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Show that $lim_{n rightarrow infty}sqrt[n]{n}=1$ using Bernoulli inequality
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Show that $lim_limits{n rightarrow infty}sqrt[n]{n}=1$ using Bernoulli inequality
I have a hint $epsilon > 0$ and $n=ncdot epsilon/epsilon$ and I need to show $$1leq sqrt[n]{n}leq frac{1}{sqrt[n]{epsilon}}(1+epsilon)$$
What I have done so far is verifying the inequality on the left.
Also $$sqrt[n]{n}= frac{1}{sqrt[n]{epsilon}}sqrt[n]{epsilon}sqrt[n]{n}$$
Thanks for taking your time
calculus limits
edited Nov 18 at 21:39
Lorenzo B.
1,6222419
asked Nov 18 at 20:41
RM777
808
add a comment |
up vote
0
down vote
favorite
Show that $lim_limits{n rightarrow infty}sqrt[n]{n}=1$ using Bernoulli inequality
I have a hint $epsilon > 0$ and $n=ncdot epsilon/epsilon$ and I need to show $$1leq sqrt[n]{n}leq frac{1}{sqrt[n]{epsilon}}(1+epsilon)$$
What I have done so far is verifying the inequality on the left.
Also $$sqrt[n]{n}= frac{1}{sqrt[n]{epsilon}}sqrt[n]{epsilon}sqrt[n]{n}$$
Thanks for taking your time
calculus limits
edited Nov 18 at 21:39
Lorenzo B.
1,6222419
asked Nov 18 at 20:41
RM777
808
I think $epsilon$ can be $leq 1$ because I have proved alread the left inequality.
– RM777
Nov 18 at 20:43
Are you forced to use Bernoulli's inequality?
– gimusi
Nov 18 at 20:56
@gimusi No it was given only as a hint in the excercise but I think other Proofs would be to high for my current understanding
– RM777
2 days ago
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Show that $lim_limits{n rightarrow infty}sqrt[n]{n}=1$ using Bernoulli inequality
I have a hint $epsilon > 0$ and $n=ncdot epsilon/epsilon$ and I need to show $$1leq sqrt[n]{n}leq frac{1}{sqrt[n]{epsilon}}(1+epsilon)$$
What I have done so far is verifying the inequality on the left.
Also $$sqrt[n]{n}= frac{1}{sqrt[n]{epsilon}}sqrt[n]{epsilon}sqrt[n]{n}$$
Thanks for taking your time
calculus limits
edited Nov 18 at 21:39
Lorenzo B.
1,6222419
asked Nov 18 at 20:41
RM777
808
Show that $lim_limits{n rightarrow infty}sqrt[n]{n}=1$ using Bernoulli inequality
I have a hint $epsilon > 0$ and $n=ncdot epsilon/epsilon$ and I need to show $$1leq sqrt[n]{n}leq frac{1}{sqrt[n]{epsilon}}(1+epsilon)$$
What I have done so far is verifying the inequality on the left.
Also $$sqrt[n]{n}= frac{1}{sqrt[n]{epsilon}}sqrt[n]{epsilon}sqrt[n]{n}$$
Thanks for taking your time
calculus limits
calculus limits
edited Nov 18 at 21:39
Lorenzo B.
1,6222419
asked Nov 18 at 20:41
RM777
808
edited Nov 18 at 21:39
Lorenzo B.
1,6222419
asked Nov 18 at 20:41
RM777
808
edited Nov 18 at 21:39
Lorenzo B.
1,6222419
edited Nov 18 at 21:39
Lorenzo B.
1,6222419
edited Nov 18 at 21:39
Lorenzo B.
1,6222419
1,6222419
asked Nov 18 at 20:41
RM777
808
asked Nov 18 at 20:41
RM777
808
asked Nov 18 at 20:41
RM777
808
808
I think $epsilon$ can be $leq 1$ because I have proved alread the left inequality.
– RM777
Nov 18 at 20:43
Are you forced to use Bernoulli's inequality?
– gimusi
Nov 18 at 20:56
@gimusi No it was given only as a hint in the excercise but I think other Proofs would be to high for my current understanding
– RM777
2 days ago
add a comment |
I think $epsilon$ can be $leq 1$ because I have proved alread the left inequality.
– RM777
Nov 18 at 20:43
Are you forced to use Bernoulli's inequality?
– gimusi
Nov 18 at 20:56
@gimusi No it was given only as a hint in the excercise but I think other Proofs would be to high for my current understanding
– RM777
2 days ago
I think $epsilon$ can be $leq 1$ because I have proved alread the left inequality.
– RM777
Nov 18 at 20:43
I think $epsilon$ can be $leq 1$ because I have proved alread the left inequality.
– RM777
Nov 18 at 20:43
Are you forced to use Bernoulli's inequality?
– gimusi
Nov 18 at 20:56
Are you forced to use Bernoulli's inequality?
– gimusi
Nov 18 at 20:56
@gimusi No it was given only as a hint in the excercise but I think other Proofs would be to high for my current understanding
– RM777
2 days ago
@gimusi No it was given only as a hint in the excercise but I think other Proofs would be to high for my current understanding
– RM777
2 days ago
add a comment |
5 Answers
5
active
oldest
votes
up vote
3
down vote
$(1+frac1{sqrt{n}})^n
ge 1+frac{n}{sqrt{n}}
gt n^{1/2}
$.
Raising to the $2/n$ power,
$n^{1/n}
=(n^{1/2})^{2/n}
lt ((1+frac1{sqrt{n}})^n)^{2/n}
= (1+frac1{sqrt{n}})^2
=1+frac{2}{sqrt{n}}+frac1{n}
lt 1+frac{3}{sqrt{n}}
to 1
$.
answered Nov 19 at 1:25
marty cohen
71.3k546123
Why is $ 1+frac{n}{sqrt{n}} gt n^{1/2} $?
