Limit of sequence: $lim_{ntoinfty}({2n+1over 3n})^n$
up vote
0
down vote
favorite
$lim_{ntoinfty}({2n+1over 3n})^n$
For sufficiently large values of $n$:
$0le ({2n+1over 3n})^nle({2over 3})^n$,
From Squeeze theorem:
$lim_{ntoinfty}({2n+1over 3n})^n=0$
Is it ok?
I also tried to do this with formula: $lim_{ntoinfty}(1+{aover n})^n=e^a$, but with no effect.
real-analysis sequences-and-series limits limits-without-lhopital
asked Nov 15 at 18:00
matematiccc
1145
add a comment |
up vote
0
down vote
favorite
$lim_{ntoinfty}({2n+1over 3n})^n$
For sufficiently large values of $n$:
$0le ({2n+1over 3n})^nle({2over 3})^n$,
From Squeeze theorem:
$lim_{ntoinfty}({2n+1over 3n})^n=0$
Is it ok?
I also tried to do this with formula: $lim_{ntoinfty}(1+{aover n})^n=e^a$, but with no effect.
real-analysis sequences-and-series limits limits-without-lhopital
asked Nov 15 at 18:00
matematiccc
1145
You never have $$frac{2n+1}{3n} leq frac{2}{3}$$ no matter how big $n$ is. Try $$frac{2n+1}{3n} leq frac{2.5}{3}$$
– Clement C.
Nov 15 at 18:03
When I changed $ frac{2}{3}$ to $ frac{2.9999}{3}$ is it ok?
– matematiccc
Nov 15 at 18:11
Yes, now it is. That is not great to change the question though, as it makes a couple answers addressing this very point obsolete.
– Clement C.
Nov 15 at 18:12
Sorry. I fixed it.
– matematiccc
Nov 15 at 18:20
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
$lim_{ntoinfty}({2n+1over 3n})^n$
For sufficiently large values of $n$:
$0le ({2n+1over 3n})^nle({2over 3})^n$,
From Squeeze theorem:
$lim_{ntoinfty}({2n+1over 3n})^n=0$
Is it ok?
I also tried to do this with formula: $lim_{ntoinfty}(1+{aover n})^n=e^a$, but with no effect.
real-analysis sequences-and-series limits limits-without-lhopital
asked Nov 15 at 18:00
matematiccc
1145
$lim_{ntoinfty}({2n+1over 3n})^n$
For sufficiently large values of $n$:
$0le ({2n+1over 3n})^nle({2over 3})^n$,
From Squeeze theorem:
$lim_{ntoinfty}({2n+1over 3n})^n=0$
Is it ok?
I also tried to do this with formula: $lim_{ntoinfty}(1+{aover n})^n=e^a$, but with no effect.
real-analysis sequences-and-series limits limits-without-lhopital
real-analysis sequences-and-series limits limits-without-lhopital
asked Nov 15 at 18:00
matematiccc
1145
asked Nov 15 at 18:00
matematiccc
1145
edited Nov 15 at 18:19
asked Nov 15 at 18:00
matematiccc
1145
asked Nov 15 at 18:00
matematiccc
1145
asked Nov 15 at 18:00
matematiccc
1145
1145
You never have $$frac{2n+1}{3n} leq frac{2}{3}$$ no matter how big $n$ is. Try $$frac{2n+1}{3n} leq frac{2.5}{3}$$
– Clement C.
Nov 15 at 18:03
When I changed $ frac{2}{3}$ to $ frac{2.9999}{3}$ is it ok?
– matematiccc
Nov 15 at 18:11
Yes, now it is. That is not great to change the question though, as it makes a couple answers addressing this very point obsolete.
– Clement C.
Nov 15 at 18:12
Sorry. I fixed it.
– matematiccc
Nov 15 at 18:20
add a comment |
You never have $$frac{2n+1}{3n} leq frac{2}{3}$$ no matter how big $n$ is. Try $$frac{2n+1}{3n} leq frac{2.5}{3}$$
– Clement C.
Nov 15 at 18:03
When I changed $ frac{2}{3}$ to $ frac{2.9999}{3}$ is it ok?
