Attempt at sequence proof $frac{n+3}{n^2 -3}$ converges to $0$
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Prove convergence of the following sequence: $$frac{n+3}{n^2 -3} rightarrow 0$$
Proof discussion:
Notice that since whenever $n>3$, we have $n^2 -3 >0$, we also know that $n+3 >0$, so $frac{n+3}{n^2 -3}>0$. This means we can drop the absolute value signs in:
$$ left|frac{n+3}{n^2 -3}-0right|=frac{n+3}{n^2 -3} $$
We now notice that for $n>3$ also $n^2 -9>0$ and $n^2 -3 > n^2 -9$ so $frac{1}{n^2 -3}< frac{1}{n^2 -9}$ we can thus write:
$$frac{n+3}{n^2 -3}<frac{n+3}{n^2 -9}=frac{(n+3)}{(n+3)(n-3)}=frac{1}{n-3} $$
To be able to complete this proof we want that $frac{1}{n-3}<epsilon$, we write $n-3>frac{1}{epsilon}$ or $n> frac{1}{epsilon} +3$. If we pick $n_0 =lceilfrac{1}{epsilon} +3rceil$, it will also be automatically larger than $3$. We can now write our proof:
Proof:
For all $epsilon>0$, we let $n_0=lceil{frac{1}{epsilon}+3 }rceil$ then for all $n>n_0$, we know that:
$$|a_n-0|=left|frac{n+3}{n^2-3} right|<frac{n+3}{n^2-9}=frac{1}{n-3}< frac{1}{frac{1}{epsilon}+3-3}=epsilon$$
And hence our sequence converges to $0$ $square$.
Is my proof okay?
real-analysis sequences-and-series proof-verification epsilon-delta
add a comment |
up vote
4
down vote
favorite
Prove convergence of the following sequence: $$frac{n+3}{n^2 -3} rightarrow 0$$
Proof discussion:
Notice that since whenever $n>3$, we have $n^2 -3 >0$, we also know that $n+3 >0$, so $frac{n+3}{n^2 -3}>0$. This means we can drop the absolute value signs in:
$$ left|frac{n+3}{n^2 -3}-0right|=frac{n+3}{n^2 -3} $$
We now notice that for $n>3$ also $n^2 -9>0$ and $n^2 -3 > n^2 -9$ so $frac{1}{n^2 -3}< frac{1}{n^2 -9}$ we can thus write:
$$frac{n+3}{n^2 -3}<frac{n+3}{n^2 -9}=frac{(n+3)}{(n+3)(n-3)}=frac{1}{n-3} $$
To be able to complete this proof we want that $frac{1}{n-3}<epsilon$, we write $n-3>frac{1}{epsilon}$ or $n> frac{1}{epsilon} +3$. If we pick $n_0 =lceilfrac{1}{epsilon} +3rceil$, it will also be automatically larger than $3$. We can now write our proof:
Proof:
For all $epsilon>0$, we let $n_0=lceil{frac{1}{epsilon}+3 }rceil$ then for all $n>n_0$, we know that:
$$|a_n-0|=left|frac{n+3}{n^2-3} right|<frac{n+3}{n^2-9}=frac{1}{n-3}< frac{1}{frac{1}{epsilon}+3-3}=epsilon$$
And hence our sequence converges to $0$ $square$.
Is my proof okay?
real-analysis sequences-and-series proof-verification epsilon-delta
1
I think you should make explicit that since $epsilon>0$, then $n_0geq 4$, so the denominator in $leftlvertfrac{n+3}{n^2-3}rightrvert$ is positive, so you can remove the absolute value sign. Otherwise, I think the proof is perfect.
– YiFan
Nov 16 at 7:16
1
Yes, perfect! Just another small mistake: twice you wrote $epsilon$ instead of $1/epsilon$. I guess it was just a type since you continued correctly.
– maxmilgram
Nov 16 at 7:17
I do make the case that the denominator is positive, It's in the first line.
– WesleyGroupshaveFeelingsToo
Nov 16 at 17:02
Maybe I should say it explicitly.