– RM777
2 days ago
$frac{n}{n^{frac{1}{2}}}=frac{n^{frac{1}{2}}n^{frac{1}{2}}}{n^{frac{1}{2}}}=n^{frac{1}{2}}$ I don't think this is what they had in mind, when they gave the hint, but i have to admit, that it is by far the shortest solution.
– Martin Erhardt
2 days ago
add a comment |
up vote
2
down vote
This is how you are taking advantage of Bernoulli in order to obtain the second inequality:
$frac{(1+epsilon)^n}{epsilon}overset{Bernoulli}{geq} frac{1+nepsilon}{epsilon} = frac{1}{epsilon}+n geq n Rightarrowfrac{(1+epsilon)}{epsilon^{frac{1}{n}}} geq n^{frac{1}{n}}$
Note that: $$lim_{epsilonrightarrow0}lim_{nrightarrowinfty}frac{(1+epsilon)}{epsilon^{frac{1}{n}}}=lim_{epsilonrightarrow 0}left (lim_{nrightarrowinfty}left (frac{1}{epsilon^{frac{1}{n}}}+frac{epsilon}{epsilon^{frac{1}{n}}}right ) right )=lim_{epsilonrightarrow 0}left (lim_{nrightarrowinfty}frac{1}{epsilon^{frac{1}{n}}}+lim_{nrightarrowinfty}frac{epsilon}{epsilon^{frac{1}{n}}} right)$$ $$=lim_{epsilonrightarrow 0}(lim_{nrightarrowinfty}epsilon^{-frac{1}{n}}+lim_{nrightarrowinfty}epsilon^{1-frac{1}{n}})=lim_{epsilonrightarrow0}(1+epsilon)=1$$
$lim_{nrightarrowinfty}epsilon^{-frac{1}{n}}=1$ because of $epsilon$ being a constant.
To conclude:
$1=lim_{epsilonrightarrow0}lim_{nrightarrowinfty}frac{(1+epsilon)}{epsilon^{frac{1}{n}}} geq lim_{epsilonrightarrow0}lim_{nrightarrowinfty}n^{frac{1}{n}}=lim_{nrightarrowinfty}n^{frac{1}{n}}$
answered Nov 18 at 21:00
Martin Erhardt
1759
Substitute $lim_{epsilon rightarrow 0}$ with $lim_{k rightarrow infty}$, $epsilon$ with $k^{-1}$ and the proof would still be correct. Just in case you are not familiar with this notion your in lecture.
– Martin Erhardt
Nov 18 at 21:59
why epsillon goes to $0$?
– RM777
2 days ago
Also the line where you say $lim_{epsilonrightarrow0}lim_{nrightarrowinfty}frac{(1+epsilon)}{epsilon^{frac{1}{n}}}=lim_{epsilonrightarrow 0}(lim_{nrightarrowinfty}epsilon^{-frac{1}{n}}+lim_{nrightarrowinfty}epsilon^{1-frac{1}{n}})=lim_{epsilonrightarrow0}(1+epsilon)=1 $ , how do you get first equality in particular $lim_{nrightarrow infty} epsilon^{1-frac{1}{n}}$?
– RM777
2 days ago
Suppose I have already proved $lim_{nrightarrow infty} sqrt[n]a = 1 , a in (0,1)$. Can this help to make it easier?
– RM777
2 days ago
Let me simply link you to math.stackexchange.com/a/488042/254056 . The cases $a in (0,1).a in [1,∞)$ are basically the same. I suppose you can just assume it is known even though the proof is not long.
– Martin Erhardt
2 days ago
add a comment |
up vote
1
down vote
You can prove it like that: $(sqrt n)^{1/n}=1+k_n$
Where $k_n$ is some positive number depending on $n$
Then $sqrt n =(1+k_n)^n geq nk_n$-Bernoulli
So $k_n leq sqrt n/n =1/sqrt n$
And then $1 leq n^{1/n} =(1+k_n)^2=1+2k_n+k_n^2<1+2/sqrt n + 1/n$
And you get what you need using squeeze theorem
answered Nov 18 at 21:19
Anton Zagrivin
1257
You introduced an extra square root for no reason.
– Trevor Gunn
Nov 18 at 21:44
add a comment |
up vote
1
down vote
Option.
$x ge 0.$
$(star)$ $(1+x)^n gt (n^2/4)x^2$.
$x=2/√n.$
$(1+2/√n)^n gt n$
$1+2/√n gt n^{1/n} gt 1.$
Proof of $(star).$
$(1+x)^n ge n(n-1)x^2/2...gt$
$(n^2/4)x^2,$ since for $n >2$:
$(n-1) > n/2$.
answered Nov 18 at 21:54
Peter Szilas
9,9392720
add a comment |
up vote
1
down vote
As shown in this answer, using Bernoulli's Inequality, $left(1+frac1nright)^{n+1}$ is decreasing.
Thus, for $nge1$,
$$
begin{align}
frac{(n+1)^n}{n^{n+1}}
&=frac1{n+1}left(1+frac1nright)^{n+1}\
&lefrac4{n+1}
end{align}
$$
Therefore, for $nge3$,
$$
begin{align}
frac{(n+1)^{frac1{n+1}}}{n^{frac1n}}
&=left(frac{(n+1)^n}{n^{n+1}}right)^{frac1{n(n+1)}}\
&leleft(frac4{n+1}right)^{frac1{n(n+1)}}\[12pt]
&le1
end{align}
$$
Thus, for $nge3$, $n^{1/n}$ is decreasing and bounded below by $1$. Therefore, $limlimits_{ntoinfty}n^{1/n}$ exists and is no less than $1$.