– matematiccc
Nov 15 at 18:11
Yes, now it is. That is not great to change the question though, as it makes a couple answers addressing this very point obsolete.
– Clement C.
Nov 15 at 18:12
Sorry. I fixed it.
– matematiccc
Nov 15 at 18:20
You never have $$frac{2n+1}{3n} leq frac{2}{3}$$ no matter how big $n$ is. Try $$frac{2n+1}{3n} leq frac{2.5}{3}$$
– Clement C.
Nov 15 at 18:03
You never have $$frac{2n+1}{3n} leq frac{2}{3}$$ no matter how big $n$ is. Try $$frac{2n+1}{3n} leq frac{2.5}{3}$$
– Clement C.
Nov 15 at 18:03
When I changed $ frac{2}{3}$ to $ frac{2.9999}{3}$ is it ok?
– matematiccc
Nov 15 at 18:11
When I changed $ frac{2}{3}$ to $ frac{2.9999}{3}$ is it ok?
– matematiccc
Nov 15 at 18:11
Yes, now it is. That is not great to change the question though, as it makes a couple answers addressing this very point obsolete.
– Clement C.
Nov 15 at 18:12
Yes, now it is. That is not great to change the question though, as it makes a couple answers addressing this very point obsolete.
– Clement C.
Nov 15 at 18:12
Sorry. I fixed it.
– matematiccc
Nov 15 at 18:20
Sorry. I fixed it.
– matematiccc
Nov 15 at 18:20
add a comment |
4 Answers
4
active
oldest
votes
up vote
3
down vote
accepted
No, it is not correct, because the inequality $left(frac{2n+1}{3n}right)^nleqslantleft(frac23right)^n$ never holds. But you can use the fact that$$lim_{ntoinfty}left(frac{2n+1}{3n}right)^n=lim_{ntoinfty}left(frac23right)^ntimeslim_{ntoinfty}left(1+frac1{2n}right)^n=0.$$
answered Nov 15 at 18:08
José Carlos Santos
142k20112208
2
I might be wrong but I don't think that it is wise to split limits this way. Doesn't always work (works this time though).
– Makina
Nov 15 at 18:12
If it didn't work this time, I would not have written it.
– José Carlos Santos
Nov 15 at 18:13
True, but for educational purposes I'd say one shouldn't write things this way.
– Makina
Nov 15 at 18:14
add a comment |
up vote
0
down vote
Hint: in the original expression take out $frac{2}{3}$ from your expression inside the brackets and then split.
answered Nov 15 at 18:05
Makina
1,006113
add a comment |
up vote
0
down vote
Note that
$${2n+1over 3n}le{2over 3}$$
is not true, use that eventually
$${2n+1over 3n}le frac45<1 iff 10n+5<12n iff 2n>5$$
As a simpler alternative by root test
$$sqrt[n]{a_n}=sqrt[n]{left({2n+1over 3n}right)^n}={2n+1over 3n}tofrac23<1implies a_n to 0$$
answered Nov 15 at 18:03
gimusi
88.5k74394
Which is not enough... since $1-1/n<1$ "eventually" (actually, always) as well. You need $< c$ for some constant $0leq c<1$.
– Clement C.
Nov 15 at 18:04
After the edit, fixing the solution, downvote removed. But you should really make sure your solution is correct before posting it (it's not the first time...).
– Clement C.
Nov 15 at 18:07
@ClementC. Yes of course! the inequalities do not hold in the limit!
– gimusi
Nov 15 at 18:10
add a comment |
up vote
0
down vote
You can use the root test to get a result quite easily. Since $$dfrac{2n+1}{3n}todfrac{2}{3}<1$$
by the ratio test
$$bigg(dfrac{2n+1}{3n}bigg)^nto0.$$
answered Nov 15 at 18:16
Melody
45211
I did not have it yet.