– WesleyGroupshaveFeelingsToo
Nov 16 at 17:03
add a comment |
up vote
4
down vote
favorite
up vote
4
down vote
favorite
Prove convergence of the following sequence: $$frac{n+3}{n^2 -3} rightarrow 0$$
Proof discussion:
Notice that since whenever $n>3$, we have $n^2 -3 >0$, we also know that $n+3 >0$, so $frac{n+3}{n^2 -3}>0$. This means we can drop the absolute value signs in:
$$ left|frac{n+3}{n^2 -3}-0right|=frac{n+3}{n^2 -3} $$
We now notice that for $n>3$ also $n^2 -9>0$ and $n^2 -3 > n^2 -9$ so $frac{1}{n^2 -3}< frac{1}{n^2 -9}$ we can thus write:
$$frac{n+3}{n^2 -3}<frac{n+3}{n^2 -9}=frac{(n+3)}{(n+3)(n-3)}=frac{1}{n-3} $$
To be able to complete this proof we want that $frac{1}{n-3}<epsilon$, we write $n-3>frac{1}{epsilon}$ or $n> frac{1}{epsilon} +3$. If we pick $n_0 =lceilfrac{1}{epsilon} +3rceil$, it will also be automatically larger than $3$. We can now write our proof:
Proof:
For all $epsilon>0$, we let $n_0=lceil{frac{1}{epsilon}+3 }rceil$ then for all $n>n_0$, we know that:
$$|a_n-0|=left|frac{n+3}{n^2-3} right|<frac{n+3}{n^2-9}=frac{1}{n-3}< frac{1}{frac{1}{epsilon}+3-3}=epsilon$$
And hence our sequence converges to $0$ $square$.
Is my proof okay?
real-analysis sequences-and-series proof-verification epsilon-delta
Prove convergence of the following sequence: $$frac{n+3}{n^2 -3} rightarrow 0$$
Proof discussion:
Notice that since whenever $n>3$, we have $n^2 -3 >0$, we also know that $n+3 >0$, so $frac{n+3}{n^2 -3}>0$. This means we can drop the absolute value signs in:
$$ left|frac{n+3}{n^2 -3}-0right|=frac{n+3}{n^2 -3} $$
We now notice that for $n>3$ also $n^2 -9>0$ and $n^2 -3 > n^2 -9$ so $frac{1}{n^2 -3}< frac{1}{n^2 -9}$ we can thus write:
$$frac{n+3}{n^2 -3}<frac{n+3}{n^2 -9}=frac{(n+3)}{(n+3)(n-3)}=frac{1}{n-3} $$
To be able to complete this proof we want that $frac{1}{n-3}<epsilon$, we write $n-3>frac{1}{epsilon}$ or $n> frac{1}{epsilon} +3$. If we pick $n_0 =lceilfrac{1}{epsilon} +3rceil$, it will also be automatically larger than $3$. We can now write our proof:
Proof:
For all $epsilon>0$, we let $n_0=lceil{frac{1}{epsilon}+3 }rceil$ then for all $n>n_0$, we know that:
$$|a_n-0|=left|frac{n+3}{n^2-3} right|<frac{n+3}{n^2-9}=frac{1}{n-3}< frac{1}{frac{1}{epsilon}+3-3}=epsilon$$
And hence our sequence converges to $0$ $square$.
Is my proof okay?
real-analysis sequences-and-series proof-verification epsilon-delta
real-analysis sequences-and-series proof-verification epsilon-delta
edited Nov 16 at 17:22
asked Nov 16 at 7:11
WesleyGroupshaveFeelingsToo
1,064321
1,064321
1
I think you should make explicit that since $epsilon>0$, then $n_0geq 4$, so the denominator in $leftlvertfrac{n+3}{n^2-3}rightrvert$ is positive, so you can remove the absolute value sign. Otherwise, I think the proof is perfect.
– YiFan
Nov 16 at 7:16
1
Yes, perfect! Just another small mistake: twice you wrote $epsilon$ instead of $1/epsilon$. I guess it was just a type since you continued correctly.
– maxmilgram
Nov 16 at 7:17
I do make the case that the denominator is positive, It's in the first line.
– WesleyGroupshaveFeelingsToo
Nov 16 at 17:02
Maybe I should say it explicitly.