Let $alpha=limlimits_{ntoinfty}n^{1/n}$, then
$$
begin{align}
alpha
&=lim_{ntoinfty}n^{frac1n}\
&=lim_{ntoinfty}(2n)^{frac1{2n}}\
&=lim_{ntoinfty}2^{frac1{2n}}lim_{ntoinfty}n^{frac1{2n}}\[3pt]
&=1cdotsqrt{alpha}
end{align}
$$
Since $alphage1$ and $alpha=sqrt{alpha}$, we get
$$
begin{align}
lim_{ntoinfty}n^{frac1n}
&=alpha\
&=1
end{align}
$$
answered Nov 18 at 23:25
robjohn♦
262k27300620
add a comment |
5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
$(1+frac1{sqrt{n}})^n
ge 1+frac{n}{sqrt{n}}
gt n^{1/2}
$.
Raising to the $2/n$ power,
$n^{1/n}
=(n^{1/2})^{2/n}
lt ((1+frac1{sqrt{n}})^n)^{2/n}
= (1+frac1{sqrt{n}})^2
=1+frac{2}{sqrt{n}}+frac1{n}
lt 1+frac{3}{sqrt{n}}
to 1
$.
answered Nov 19 at 1:25
marty cohen
71.3k546123
Why is $ 1+frac{n}{sqrt{n}} gt n^{1/2} $?
– RM777
2 days ago
$frac{n}{n^{frac{1}{2}}}=frac{n^{frac{1}{2}}n^{frac{1}{2}}}{n^{frac{1}{2}}}=n^{frac{1}{2}}$ I don't think this is what they had in mind, when they gave the hint, but i have to admit, that it is by far the shortest solution.
– Martin Erhardt
2 days ago
add a comment |
up vote
3
down vote
$(1+frac1{sqrt{n}})^n
ge 1+frac{n}{sqrt{n}}
gt n^{1/2}
$.
Raising to the $2/n$ power,
$n^{1/n}
=(n^{1/2})^{2/n}
lt ((1+frac1{sqrt{n}})^n)^{2/n}
= (1+frac1{sqrt{n}})^2
=1+frac{2}{sqrt{n}}+frac1{n}
lt 1+frac{3}{sqrt{n}}
to 1
$.
answered Nov 19 at 1:25
marty cohen
71.3k546123
Why is $ 1+frac{n}{sqrt{n}} gt n^{1/2} $?
– RM777
2 days ago
$frac{n}{n^{frac{1}{2}}}=frac{n^{frac{1}{2}}n^{frac{1}{2}}}{n^{frac{1}{2}}}=n^{frac{1}{2}}$ I don't think this is what they had in mind, when they gave the hint, but i have to admit, that it is by far the shortest solution.
– Martin Erhardt
2 days ago
add a comment |
up vote
3
down vote
up vote
3
down vote
$(1+frac1{sqrt{n}})^n
ge 1+frac{n}{sqrt{n}}
gt n^{1/2}
$.
Raising to the $2/n$ power,
$n^{1/n}
=(n^{1/2})^{2/n}
lt ((1+frac1{sqrt{n}})^n)^{2/n}
= (1+frac1{sqrt{n}})^2
=1+frac{2}{sqrt{n}}+frac1{n}
lt 1+frac{3}{sqrt{n}}
to 1
$.
answered Nov 19 at 1:25
marty cohen
71.3k546123
$(1+frac1{sqrt{n}})^n
ge 1+frac{n}{sqrt{n}}
gt n^{1/2}
$.
Raising to the $2/n$ power,
$n^{1/n}
=(n^{1/2})^{2/n}
lt ((1+frac1{sqrt{n}})^n)^{2/n}
= (1+frac1{sqrt{n}})^2
=1+frac{2}{sqrt{n}}+frac1{n}
lt 1+frac{3}{sqrt{n}}
to 1
$.
answered Nov 19 at 1:25
marty cohen
71.3k546123
answered Nov 19 at 1:25
marty cohen
71.3k546123
answered Nov 19 at 1:25
marty cohen
71.3k546123
answered Nov 19 at 1:25
marty cohen
71.3k546123
71.3k546123
Why is $ 1+frac{n}{sqrt{n}} gt n^{1/2} $?
– RM777
2 days ago
$frac{n}{n^{frac{1}{2}}}=frac{n^{frac{1}{2}}n^{frac{1}{2}}}{n^{frac{1}{2}}}=n^{frac{1}{2}}$ I don't think this is what they had in mind, when they gave the hint, but i have to admit, that it is by far the shortest solution.
– Martin Erhardt
2 days ago
add a comment |
Why is $ 1+frac{n}{sqrt{n}} gt n^{1/2} $?
– RM777
2 days ago
$frac{n}{n^{frac{1}{2}}}=frac{n^{frac{1}{2}}n^{frac{1}{2}}}{n^{frac{1}{2}}}=n^{frac{1}{2}}$ I don't think this is what they had in mind, when they gave the hint, but i have to admit, that it is by far the shortest solution.
– Martin Erhardt
2 days ago
Why is $ 1+frac{n}{sqrt{n}} gt n^{1/2} $?
– RM777
2 days ago
Why is $ 1+frac{n}{sqrt{n}} gt n^{1/2} $?
– RM777
2 days ago
$frac{n}{n^{frac{1}{2}}}=frac{n^{frac{1}{2}}n^{frac{1}{2}}}{n^{frac{1}{2}}}=n^{frac{1}{2}}$ I don't think this is what they had in mind, when they gave the hint, but i have to admit, that it is by far the shortest solution.
