to 1^k = 1$. This works for integer $k$ (induction). In general you can use continuity of the power-function.
– Winther
yesterday
You can use the basic result: if $a_nto a$ and $b_nto b$ then $a_nb_n to ab$ to get $sqrt[n]{n}^k to 1^k = 1$. This works for integer $k$ (induction). In general you can use continuity of the power-function.
– Winther
yesterday
These answers actually show that if $p(n)$ is a polynomial in $n$, then $(p(n))^{1/n} to 1$.
– JavaMan
yesterday
These answers actually show that if $p(n)$ is a polynomial in $n$, then $(p(n))^{1/n} to 1$.
– JavaMan
yesterday
add a comment |
3 Answers
3
active
oldest
votes
up vote
3
down vote
accepted
Yes your idea is correct indeed we have that eventually
$$sqrt[n]{25n}le sqrt[n]{25n+n^3}le sqrt[n]{2n^3}$$
and
- $sqrt[n]{25n}=sqrt[n]{25}sqrt[n]{n}to 1$
- $sqrt[n]{2n^3}=(sqrt[n]{2})^3(sqrt[n]{n})^3 to 1$
answered yesterday
gimusi
85.5k74294
add a comment |
up vote
0
down vote
use $limsup_{n rightarrow infty} |a_n|^{1/n}$ to help you find the answer.
edited yesterday
Tianlalu
2,549632
answered yesterday
pfmr1995
113
New contributor
pfmr1995 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
up vote
0
down vote
You could also have good approximations for finite values of $n$.
$$a_n=sqrt[n]{kn+n^3}implies log(a_n)=frac 1 n log(kn+n^3)=frac 1 n left(3 log(n)+log left(1+frac{k}{n^2}right) right)$$ Now, using Taylor expansions for large values of $n$
$$log(a_n)=frac{3 log
left({n}right)}{n}+frac{k}{n^3}+Oleft(frac{1}{n^5}right)$$ Continuing with Taylor
$$a_n=e^{log(a_n)}=1+frac{3 log left({n}right)}{n}+frac{9 log
^2left({n}right)}{2 n^2}+frac{6 k+27 log
^3left({n}right)}{6 n^3}+Oleft(frac{1}{n^4}right)$$ which, for sure, shows the limit and how "far" is $k$ appearing in the approximation.
answered 23 hours ago
Claude Leibovici
116k1156131
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
Yes your idea is correct indeed we have that eventually
$$sqrt[n]{25n}le sqrt[n]{25n+n^3}le sqrt[n]{2n^3}$$
and
- $sqrt[n]{25n}=sqrt[n]{25}sqrt[n]{n}to 1$
- $sqrt[n]{2n^3}=(sqrt[n]{2})^3(sqrt[n]{n})^3 to 1$
answered yesterday
gimusi
85.5k74294
add a comment |
up vote
3
down vote
accepted
Yes your idea is correct indeed we have that eventually
$$sqrt[n]{25n}le sqrt[n]{25n+n^3}le sqrt[n]{2n^3}$$
and
- $sqrt[n]{25n}=sqrt[n]{25}sqrt[n]{n}to 1$
- $sqrt[n]{2n^3}=(sqrt[n]{2})^3(sqrt[n]{n})^3 to 1$
answered yesterday
gimusi
85.5k74294
add a comment |
up vote
3
down vote
accepted
up vote
3
down vote
accepted
Yes your idea is correct indeed we have that eventually
$$sqrt[n]{25n}le sqrt[n]{25n+n^3}le sqrt[n]{2n^3}$$
and
- $sqrt[n]{25n}=sqrt[n]{25}sqrt[n]{n}to 1$
- $sqrt[n]{2n^3}=(sqrt[n]{2})^3(sqrt[n]{n})^3 to 1$
answered yesterday
gimusi
85.5k74294
Yes your idea is correct indeed we have that eventually
$$sqrt[n]{25n}le sqrt[n]{25n+n^3}le sqrt[n]{2n^3}$$
and
- $sqrt[n]{25n}=sqrt[n]{25}sqrt[n]{n}to 1$
- $sqrt[n]{2n^3}=(sqrt[n]{2})^3(sqrt[n]{n})^3 to 1$
answered yesterday
gimusi
85.5k74294
answered yesterday
gimusi
85.5k74294
answered yesterday
gimusi
85.5k74294
answered yesterday
gimusi
85.5k74294
85.5k74294
add a comment |
add a comment |
up vote
0
down vote
use $limsup_{n rightarrow infty} |a_n|^{1/n}$ to help you find the answer.
edited yesterday
Tianlalu
2,549632
answered yesterday
pfmr1995
113
New contributor
pfmr1995 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
up vote
0
down vote
use $limsup_{n rightarrow infty} |a_n|^{1/n}$ to help you find the answer.
edited yesterday
Tianlalu
2,549632
answered yesterday
pfmr1995
113
New contributor
pfmr1995 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
up vote
0
down vote
up vote
0
down vote
use $limsup_{n rightarrow infty} |a_n|^{1/n}$ to help you find the answer.
