would the proof to show that $1/(n+n^2)$ converges to 0 be the same as for $1/n$
I have started this by doing $mid1/(n+n^2) - 0mid< epsilon$ and I have fixed $epsilon > 0$ which then leads me to $n+n^2 > 1/epsilon$ however I'm not sure if I should leave it like that or make it so only 1 stays on that side. I carried on otherwise and left it like that and by the archimedian property chose a value $N$ where $N>1/epsilon.$ Then for any $n > N$ I have $1/(n+n^2) < 1/N < epsilon.$
real-analysis
edited Nov 27 '18 at 10:37
user376343
2,9132823
asked Nov 27 '18 at 10:29
Micheal smithMicheal smith
122
add a comment |
I have started this by doing $mid1/(n+n^2) - 0mid< epsilon$ and I have fixed $epsilon > 0$ which then leads me to $n+n^2 > 1/epsilon$ however I'm not sure if I should leave it like that or make it so only 1 stays on that side. I carried on otherwise and left it like that and by the archimedian property chose a value $N$ where $N>1/epsilon.$ Then for any $n > N$ I have $1/(n+n^2) < 1/N < epsilon.$
real-analysis
edited Nov 27 '18 at 10:37
user376343
2,9132823
asked Nov 27 '18 at 10:29
Micheal smithMicheal smith
122
' only 1 n stays on that side '
– Micheal smith
Nov 27 '18 at 10:30
add a comment |
I have started this by doing $mid1/(n+n^2) - 0mid< epsilon$ and I have fixed $epsilon > 0$ which then leads me to $n+n^2 > 1/epsilon$ however I'm not sure if I should leave it like that or make it so only 1 stays on that side. I carried on otherwise and left it like that and by the archimedian property chose a value $N$ where $N>1/epsilon.$ Then for any $n > N$ I have $1/(n+n^2) < 1/N < epsilon.$
real-analysis
edited Nov 27 '18 at 10:37
user376343
2,9132823
asked Nov 27 '18 at 10:29
Micheal smithMicheal smith
122
I have started this by doing $mid1/(n+n^2) - 0mid< epsilon$ and I have fixed $epsilon > 0$ which then leads me to $n+n^2 > 1/epsilon$ however I'm not sure if I should leave it like that or make it so only 1 stays on that side. I carried on otherwise and left it like that and by the archimedian property chose a value $N$ where $N>1/epsilon.$ Then for any $n > N$ I have $1/(n+n^2) < 1/N < epsilon.$
real-analysis
real-analysis
edited Nov 27 '18 at 10:37
user376343
2,9132823
asked Nov 27 '18 at 10:29
Micheal smithMicheal smith
122
edited Nov 27 '18 at 10:37
user376343
2,9132823
asked Nov 27 '18 at 10:29
Micheal smithMicheal smith
122
edited Nov 27 '18 at 10:37
user376343
2,9132823
edited Nov 27 '18 at 10:37
user376343
2,9132823
edited Nov 27 '18 at 10:37
user376343
2,9132823
2,9132823
asked Nov 27 '18 at 10:29
Micheal smithMicheal smith
122
asked Nov 27 '18 at 10:29
Micheal smithMicheal smith
122
asked Nov 27 '18 at 10:29
Micheal smithMicheal smith
122
122
' only 1 n stays on that side '
– Micheal smith
Nov 27 '18 at 10:30
add a comment |
' only 1 n stays on that side '
– Micheal smith
Nov 27 '18 at 10:30
' only 1 n stays on that side '
– Micheal smith
Nov 27 '18 at 10:30
' only 1 n stays on that side '
– Micheal smith
Nov 27 '18 at 10:30
add a comment |
2 Answers
2
active
oldest
votes
Yes, the proof that $frac1{n+n^2}$ converges can be very similar to the one that $n$ converges. In fact, it will be even simpler, since, in the original proof, you needed to find an $N$ such that $n>frac1epsilon$ for $n>N$, and in this proof, you need to find and $N'$ such that $n+n^2>frac1epsilon$ for $n>N'$. Clearly, since $n+n^2>n$, simply taking the same $N'$ as in the original proof would be sufficient, but even smaller values of $N'$ would also work (but are not required).
An alternative way of proving convergence is to use the fact that $$0leq frac{1}{n+n^2} leq frac 1n$$ and using the squeeze theorem.
answered Nov 27 '18 at 10:36
5xum5xum
89.6k393161
add a comment |
What we could use, if we don't need to use a direct approach, is that
$$frac1{n+n^2} le frac1n$$
and then invoke the squeeze theorem.
Otherwise your way seems fine, from here we can use that
$$n+n^2 >n >frac1{epsilon}$$
and then take $N>frac1{epsilon}$.
answered Nov 27 '18 at 10:32
gimusigimusi
1
im trying to use a direct approach to see is this is provable the same way if you prove 1/n , using the squeeze theorem will not be the same way as in most cases you wont use the squeeze theorem to prove 1/n converges to 0. im just wondering is it ok to leave it as n + n^2 or is that not possible ?
– Micheal smith
Nov 27 '18 at 10:38
sorry did not see the continued post you made before i typed that comment in
– Micheal smith
Nov 27 '18 at 10:38
@Michealsmith Yes indeed after editing I see better your work on that and you are almost done.