– Martin Erhardt
2 days ago
$frac{n}{n^{frac{1}{2}}}=frac{n^{frac{1}{2}}n^{frac{1}{2}}}{n^{frac{1}{2}}}=n^{frac{1}{2}}$ I don't think this is what they had in mind, when they gave the hint, but i have to admit, that it is by far the shortest solution.
– Martin Erhardt
2 days ago
add a comment |
up vote
2
down vote
This is how you are taking advantage of Bernoulli in order to obtain the second inequality:
$frac{(1+epsilon)^n}{epsilon}overset{Bernoulli}{geq} frac{1+nepsilon}{epsilon} = frac{1}{epsilon}+n geq n Rightarrowfrac{(1+epsilon)}{epsilon^{frac{1}{n}}} geq n^{frac{1}{n}}$
Note that: $$lim_{epsilonrightarrow0}lim_{nrightarrowinfty}frac{(1+epsilon)}{epsilon^{frac{1}{n}}}=lim_{epsilonrightarrow 0}left (lim_{nrightarrowinfty}left (frac{1}{epsilon^{frac{1}{n}}}+frac{epsilon}{epsilon^{frac{1}{n}}}right ) right )=lim_{epsilonrightarrow 0}left (lim_{nrightarrowinfty}frac{1}{epsilon^{frac{1}{n}}}+lim_{nrightarrowinfty}frac{epsilon}{epsilon^{frac{1}{n}}} right)$$ $$=lim_{epsilonrightarrow 0}(lim_{nrightarrowinfty}epsilon^{-frac{1}{n}}+lim_{nrightarrowinfty}epsilon^{1-frac{1}{n}})=lim_{epsilonrightarrow0}(1+epsilon)=1$$
$lim_{nrightarrowinfty}epsilon^{-frac{1}{n}}=1$ because of $epsilon$ being a constant.
To conclude:
$1=lim_{epsilonrightarrow0}lim_{nrightarrowinfty}frac{(1+epsilon)}{epsilon^{frac{1}{n}}} geq lim_{epsilonrightarrow0}lim_{nrightarrowinfty}n^{frac{1}{n}}=lim_{nrightarrowinfty}n^{frac{1}{n}}$
answered Nov 18 at 21:00
Martin Erhardt
1759
Substitute $lim_{epsilon rightarrow 0}$ with $lim_{k rightarrow infty}$, $epsilon$ with $k^{-1}$ and the proof would still be correct. Just in case you are not familiar with this notion your in lecture.
– Martin Erhardt
Nov 18 at 21:59
why epsillon goes to $0$?
– RM777
2 days ago
Also the line where you say $lim_{epsilonrightarrow0}lim_{nrightarrowinfty}frac{(1+epsilon)}{epsilon^{frac{1}{n}}}=lim_{epsilonrightarrow 0}(lim_{nrightarrowinfty}epsilon^{-frac{1}{n}}+lim_{nrightarrowinfty}epsilon^{1-frac{1}{n}})=lim_{epsilonrightarrow0}(1+epsilon)=1 $ , how do you get first equality in particular $lim_{nrightarrow infty} epsilon^{1-frac{1}{n}}$?
– RM777
2 days ago
Suppose I have already proved $lim_{nrightarrow infty} sqrt[n]a = 1 , a in (0,1)$. Can this help to make it easier?
– RM777
2 days ago
Let me simply link you to math.stackexchange.com/a/488042/254056 . The cases $a in (0,1).a in [1,∞)$ are basically the same. I suppose you can just assume it is known even though the proof is not long.
– Martin Erhardt
2 days ago
add a comment |
up vote
2
down vote
This is how you are taking advantage of Bernoulli in order to obtain the second inequality:
$frac{(1+epsilon)^n}{epsilon}overset{Bernoulli}{geq} frac{1+nepsilon}{epsilon} = frac{1}{epsilon}+n geq n Rightarrowfrac{(1+epsilon)}{epsilon^{frac{1}{n}}} geq n^{frac{1}{n}}$
Note that: $$lim_{epsilonrightarrow0}lim_{nrightarrowinfty}frac{(1+epsilon)}{epsilon^{frac{1}{n}}}=lim_{epsilonrightarrow 0}left (lim_{nrightarrowinfty}left (frac{1}{epsilon^{frac{1}{n}}}+frac{epsilon}{epsilon^{frac{1}{n}}}right ) right )=lim_{epsilonrightarrow 0}left (lim_{nrightarrowinfty}frac{1}{epsilon^{frac{1}{n}}}+lim_{nrightarrowinfty}frac{epsilon}{epsilon^{frac{1}{n}}} right)$$ $$=lim_{epsilonrightarrow 0}(lim_{nrightarrowinfty}epsilon^{-frac{1}{n}}+lim_{nrightarrowinfty}epsilon^{1-frac{1}{n}})=lim_{epsilonrightarrow0}(1+epsilon)=1$$
$lim_{nrightarrowinfty}epsilon^{-frac{1}{n}}=1$ because of $epsilon$ being a constant.
To conclude:
$1=lim_{epsilonrightarrow0}lim_{nrightarrowinfty}frac{(1+epsilon)}{epsilon^{frac{1}{n}}} geq lim_{epsilonrightarrow0}lim_{nrightarrowinfty}n^{frac{1}{n}}=lim_{nrightarrowinfty}n^{frac{1}{n}}$
answered Nov 18 at 21:00
Martin Erhardt
1759
Substitute $lim_{epsilon rightarrow 0}$ with $lim_{k rightarrow infty}$, $epsilon$ with $k^{-1}$ and the proof would still be correct. Just in case you are not familiar with this notion your in lecture.