– matematiccc
Nov 15 at 18:18
add a comment |
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
No, it is not correct, because the inequality $left(frac{2n+1}{3n}right)^nleqslantleft(frac23right)^n$ never holds. But you can use the fact that$$lim_{ntoinfty}left(frac{2n+1}{3n}right)^n=lim_{ntoinfty}left(frac23right)^ntimeslim_{ntoinfty}left(1+frac1{2n}right)^n=0.$$
answered Nov 15 at 18:08
José Carlos Santos
142k20112208
2
I might be wrong but I don't think that it is wise to split limits this way. Doesn't always work (works this time though).
– Makina
Nov 15 at 18:12
If it didn't work this time, I would not have written it.
– José Carlos Santos
Nov 15 at 18:13
True, but for educational purposes I'd say one shouldn't write things this way.
– Makina
Nov 15 at 18:14
add a comment |
up vote
3
down vote
accepted
No, it is not correct, because the inequality $left(frac{2n+1}{3n}right)^nleqslantleft(frac23right)^n$ never holds. But you can use the fact that$$lim_{ntoinfty}left(frac{2n+1}{3n}right)^n=lim_{ntoinfty}left(frac23right)^ntimeslim_{ntoinfty}left(1+frac1{2n}right)^n=0.$$
answered Nov 15 at 18:08
José Carlos Santos
142k20112208
2
I might be wrong but I don't think that it is wise to split limits this way. Doesn't always work (works this time though).
– Makina
Nov 15 at 18:12
If it didn't work this time, I would not have written it.
– José Carlos Santos
Nov 15 at 18:13
True, but for educational purposes I'd say one shouldn't write things this way.
– Makina
Nov 15 at 18:14
add a comment |
up vote
3
down vote
accepted
up vote
3
down vote
accepted
No, it is not correct, because the inequality $left(frac{2n+1}{3n}right)^nleqslantleft(frac23right)^n$ never holds. But you can use the fact that$$lim_{ntoinfty}left(frac{2n+1}{3n}right)^n=lim_{ntoinfty}left(frac23right)^ntimeslim_{ntoinfty}left(1+frac1{2n}right)^n=0.$$
answered Nov 15 at 18:08
José Carlos Santos
142k20112208
No, it is not correct, because the inequality $left(frac{2n+1}{3n}right)^nleqslantleft(frac23right)^n$ never holds. But you can use the fact that$$lim_{ntoinfty}left(frac{2n+1}{3n}right)^n=lim_{ntoinfty}left(frac23right)^ntimeslim_{ntoinfty}left(1+frac1{2n}right)^n=0.$$
answered Nov 15 at 18:08
José Carlos Santos
142k20112208
answered Nov 15 at 18:08
José Carlos Santos
142k20112208
answered Nov 15 at 18:08
José Carlos Santos
142k20112208
answered Nov 15 at 18:08
José Carlos Santos
142k20112208
142k20112208
2
I might be wrong but I don't think that it is wise to split limits this way. Doesn't always work (works this time though).
– Makina
Nov 15 at 18:12
If it didn't work this time, I would not have written it.
– José Carlos Santos
Nov 15 at 18:13
True, but for educational purposes I'd say one shouldn't write things this way.
– Makina
Nov 15 at 18:14
add a comment |
2
I might be wrong but I don't think that it is wise to split limits this way. Doesn't always work (works this time though).
– Makina
Nov 15 at 18:12
If it didn't work this time, I would not have written it.
– José Carlos Santos
Nov 15 at 18:13
True, but for educational purposes I'd say one shouldn't write things this way.
– Makina
Nov 15 at 18:14
2
2
I might be wrong but I don't think that it is wise to split limits this way. Doesn't always work (works this time though).
– Makina
Nov 15 at 18:12
I might be wrong but I don't think that it is wise to split limits this way. Doesn't always work (works this time though).
– Makina
Nov 15 at 18:12
If it didn't work this time, I would not have written it.
– José Carlos Santos
Nov 15 at 18:13
If it didn't work this time, I would not have written it.
– José Carlos Santos
Nov 15 at 18:13
True, but for educational purposes I'd say one shouldn't write things this way.
– Makina
Nov 15 at 18:14
True, but for educational purposes I'd say one shouldn't write things this way.