– WesleyGroupshaveFeelingsToo
Nov 16 at 17:03
add a comment |
1
I think you should make explicit that since $epsilon>0$, then $n_0geq 4$, so the denominator in $leftlvertfrac{n+3}{n^2-3}rightrvert$ is positive, so you can remove the absolute value sign. Otherwise, I think the proof is perfect.
– YiFan
Nov 16 at 7:16
1
Yes, perfect! Just another small mistake: twice you wrote $epsilon$ instead of $1/epsilon$. I guess it was just a type since you continued correctly.
– maxmilgram
Nov 16 at 7:17
I do make the case that the denominator is positive, It's in the first line.
– WesleyGroupshaveFeelingsToo
Nov 16 at 17:02
Maybe I should say it explicitly.
– WesleyGroupshaveFeelingsToo
Nov 16 at 17:03
1
1
I think you should make explicit that since $epsilon>0$, then $n_0geq 4$, so the denominator in $leftlvertfrac{n+3}{n^2-3}rightrvert$ is positive, so you can remove the absolute value sign. Otherwise, I think the proof is perfect.
– YiFan
Nov 16 at 7:16
I think you should make explicit that since $epsilon>0$, then $n_0geq 4$, so the denominator in $leftlvertfrac{n+3}{n^2-3}rightrvert$ is positive, so you can remove the absolute value sign. Otherwise, I think the proof is perfect.
– YiFan
Nov 16 at 7:16
1
1
Yes, perfect! Just another small mistake: twice you wrote $epsilon$ instead of $1/epsilon$. I guess it was just a type since you continued correctly.
– maxmilgram
Nov 16 at 7:17
Yes, perfect! Just another small mistake: twice you wrote $epsilon$ instead of $1/epsilon$. I guess it was just a type since you continued correctly.
– maxmilgram
Nov 16 at 7:17
I do make the case that the denominator is positive, It's in the first line.
– WesleyGroupshaveFeelingsToo
Nov 16 at 17:02
I do make the case that the denominator is positive, It's in the first line.
– WesleyGroupshaveFeelingsToo
Nov 16 at 17:02
Maybe I should say it explicitly.
– WesleyGroupshaveFeelingsToo
Nov 16 at 17:03
Maybe I should say it explicitly.
– WesleyGroupshaveFeelingsToo
Nov 16 at 17:03
add a comment |
2 Answers
2
active
oldest
votes
up vote
2
down vote
accepted
Yup, the proof is correct.
Also, "we write $n-3>color{blue}{frac1{epsilon}}$"
Why is this an inequality and not an equality? we can just cancel the $n+3$ right?
– WesleyGroupshaveFeelingsToo
Nov 16 at 16:58
1
oops, you are right. I made a mistake.
– Siong Thye Goh
Nov 16 at 17:01
Thank you. I did do a typo :) you were right about that
– WesleyGroupshaveFeelingsToo
Nov 16 at 17:03
1
huzzah! I've done it :D
– WesleyGroupshaveFeelingsToo
Nov 16 at 17:14
add a comment |
up vote
2
down vote
Alternatively, decompose
$$frac{n+3}{n^2-3}=frac1{n-sqrt3}+(3-sqrt3)frac1{n^2-3}$$
and the two terms are of the form $dfrac1infty$.
That could work later on in the course, right now I am not yet allowed to use this.
– WesleyGroupshaveFeelingsToo
Nov 16 at 17:41
I realised this is okay since sequences map to $mathbb{R}$, so you can write it in this manner.
– WesleyGroupshaveFeelingsToo
Nov 16 at 17:42
@WesleyGroupshaveFeelingsToo: isn't $epsilon$ in $mathbb R$ ?
– Yves Daoust
Nov 16 at 17:43
Yes, of course :)
– WesleyGroupshaveFeelingsToo
Nov 16 at 17:48
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Yup, the proof is correct.
Also, "we write $n-3>color{blue}{frac1{epsilon}}$"
Why is this an inequality and not an equality? we can just cancel the $n+3$ right?
– WesleyGroupshaveFeelingsToo
Nov 16 at 16:58
1
oops, you are right. I made a mistake.