– Martin Erhardt
Nov 18 at 21:59
why epsillon goes to $0$?
– RM777
2 days ago
Also the line where you say $lim_{epsilonrightarrow0}lim_{nrightarrowinfty}frac{(1+epsilon)}{epsilon^{frac{1}{n}}}=lim_{epsilonrightarrow 0}(lim_{nrightarrowinfty}epsilon^{-frac{1}{n}}+lim_{nrightarrowinfty}epsilon^{1-frac{1}{n}})=lim_{epsilonrightarrow0}(1+epsilon)=1 $ , how do you get first equality in particular $lim_{nrightarrow infty} epsilon^{1-frac{1}{n}}$?
– RM777
2 days ago
Suppose I have already proved $lim_{nrightarrow infty} sqrt[n]a = 1 , a in (0,1)$. Can this help to make it easier?
– RM777
2 days ago
Let me simply link you to math.stackexchange.com/a/488042/254056 . The cases $a in (0,1).a in [1,∞)$ are basically the same. I suppose you can just assume it is known even though the proof is not long.
– Martin Erhardt
2 days ago
add a comment |
up vote
2
down vote
up vote
2
down vote
This is how you are taking advantage of Bernoulli in order to obtain the second inequality:
$frac{(1+epsilon)^n}{epsilon}overset{Bernoulli}{geq} frac{1+nepsilon}{epsilon} = frac{1}{epsilon}+n geq n Rightarrowfrac{(1+epsilon)}{epsilon^{frac{1}{n}}} geq n^{frac{1}{n}}$
Note that: $$lim_{epsilonrightarrow0}lim_{nrightarrowinfty}frac{(1+epsilon)}{epsilon^{frac{1}{n}}}=lim_{epsilonrightarrow 0}left (lim_{nrightarrowinfty}left (frac{1}{epsilon^{frac{1}{n}}}+frac{epsilon}{epsilon^{frac{1}{n}}}right ) right )=lim_{epsilonrightarrow 0}left (lim_{nrightarrowinfty}frac{1}{epsilon^{frac{1}{n}}}+lim_{nrightarrowinfty}frac{epsilon}{epsilon^{frac{1}{n}}} right)$$ $$=lim_{epsilonrightarrow 0}(lim_{nrightarrowinfty}epsilon^{-frac{1}{n}}+lim_{nrightarrowinfty}epsilon^{1-frac{1}{n}})=lim_{epsilonrightarrow0}(1+epsilon)=1$$
$lim_{nrightarrowinfty}epsilon^{-frac{1}{n}}=1$ because of $epsilon$ being a constant.
To conclude:
$1=lim_{epsilonrightarrow0}lim_{nrightarrowinfty}frac{(1+epsilon)}{epsilon^{frac{1}{n}}} geq lim_{epsilonrightarrow0}lim_{nrightarrowinfty}n^{frac{1}{n}}=lim_{nrightarrowinfty}n^{frac{1}{n}}$
answered Nov 18 at 21:00
Martin Erhardt
1759
This is how you are taking advantage of Bernoulli in order to obtain the second inequality:
$frac{(1+epsilon)^n}{epsilon}overset{Bernoulli}{geq} frac{1+nepsilon}{epsilon} = frac{1}{epsilon}+n geq n Rightarrowfrac{(1+epsilon)}{epsilon^{frac{1}{n}}} geq n^{frac{1}{n}}$
Note that: $$lim_{epsilonrightarrow0}lim_{nrightarrowinfty}frac{(1+epsilon)}{epsilon^{frac{1}{n}}}=lim_{epsilonrightarrow 0}left (lim_{nrightarrowinfty}left (frac{1}{epsilon^{frac{1}{n}}}+frac{epsilon}{epsilon^{frac{1}{n}}}right ) right )=lim_{epsilonrightarrow 0}left (lim_{nrightarrowinfty}frac{1}{epsilon^{frac{1}{n}}}+lim_{nrightarrowinfty}frac{epsilon}{epsilon^{frac{1}{n}}} right)$$ $$=lim_{epsilonrightarrow 0}(lim_{nrightarrowinfty}epsilon^{-frac{1}{n}}+lim_{nrightarrowinfty}epsilon^{1-frac{1}{n}})=lim_{epsilonrightarrow0}(1+epsilon)=1$$
$lim_{nrightarrowinfty}epsilon^{-frac{1}{n}}=1$ because of $epsilon$ being a constant.
To conclude:
$1=lim_{epsilonrightarrow0}lim_{nrightarrowinfty}frac{(1+epsilon)}{epsilon^{frac{1}{n}}} geq lim_{epsilonrightarrow0}lim_{nrightarrowinfty}n^{frac{1}{n}}=lim_{nrightarrowinfty}n^{frac{1}{n}}$
answered Nov 18 at 21:00
Martin Erhardt
1759
edited 2 days ago
answered Nov 18 at 21:00
Martin Erhardt
1759
answered Nov 18 at 21:00
Martin Erhardt
1759
answered Nov 18 at 21:00
Martin Erhardt
1759
1759
Substitute $lim_{epsilon rightarrow 0}$ with $lim_{k rightarrow infty}$, $epsilon$ with $k^{-1}$ and the proof would still be correct. Just in case you are not familiar with this notion your in lecture.
– Martin Erhardt
Nov 18 at 21:59
why epsillon goes to $0$?