– Makina
Nov 15 at 18:14
add a comment |
up vote
0
down vote
Hint: in the original expression take out $frac{2}{3}$ from your expression inside the brackets and then split.
answered Nov 15 at 18:05
Makina
1,006113
add a comment |
up vote
0
down vote
Hint: in the original expression take out $frac{2}{3}$ from your expression inside the brackets and then split.
answered Nov 15 at 18:05
Makina
1,006113
add a comment |
up vote
0
down vote
up vote
0
down vote
Hint: in the original expression take out $frac{2}{3}$ from your expression inside the brackets and then split.
answered Nov 15 at 18:05
Makina
1,006113
Hint: in the original expression take out $frac{2}{3}$ from your expression inside the brackets and then split.
answered Nov 15 at 18:05
Makina
1,006113
answered Nov 15 at 18:05
Makina
1,006113
answered Nov 15 at 18:05
Makina
1,006113
answered Nov 15 at 18:05
Makina
1,006113
1,006113
add a comment |
add a comment |
up vote
0
down vote
Note that
$${2n+1over 3n}le{2over 3}$$
is not true, use that eventually
$${2n+1over 3n}le frac45<1 iff 10n+5<12n iff 2n>5$$
As a simpler alternative by root test
$$sqrt[n]{a_n}=sqrt[n]{left({2n+1over 3n}right)^n}={2n+1over 3n}tofrac23<1implies a_n to 0$$
answered Nov 15 at 18:03
gimusi
88.5k74394
Which is not enough... since $1-1/n<1$ "eventually" (actually, always) as well. You need $< c$ for some constant $0leq c<1$.
– Clement C.
Nov 15 at 18:04
After the edit, fixing the solution, downvote removed. But you should really make sure your solution is correct before posting it (it's not the first time...).
– Clement C.
Nov 15 at 18:07
@ClementC. Yes of course! the inequalities do not hold in the limit!
– gimusi
Nov 15 at 18:10
add a comment |
up vote
0
down vote
Note that
$${2n+1over 3n}le{2over 3}$$
is not true, use that eventually
$${2n+1over 3n}le frac45<1 iff 10n+5<12n iff 2n>5$$
As a simpler alternative by root test
$$sqrt[n]{a_n}=sqrt[n]{left({2n+1over 3n}right)^n}={2n+1over 3n}tofrac23<1implies a_n to 0$$
answered Nov 15 at 18:03
gimusi
88.5k74394
Which is not enough... since $1-1/n<1$ "eventually" (actually, always) as well. You need $< c$ for some constant $0leq c<1$.
– Clement C.
Nov 15 at 18:04
After the edit, fixing the solution, downvote removed. But you should really make sure your solution is correct before posting it (it's not the first time...).
– Clement C.
Nov 15 at 18:07
@ClementC. Yes of course! the inequalities do not hold in the limit!
– gimusi
Nov 15 at 18:10
add a comment |
up vote
0
down vote
up vote
0
down vote
Note that
$${2n+1over 3n}le{2over 3}$$
is not true, use that eventually
$${2n+1over 3n}le frac45<1 iff 10n+5<12n iff 2n>5$$
As a simpler alternative by root test
$$sqrt[n]{a_n}=sqrt[n]{left({2n+1over 3n}right)^n}={2n+1over 3n}tofrac23<1implies a_n to 0$$
answered Nov 15 at 18:03
gimusi
88.5k74394
Note that
$${2n+1over 3n}le{2over 3}$$
is not true, use that eventually
$${2n+1over 3n}le frac45<1 iff 10n+5<12n iff 2n>5$$
As a simpler alternative by root test
$$sqrt[n]{a_n}=sqrt[n]{left({2n+1over 3n}right)^n}={2n+1over 3n}tofrac23<1implies a_n to 0$$
answered Nov 15 at 18:03
gimusi
88.5k74394
edited Nov 15 at 18:12
answered Nov 15 at 18:03
gimusi
88.5k74394
answered Nov 15 at 18:03
gimusi
88.5k74394
answered Nov 15 at 18:03
gimusi
88.5k74394
88.5k74394
Which is not enough... since $1-1/n<1$ "eventually" (actually, always) as well. You need $< c$ for some constant $0leq c<1$.
– Clement C.