– Siong Thye Goh
Nov 16 at 17:01
Thank you. I did do a typo :) you were right about that
– WesleyGroupshaveFeelingsToo
Nov 16 at 17:03
1
huzzah! I've done it :D
– WesleyGroupshaveFeelingsToo
Nov 16 at 17:14
add a comment |
up vote
2
down vote
accepted
Yup, the proof is correct.
Also, "we write $n-3>color{blue}{frac1{epsilon}}$"
Why is this an inequality and not an equality? we can just cancel the $n+3$ right?
– WesleyGroupshaveFeelingsToo
Nov 16 at 16:58
1
oops, you are right. I made a mistake.
– Siong Thye Goh
Nov 16 at 17:01
Thank you. I did do a typo :) you were right about that
– WesleyGroupshaveFeelingsToo
Nov 16 at 17:03
1
huzzah! I've done it :D
– WesleyGroupshaveFeelingsToo
Nov 16 at 17:14
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Yup, the proof is correct.
Also, "we write $n-3>color{blue}{frac1{epsilon}}$"
Yup, the proof is correct.
Also, "we write $n-3>color{blue}{frac1{epsilon}}$"
edited Nov 16 at 17:02
answered Nov 16 at 7:16
Siong Thye Goh
94.9k1462115
94.9k1462115
Why is this an inequality and not an equality? we can just cancel the $n+3$ right?
– WesleyGroupshaveFeelingsToo
Nov 16 at 16:58
1
oops, you are right. I made a mistake.
– Siong Thye Goh
Nov 16 at 17:01
Thank you. I did do a typo :) you were right about that
– WesleyGroupshaveFeelingsToo
Nov 16 at 17:03
1
huzzah! I've done it :D
– WesleyGroupshaveFeelingsToo
Nov 16 at 17:14
add a comment |
Why is this an inequality and not an equality? we can just cancel the $n+3$ right?
– WesleyGroupshaveFeelingsToo
Nov 16 at 16:58
1
oops, you are right. I made a mistake.
– Siong Thye Goh
Nov 16 at 17:01
Thank you. I did do a typo :) you were right about that
– WesleyGroupshaveFeelingsToo
Nov 16 at 17:03
1
huzzah! I've done it :D
– WesleyGroupshaveFeelingsToo
Nov 16 at 17:14
Why is this an inequality and not an equality? we can just cancel the $n+3$ right?
– WesleyGroupshaveFeelingsToo
Nov 16 at 16:58
Why is this an inequality and not an equality? we can just cancel the $n+3$ right?
– WesleyGroupshaveFeelingsToo
Nov 16 at 16:58
1
1
oops, you are right. I made a mistake.
– Siong Thye Goh
Nov 16 at 17:01
oops, you are right. I made a mistake.
– Siong Thye Goh
Nov 16 at 17:01
Thank you. I did do a typo :) you were right about that
– WesleyGroupshaveFeelingsToo
Nov 16 at 17:03
Thank you. I did do a typo :) you were right about that
– WesleyGroupshaveFeelingsToo
Nov 16 at 17:03
1
1
huzzah! I've done it :D
– WesleyGroupshaveFeelingsToo
Nov 16 at 17:14
huzzah! I've done it :D
– WesleyGroupshaveFeelingsToo
Nov 16 at 17:14
add a comment |
up vote
2
down vote
Alternatively, decompose
$$frac{n+3}{n^2-3}=frac1{n-sqrt3}+(3-sqrt3)frac1{n^2-3}$$
and the two terms are of the form $dfrac1infty$.
That could work later on in the course, right now I am not yet allowed to use this.
– WesleyGroupshaveFeelingsToo
Nov 16 at 17:41
I realised this is okay since sequences map to $mathbb{R}$, so you can write it in this manner.
– WesleyGroupshaveFeelingsToo
Nov 16 at 17:42
@WesleyGroupshaveFeelingsToo: isn't $epsilon$ in $mathbb R$ ?
– Yves Daoust
Nov 16 at 17:43
Yes, of course :)
– WesleyGroupshaveFeelingsToo
Nov 16 at 17:48
add a comment |
up vote
2
down vote
Alternatively, decompose
$$frac{n+3}{n^2-3}=frac1{n-sqrt3}+(3-sqrt3)frac1{n^2-3}$$
and the two terms are of the form $dfrac1infty$.