– RM777
2 days ago
Also the line where you say $lim_{epsilonrightarrow0}lim_{nrightarrowinfty}frac{(1+epsilon)}{epsilon^{frac{1}{n}}}=lim_{epsilonrightarrow 0}(lim_{nrightarrowinfty}epsilon^{-frac{1}{n}}+lim_{nrightarrowinfty}epsilon^{1-frac{1}{n}})=lim_{epsilonrightarrow0}(1+epsilon)=1 $ , how do you get first equality in particular $lim_{nrightarrow infty} epsilon^{1-frac{1}{n}}$?
– RM777
2 days ago
Suppose I have already proved $lim_{nrightarrow infty} sqrt[n]a = 1 , a in (0,1)$. Can this help to make it easier?
– RM777
2 days ago
Let me simply link you to math.stackexchange.com/a/488042/254056 . The cases $a in (0,1).a in [1,∞)$ are basically the same. I suppose you can just assume it is known even though the proof is not long.
– Martin Erhardt
2 days ago
add a comment |
Substitute $lim_{epsilon rightarrow 0}$ with $lim_{k rightarrow infty}$, $epsilon$ with $k^{-1}$ and the proof would still be correct. Just in case you are not familiar with this notion your in lecture.
– Martin Erhardt
Nov 18 at 21:59
why epsillon goes to $0$?
– RM777
2 days ago
Also the line where you say $lim_{epsilonrightarrow0}lim_{nrightarrowinfty}frac{(1+epsilon)}{epsilon^{frac{1}{n}}}=lim_{epsilonrightarrow 0}(lim_{nrightarrowinfty}epsilon^{-frac{1}{n}}+lim_{nrightarrowinfty}epsilon^{1-frac{1}{n}})=lim_{epsilonrightarrow0}(1+epsilon)=1 $ , how do you get first equality in particular $lim_{nrightarrow infty} epsilon^{1-frac{1}{n}}$?
– RM777
2 days ago
Suppose I have already proved $lim_{nrightarrow infty} sqrt[n]a = 1 , a in (0,1)$. Can this help to make it easier?
– RM777
2 days ago
Let me simply link you to math.stackexchange.com/a/488042/254056 . The cases $a in (0,1).a in [1,∞)$ are basically the same. I suppose you can just assume it is known even though the proof is not long.
– Martin Erhardt
2 days ago
Substitute $lim_{epsilon rightarrow 0}$ with $lim_{k rightarrow infty}$, $epsilon$ with $k^{-1}$ and the proof would still be correct. Just in case you are not familiar with this notion your in lecture.
– Martin Erhardt
Nov 18 at 21:59
Substitute $lim_{epsilon rightarrow 0}$ with $lim_{k rightarrow infty}$, $epsilon$ with $k^{-1}$ and the proof would still be correct. Just in case you are not familiar with this notion your in lecture.
– Martin Erhardt
Nov 18 at 21:59
why epsillon goes to $0$?
– RM777
2 days ago
why epsillon goes to $0$?
– RM777
2 days ago
Also the line where you say $lim_{epsilonrightarrow0}lim_{nrightarrowinfty}frac{(1+epsilon)}{epsilon^{frac{1}{n}}}=lim_{epsilonrightarrow 0}(lim_{nrightarrowinfty}epsilon^{-frac{1}{n}}+lim_{nrightarrowinfty}epsilon^{1-frac{1}{n}})=lim_{epsilonrightarrow0}(1+epsilon)=1 $ , how do you get first equality in particular $lim_{nrightarrow infty} epsilon^{1-frac{1}{n}}$?
– RM777
2 days ago
Also the line where you say $lim_{epsilonrightarrow0}lim_{nrightarrowinfty}frac{(1+epsilon)}{epsilon^{frac{1}{n}}}=lim_{epsilonrightarrow 0}(lim_{nrightarrowinfty}epsilon^{-frac{1}{n}}+lim_{nrightarrowinfty}epsilon^{1-frac{1}{n}})=lim_{epsilonrightarrow0}(1+epsilon)=1 $ , how do you get first equality in particular $lim_{nrightarrow infty} epsilon^{1-frac{1}{n}}$?
– RM777
2 days ago
Suppose I have already proved $lim_{nrightarrow infty} sqrt[n]a = 1 , a in (0,1)$. Can this help to make it easier?
– RM777
2 days ago
Suppose I have already proved $lim_{nrightarrow infty} sqrt[n]a = 1 , a in (0,1)$. Can this help to make it easier?
– RM777
2 days ago
Let me simply link you to math.stackexchange.com/a/488042/254056 . The cases $a in (0,1).a in [1,∞)$ are basically the same. I suppose you can just assume it is known even though the proof is not long.
– Martin Erhardt
2 days ago
Let me simply link you to math.stackexchange.com/a/488042/254056 . The cases $a in (0,1).a in [1,∞)$ are basically the same. I suppose you can just assume it is known even though the proof is not long.
– Martin Erhardt
2 days ago
add a comment |
up vote
1
down vote
You can prove it like that: $(sqrt n)^{1/n}=1+k_n$
Where $k_n$ is some positive number depending on $n$
Then $sqrt n =(1+k_n)^n geq nk_n$-Bernoulli
So $k_n leq sqrt n/n =1/sqrt n$
And then $1 leq n^{1/n} =(1+k_n)^2=1+2k_n+k_n^2<1+2/sqrt n + 1/n$
And you get what you need using squeeze theorem
answered Nov 18 at 21:19
Anton Zagrivin
1257
You introduced an extra square root for no reason.
– Trevor Gunn
Nov 18 at 21:44
add a comment |
up vote
1
down vote
You can prove it like that: $(sqrt n)^{1/n}=1+k_n$
Where $k_n$ is some positive number depending on $n$
Then $sqrt n =(1+k_n)^n geq nk_n$-Bernoulli
So $k_n leq sqrt n/n =1/sqrt n$
And then $1 leq n^{1/n} =(1+k_n)^2=1+2k_n+k_n^2<1+2/sqrt n + 1/n$
And you get what you need using squeeze theorem
answered Nov 18 at 21:19
Anton Zagrivin
1257
You introduced an extra square root for no reason.