Nov 15 at 18:04
After the edit, fixing the solution, downvote removed. But you should really make sure your solution is correct before posting it (it's not the first time...).
– Clement C.
Nov 15 at 18:07
@ClementC. Yes of course! the inequalities do not hold in the limit!
– gimusi
Nov 15 at 18:10
add a comment |
Which is not enough... since $1-1/n<1$ "eventually" (actually, always) as well. You need $< c$ for some constant $0leq c<1$.
– Clement C.
Nov 15 at 18:04
After the edit, fixing the solution, downvote removed. But you should really make sure your solution is correct before posting it (it's not the first time...).
– Clement C.
Nov 15 at 18:07
@ClementC. Yes of course! the inequalities do not hold in the limit!
– gimusi
Nov 15 at 18:10
Which is not enough... since $1-1/n<1$ "eventually" (actually, always) as well. You need $< c$ for some constant $0leq c<1$.
– Clement C.
Nov 15 at 18:04
Which is not enough... since $1-1/n<1$ "eventually" (actually, always) as well. You need $< c$ for some constant $0leq c<1$.
– Clement C.
Nov 15 at 18:04
After the edit, fixing the solution, downvote removed. But you should really make sure your solution is correct before posting it (it's not the first time...).
– Clement C.
Nov 15 at 18:07
After the edit, fixing the solution, downvote removed. But you should really make sure your solution is correct before posting it (it's not the first time...).
– Clement C.
Nov 15 at 18:07
@ClementC. Yes of course! the inequalities do not hold in the limit!
– gimusi
Nov 15 at 18:10
@ClementC. Yes of course! the inequalities do not hold in the limit!
– gimusi
Nov 15 at 18:10
add a comment |
up vote
0
down vote
You can use the root test to get a result quite easily. Since $$dfrac{2n+1}{3n}todfrac{2}{3}<1$$
by the ratio test
$$bigg(dfrac{2n+1}{3n}bigg)^nto0.$$
answered Nov 15 at 18:16
Melody
45211
I did not have it yet.
– matematiccc
Nov 15 at 18:18
add a comment |
up vote
0
down vote
You can use the root test to get a result quite easily. Since $$dfrac{2n+1}{3n}todfrac{2}{3}<1$$
by the ratio test
$$bigg(dfrac{2n+1}{3n}bigg)^nto0.$$
answered Nov 15 at 18:16
Melody
45211
I did not have it yet.
– matematiccc
Nov 15 at 18:18
add a comment |
up vote
0
down vote
up vote
0
down vote
You can use the root test to get a result quite easily. Since $$dfrac{2n+1}{3n}todfrac{2}{3}<1$$
by the ratio test
$$bigg(dfrac{2n+1}{3n}bigg)^nto0.$$
answered Nov 15 at 18:16
Melody
45211
You can use the root test to get a result quite easily. Since $$dfrac{2n+1}{3n}todfrac{2}{3}<1$$
by the ratio test
$$bigg(dfrac{2n+1}{3n}bigg)^nto0.$$
answered Nov 15 at 18:16
Melody
45211
answered Nov 15 at 18:16
Melody
45211
answered Nov 15 at 18:16
Melody
45211
answered Nov 15 at 18:16
Melody
45211
45211
I did not have it yet.
– matematiccc
Nov 15 at 18:18
add a comment |
I did not have it yet.
– matematiccc
Nov 15 at 18:18
I did not have it yet.
– matematiccc
Nov 15 at 18:18
I did not have it yet.
– matematiccc
Nov 15 at 18:18
add a comment |
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You never have $$frac{2n+1}{3n} leq frac{2}{3}$$ no matter how big $n$ is. Try $$frac{2n+1}{3n} leq frac{2.5}{3}$$
– Clement C.
Nov 15 at 18:03
When I changed $ frac{2}{3}$ to $ frac{2.9999}{3}$ is it ok?
– matematiccc
Nov 15 at 18:11
Yes, now it is. That is not great to change the question though, as it makes a couple answers addressing this very point obsolete.
– Clement C.
Nov 15 at 18:12
Sorry. I fixed it.
– matematiccc
Nov 15 at 18:20