edited yesterday
Tianlalu
2,549632
answered yesterday
pfmr1995
113
New contributor
pfmr1995 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
use $limsup_{n rightarrow infty} |a_n|^{1/n}$ to help you find the answer.
edited yesterday
Tianlalu
2,549632
answered yesterday
pfmr1995
113
New contributor
pfmr1995 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited yesterday
Tianlalu
2,549632
edited yesterday
Tianlalu
2,549632
edited yesterday
Tianlalu
2,549632
2,549632
answered yesterday
pfmr1995
113
New contributor
pfmr1995 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered yesterday
pfmr1995
113
answered yesterday
pfmr1995
113
113
New contributor
pfmr1995 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
pfmr1995 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
pfmr1995 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
up vote
0
down vote
You could also have good approximations for finite values of $n$.
$$a_n=sqrt[n]{kn+n^3}implies log(a_n)=frac 1 n log(kn+n^3)=frac 1 n left(3 log(n)+log left(1+frac{k}{n^2}right) right)$$ Now, using Taylor expansions for large values of $n$
$$log(a_n)=frac{3 log
left({n}right)}{n}+frac{k}{n^3}+Oleft(frac{1}{n^5}right)$$ Continuing with Taylor
$$a_n=e^{log(a_n)}=1+frac{3 log left({n}right)}{n}+frac{9 log
^2left({n}right)}{2 n^2}+frac{6 k+27 log
^3left({n}right)}{6 n^3}+Oleft(frac{1}{n^4}right)$$ which, for sure, shows the limit and how "far" is $k$ appearing in the approximation.
answered 23 hours ago
Claude Leibovici
116k1156131
add a comment |
up vote
0
down vote
You could also have good approximations for finite values of $n$.
$$a_n=sqrt[n]{kn+n^3}implies log(a_n)=frac 1 n log(kn+n^3)=frac 1 n left(3 log(n)+log left(1+frac{k}{n^2}right) right)$$ Now, using Taylor expansions for large values of $n$
$$log(a_n)=frac{3 log
left({n}right)}{n}+frac{k}{n^3}+Oleft(frac{1}{n^5}right)$$ Continuing with Taylor
$$a_n=e^{log(a_n)}=1+frac{3 log left({n}right)}{n}+frac{9 log
^2left({n}right)}{2 n^2}+frac{6 k+27 log
^3left({n}right)}{6 n^3}+Oleft(frac{1}{n^4}right)$$ which, for sure, shows the limit and how "far" is $k$ appearing in the approximation.
answered 23 hours ago
Claude Leibovici
116k1156131
add a comment |
up vote
0
down vote
up vote
0
down vote
You could also have good approximations for finite values of $n$.
$$a_n=sqrt[n]{kn+n^3}implies log(a_n)=frac 1 n log(kn+n^3)=frac 1 n left(3 log(n)+log left(1+frac{k}{n^2}right) right)$$ Now, using Taylor expansions for large values of $n$
$$log(a_n)=frac{3 log
left({n}right)}{n}+frac{k}{n^3}+Oleft(frac{1}{n^5}right)$$ Continuing with Taylor
$$a_n=e^{log(a_n)}=1+frac{3 log left({n}right)}{n}+frac{9 log
^2left({n}right)}{2 n^2}+frac{6 k+27 log
^3left({n}right)}{6 n^3}+Oleft(frac{1}{n^4}right)$$ which, for sure, shows the limit and how "far" is $k$ appearing in the approximation.
answered 23 hours ago
Claude Leibovici
116k1156131
You could also have good approximations for finite values of $n$.
$$a_n=sqrt[n]{kn+n^3}implies log(a_n)=frac 1 n log(kn+n^3)=frac 1 n left(3 log(n)+log left(1+frac{k}{n^2}right) right)$$ Now, using Taylor expansions for large values of $n$
$$log(a_n)=frac{3 log
left({n}right)}{n}+frac{k}{n^3}+Oleft(frac{1}{n^5}right)$$ Continuing with Taylor
$$a_n=e^{log(a_n)}=1+frac{3 log left({n}right)}{n}+frac{9 log
^2left({n}right)}{2 n^2}+frac{6 k+27 log
^3left({n}right)}{6 n^3}+Oleft(frac{1}{n^4}right)$$ which, for sure, shows the limit and how "far" is $k$ appearing in the approximation.
answered 23 hours ago
Claude Leibovici
116k1156131
answered 23 hours ago
Claude Leibovici
116k1156131
answered 23 hours ago
Claude Leibovici
116k1156131
answered 23 hours ago
Claude Leibovici
116k1156131
116k1156131
add a comment |
add a comment |
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$n^{3/n}=(n/3)^{3/n}cdot 3^{3/n}to 1$.
– user254433
yesterday
1
You can use the basic result: if $a_nto a$ and $b_nto b$ then $a_nb_n to ab$ to get $sqrt[n]{n}^k to 1^k = 1$. This works for integer $k$ (induction). In general you can use continuity of the power-function.
– Winther
yesterday
These answers actually show that if $p(n)$ is a polynomial in $n$, then $(p(n))^{1/n} to 1$.
– JavaMan
yesterday