– Trevor Gunn
Nov 18 at 21:44
add a comment |
up vote
1
down vote
up vote
1
down vote
You can prove it like that: $(sqrt n)^{1/n}=1+k_n$
Where $k_n$ is some positive number depending on $n$
Then $sqrt n =(1+k_n)^n geq nk_n$-Bernoulli
So $k_n leq sqrt n/n =1/sqrt n$
And then $1 leq n^{1/n} =(1+k_n)^2=1+2k_n+k_n^2<1+2/sqrt n + 1/n$
And you get what you need using squeeze theorem
answered Nov 18 at 21:19
Anton Zagrivin
1257
You can prove it like that: $(sqrt n)^{1/n}=1+k_n$
Where $k_n$ is some positive number depending on $n$
Then $sqrt n =(1+k_n)^n geq nk_n$-Bernoulli
So $k_n leq sqrt n/n =1/sqrt n$
And then $1 leq n^{1/n} =(1+k_n)^2=1+2k_n+k_n^2<1+2/sqrt n + 1/n$
And you get what you need using squeeze theorem
answered Nov 18 at 21:19
Anton Zagrivin
1257
answered Nov 18 at 21:19
Anton Zagrivin
1257
answered Nov 18 at 21:19
Anton Zagrivin
1257
answered Nov 18 at 21:19
Anton Zagrivin
1257
1257
You introduced an extra square root for no reason.
– Trevor Gunn
Nov 18 at 21:44
add a comment |
You introduced an extra square root for no reason.
– Trevor Gunn
Nov 18 at 21:44
You introduced an extra square root for no reason.
– Trevor Gunn
Nov 18 at 21:44
You introduced an extra square root for no reason.
– Trevor Gunn
Nov 18 at 21:44
add a comment |
up vote
1
down vote
Option.
$x ge 0.$
$(star)$ $(1+x)^n gt (n^2/4)x^2$.
$x=2/√n.$
$(1+2/√n)^n gt n$
$1+2/√n gt n^{1/n} gt 1.$
Proof of $(star).$
$(1+x)^n ge n(n-1)x^2/2...gt$
$(n^2/4)x^2,$ since for $n >2$:
$(n-1) > n/2$.
answered Nov 18 at 21:54
Peter Szilas
9,9392720
add a comment |
up vote
1
down vote
Option.
$x ge 0.$
$(star)$ $(1+x)^n gt (n^2/4)x^2$.
$x=2/√n.$
$(1+2/√n)^n gt n$
$1+2/√n gt n^{1/n} gt 1.$
Proof of $(star).$
$(1+x)^n ge n(n-1)x^2/2...gt$
$(n^2/4)x^2,$ since for $n >2$:
$(n-1) > n/2$.
answered Nov 18 at 21:54
Peter Szilas
9,9392720
add a comment |
up vote
1
down vote
up vote
1
down vote
Option.
$x ge 0.$
$(star)$ $(1+x)^n gt (n^2/4)x^2$.
$x=2/√n.$
$(1+2/√n)^n gt n$
$1+2/√n gt n^{1/n} gt 1.$
Proof of $(star).$
$(1+x)^n ge n(n-1)x^2/2...gt$
$(n^2/4)x^2,$ since for $n >2$:
$(n-1) > n/2$.
answered Nov 18 at 21:54
Peter Szilas
9,9392720
Option.
$x ge 0.$
$(star)$ $(1+x)^n gt (n^2/4)x^2$.
$x=2/√n.$
$(1+2/√n)^n gt n$
$1+2/√n gt n^{1/n} gt 1.$
Proof of $(star).$
$(1+x)^n ge n(n-1)x^2/2...gt$
$(n^2/4)x^2,$ since for $n >2$:
$(n-1) > n/2$.
answered Nov 18 at 21:54
Peter Szilas
9,9392720
answered Nov 18 at 21:54
Peter Szilas
9,9392720
answered Nov 18 at 21:54
Peter Szilas
9,9392720
answered Nov 18 at 21:54
Peter Szilas
9,9392720
9,9392720
add a comment |
add a comment |
up vote
1
down vote
As shown in this answer, using Bernoulli's Inequality, $left(1+frac1nright)^{n+1}$ is decreasing.
Thus, for $nge1$,
$$
begin{align}
frac{(n+1)^n}{n^{n+1}}
&=frac1{n+1}left(1+frac1nright)^{n+1}\
&lefrac4{n+1}
end{align}
$$
Therefore, for $nge3$,
$$
begin{align}
frac{(n+1)^{frac1{n+1}}}{n^{frac1n}}
&=left(frac{(n+1)^n}{n^{n+1}}right)^{frac1{n(n+1)}}\
&leleft(frac4{n+1}right)^{frac1{n(n+1)}}\[12pt]
&le1
end{align}
$$
Thus, for $nge3$, $n^{1/n}$ is decreasing and bounded below by $1$. Therefore, $limlimits_{ntoinfty}n^{1/n}$ exists and is no less than $1$.
Let $alpha=limlimits_{ntoinfty}n^{1/n}$, then
$$
begin{align}
alpha
&=lim_{ntoinfty}n^{frac1n}\
&=lim_{ntoinfty}(2n)^{frac1{2n}}\
&=lim_{ntoinfty}2^{frac1{2n}}lim_{ntoinfty}n^{frac1{2n}}\[3pt]
&=1cdotsqrt{alpha}
end{align}
$$
Since $alphage1$ and $alpha=sqrt{alpha}$, we get
$$
begin{align}
lim_{ntoinfty}n^{frac1n}
&=alpha\
&=1
end{align}
$$
answered Nov 18 at 23:25
robjohn♦
262k27300620
add a comment |
up vote
1
down vote
As shown in this answer, using Bernoulli's Inequality, $left(1+frac1nright)^{n+1}$ is decreasing.