That could work later on in the course, right now I am not yet allowed to use this.
– WesleyGroupshaveFeelingsToo
Nov 16 at 17:41
I realised this is okay since sequences map to $mathbb{R}$, so you can write it in this manner.
– WesleyGroupshaveFeelingsToo
Nov 16 at 17:42
@WesleyGroupshaveFeelingsToo: isn't $epsilon$ in $mathbb R$ ?
– Yves Daoust
Nov 16 at 17:43
Yes, of course :)
– WesleyGroupshaveFeelingsToo
Nov 16 at 17:48
add a comment |
up vote
2
down vote
up vote
2
down vote
Alternatively, decompose
$$frac{n+3}{n^2-3}=frac1{n-sqrt3}+(3-sqrt3)frac1{n^2-3}$$
and the two terms are of the form $dfrac1infty$.
Alternatively, decompose
$$frac{n+3}{n^2-3}=frac1{n-sqrt3}+(3-sqrt3)frac1{n^2-3}$$
and the two terms are of the form $dfrac1infty$.
answered Nov 16 at 17:28
Yves Daoust
122k668217
122k668217
That could work later on in the course, right now I am not yet allowed to use this.
– WesleyGroupshaveFeelingsToo
Nov 16 at 17:41
I realised this is okay since sequences map to $mathbb{R}$, so you can write it in this manner.
– WesleyGroupshaveFeelingsToo
Nov 16 at 17:42
@WesleyGroupshaveFeelingsToo: isn't $epsilon$ in $mathbb R$ ?
– Yves Daoust
Nov 16 at 17:43
Yes, of course :)
– WesleyGroupshaveFeelingsToo
Nov 16 at 17:48
add a comment |
That could work later on in the course, right now I am not yet allowed to use this.
– WesleyGroupshaveFeelingsToo
Nov 16 at 17:41
I realised this is okay since sequences map to $mathbb{R}$, so you can write it in this manner.
– WesleyGroupshaveFeelingsToo
Nov 16 at 17:42
@WesleyGroupshaveFeelingsToo: isn't $epsilon$ in $mathbb R$ ?
– Yves Daoust
Nov 16 at 17:43
Yes, of course :)
– WesleyGroupshaveFeelingsToo
Nov 16 at 17:48
That could work later on in the course, right now I am not yet allowed to use this.
– WesleyGroupshaveFeelingsToo
Nov 16 at 17:41
That could work later on in the course, right now I am not yet allowed to use this.
– WesleyGroupshaveFeelingsToo
Nov 16 at 17:41
I realised this is okay since sequences map to $mathbb{R}$, so you can write it in this manner.
– WesleyGroupshaveFeelingsToo
Nov 16 at 17:42
I realised this is okay since sequences map to $mathbb{R}$, so you can write it in this manner.
– WesleyGroupshaveFeelingsToo
Nov 16 at 17:42
@WesleyGroupshaveFeelingsToo: isn't $epsilon$ in $mathbb R$ ?
– Yves Daoust
Nov 16 at 17:43
@WesleyGroupshaveFeelingsToo: isn't $epsilon$ in $mathbb R$ ?
– Yves Daoust
Nov 16 at 17:43
Yes, of course :)
– WesleyGroupshaveFeelingsToo
Nov 16 at 17:48
Yes, of course :)
– WesleyGroupshaveFeelingsToo
Nov 16 at 17:48
add a comment |
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1
I think you should make explicit that since $epsilon>0$, then $n_0geq 4$, so the denominator in $leftlvertfrac{n+3}{n^2-3}rightrvert$ is positive, so you can remove the absolute value sign. Otherwise, I think the proof is perfect.
– YiFan
Nov 16 at 7:16
1
Yes, perfect! Just another small mistake: twice you wrote $epsilon$ instead of $1/epsilon$. I guess it was just a type since you continued correctly.
– maxmilgram
Nov 16 at 7:17
I do make the case that the denominator is positive, It's in the first line.
– WesleyGroupshaveFeelingsToo
Nov 16 at 17:02
Maybe I should say it explicitly.
– WesleyGroupshaveFeelingsToo
Nov 16 at 17:03