Thus, for $nge1$,
$$
begin{align}
frac{(n+1)^n}{n^{n+1}}
&=frac1{n+1}left(1+frac1nright)^{n+1}\
&lefrac4{n+1}
end{align}
$$
Therefore, for $nge3$,
$$
begin{align}
frac{(n+1)^{frac1{n+1}}}{n^{frac1n}}
&=left(frac{(n+1)^n}{n^{n+1}}right)^{frac1{n(n+1)}}\
&leleft(frac4{n+1}right)^{frac1{n(n+1)}}\[12pt]
&le1
end{align}
$$
Thus, for $nge3$, $n^{1/n}$ is decreasing and bounded below by $1$. Therefore, $limlimits_{ntoinfty}n^{1/n}$ exists and is no less than $1$.
Let $alpha=limlimits_{ntoinfty}n^{1/n}$, then
$$
begin{align}
alpha
&=lim_{ntoinfty}n^{frac1n}\
&=lim_{ntoinfty}(2n)^{frac1{2n}}\
&=lim_{ntoinfty}2^{frac1{2n}}lim_{ntoinfty}n^{frac1{2n}}\[3pt]
&=1cdotsqrt{alpha}
end{align}
$$
Since $alphage1$ and $alpha=sqrt{alpha}$, we get
$$
begin{align}
lim_{ntoinfty}n^{frac1n}
&=alpha\
&=1
end{align}
$$
answered Nov 18 at 23:25
robjohn♦
262k27300620
add a comment |
up vote
1
down vote
up vote
1
down vote
As shown in this answer, using Bernoulli's Inequality, $left(1+frac1nright)^{n+1}$ is decreasing.
Thus, for $nge1$,
$$
begin{align}
frac{(n+1)^n}{n^{n+1}}
&=frac1{n+1}left(1+frac1nright)^{n+1}\
&lefrac4{n+1}
end{align}
$$
Therefore, for $nge3$,
$$
begin{align}
frac{(n+1)^{frac1{n+1}}}{n^{frac1n}}
&=left(frac{(n+1)^n}{n^{n+1}}right)^{frac1{n(n+1)}}\
&leleft(frac4{n+1}right)^{frac1{n(n+1)}}\[12pt]
&le1
end{align}
$$
Thus, for $nge3$, $n^{1/n}$ is decreasing and bounded below by $1$. Therefore, $limlimits_{ntoinfty}n^{1/n}$ exists and is no less than $1$.
Let $alpha=limlimits_{ntoinfty}n^{1/n}$, then
$$
begin{align}
alpha
&=lim_{ntoinfty}n^{frac1n}\
&=lim_{ntoinfty}(2n)^{frac1{2n}}\
&=lim_{ntoinfty}2^{frac1{2n}}lim_{ntoinfty}n^{frac1{2n}}\[3pt]
&=1cdotsqrt{alpha}
end{align}
$$
Since $alphage1$ and $alpha=sqrt{alpha}$, we get
$$
begin{align}
lim_{ntoinfty}n^{frac1n}
&=alpha\
&=1
end{align}
$$
answered Nov 18 at 23:25
robjohn♦
262k27300620
As shown in this answer, using Bernoulli's Inequality, $left(1+frac1nright)^{n+1}$ is decreasing.
Thus, for $nge1$,
$$
begin{align}
frac{(n+1)^n}{n^{n+1}}
&=frac1{n+1}left(1+frac1nright)^{n+1}\
&lefrac4{n+1}
end{align}
$$
Therefore, for $nge3$,
$$
begin{align}
frac{(n+1)^{frac1{n+1}}}{n^{frac1n}}
&=left(frac{(n+1)^n}{n^{n+1}}right)^{frac1{n(n+1)}}\
&leleft(frac4{n+1}right)^{frac1{n(n+1)}}\[12pt]
&le1
end{align}
$$
Thus, for $nge3$, $n^{1/n}$ is decreasing and bounded below by $1$. Therefore, $limlimits_{ntoinfty}n^{1/n}$ exists and is no less than $1$.
Let $alpha=limlimits_{ntoinfty}n^{1/n}$, then
$$
begin{align}
alpha
&=lim_{ntoinfty}n^{frac1n}\
&=lim_{ntoinfty}(2n)^{frac1{2n}}\
&=lim_{ntoinfty}2^{frac1{2n}}lim_{ntoinfty}n^{frac1{2n}}\[3pt]
&=1cdotsqrt{alpha}
end{align}
$$
Since $alphage1$ and $alpha=sqrt{alpha}$, we get
$$
begin{align}
lim_{ntoinfty}n^{frac1n}
&=alpha\
&=1
end{align}
$$
answered Nov 18 at 23:25
robjohn♦
262k27300620
answered Nov 18 at 23:25
robjohn♦
262k27300620
answered Nov 18 at 23:25
robjohn♦
262k27300620
answered Nov 18 at 23:25
robjohn♦
262k27300620
262k27300620
add a comment |
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I think $epsilon$ can be $leq 1$ because I have proved alread the left inequality.
– RM777
Nov 18 at 20:43
Are you forced to use Bernoulli's inequality?
– gimusi
Nov 18 at 20:56
@gimusi No it was given only as a hint in the excercise but I think other Proofs would be to high for my current understanding
– RM777
2 days ago