– gimusi
Nov 27 '18 at 10:40
add a comment |
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2 Answers
2
active
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2 Answers
2
active
oldest
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active
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active
oldest
votes
Yes, the proof that $frac1{n+n^2}$ converges can be very similar to the one that $n$ converges. In fact, it will be even simpler, since, in the original proof, you needed to find an $N$ such that $n>frac1epsilon$ for $n>N$, and in this proof, you need to find and $N'$ such that $n+n^2>frac1epsilon$ for $n>N'$. Clearly, since $n+n^2>n$, simply taking the same $N'$ as in the original proof would be sufficient, but even smaller values of $N'$ would also work (but are not required).
An alternative way of proving convergence is to use the fact that $$0leq frac{1}{n+n^2} leq frac 1n$$ and using the squeeze theorem.
answered Nov 27 '18 at 10:36
5xum5xum
89.6k393161
add a comment |
Yes, the proof that $frac1{n+n^2}$ converges can be very similar to the one that $n$ converges. In fact, it will be even simpler, since, in the original proof, you needed to find an $N$ such that $n>frac1epsilon$ for $n>N$, and in this proof, you need to find and $N'$ such that $n+n^2>frac1epsilon$ for $n>N'$. Clearly, since $n+n^2>n$, simply taking the same $N'$ as in the original proof would be sufficient, but even smaller values of $N'$ would also work (but are not required).
An alternative way of proving convergence is to use the fact that $$0leq frac{1}{n+n^2} leq frac 1n$$ and using the squeeze theorem.
answered Nov 27 '18 at 10:36
5xum5xum
89.6k393161
add a comment |
Yes, the proof that $frac1{n+n^2}$ converges can be very similar to the one that $n$ converges. In fact, it will be even simpler, since, in the original proof, you needed to find an $N$ such that $n>frac1epsilon$ for $n>N$, and in this proof, you need to find and $N'$ such that $n+n^2>frac1epsilon$ for $n>N'$. Clearly, since $n+n^2>n$, simply taking the same $N'$ as in the original proof would be sufficient, but even smaller values of $N'$ would also work (but are not required).
An alternative way of proving convergence is to use the fact that $$0leq frac{1}{n+n^2} leq frac 1n$$ and using the squeeze theorem.
answered Nov 27 '18 at 10:36
5xum5xum
89.6k393161
Yes, the proof that $frac1{n+n^2}$ converges can be very similar to the one that $n$ converges. In fact, it will be even simpler, since, in the original proof, you needed to find an $N$ such that $n>frac1epsilon$ for $n>N$, and in this proof, you need to find and $N'$ such that $n+n^2>frac1epsilon$ for $n>N'$. Clearly, since $n+n^2>n$, simply taking the same $N'$ as in the original proof would be sufficient, but even smaller values of $N'$ would also work (but are not required).
An alternative way of proving convergence is to use the fact that $$0leq frac{1}{n+n^2} leq frac 1n$$ and using the squeeze theorem.
answered Nov 27 '18 at 10:36
5xum5xum
89.6k393161
answered Nov 27 '18 at 10:36
5xum5xum
89.6k393161
answered Nov 27 '18 at 10:36
5xum5xum
89.6k393161
answered Nov 27 '18 at 10:36
5xum5xum
89.6k393161
89.6k393161
add a comment |
add a comment |
What we could use, if we don't need to use a direct approach, is that
$$frac1{n+n^2} le frac1n$$
and then invoke the squeeze theorem.
Otherwise your way seems fine, from here we can use that
$$n+n^2 >n >frac1{epsilon}$$
and then take $N>frac1{epsilon}$.
answered Nov 27 '18 at 10:32
gimusigimusi
1
im trying to use a direct approach to see is this is provable the same way if you prove 1/n , using the squeeze theorem will not be the same way as in most cases you wont use the squeeze theorem to prove 1/n converges to 0. im just wondering is it ok to leave it as n + n^2 or is that not possible ?
– Micheal smith
Nov 27 '18 at 10:38
sorry did not see the continued post you made before i typed that comment in
– Micheal smith
Nov 27 '18 at 10:38
@Michealsmith Yes indeed after editing I see better your work on that and you are almost done.
– gimusi
Nov 27 '18 at 10:40
add a comment |
What we could use, if we don't need to use a direct approach, is that
$$frac1{n+n^2} le frac1n$$
and then invoke the squeeze theorem.
Otherwise your way seems fine, from here we can use that
$$n+n^2 >n >frac1{epsilon}$$
and then take $N>frac1{epsilon}$.
answered Nov 27 '18 at 10:32
gimusigimusi
1
im trying to use a direct approach to see is this is provable the same way if you prove 1/n , using the squeeze theorem will not be the same way as in most cases you wont use the squeeze theorem to prove 1/n converges to 0. im just wondering is it ok to leave it as n + n^2 or is that not possible ?
– Micheal smith
Nov 27 '18 at 10:38
sorry did not see the continued post you made before i typed that comment in
– Micheal smith
Nov 27 '18 at 10:38
@Michealsmith Yes indeed after editing I see better your work on that and you are almost done.
– gimusi
Nov 27 '18 at 10:40
add a comment |
What we could use, if we don't need to use a direct approach, is that
$$frac1{n+n^2} le frac1n$$
and then invoke the squeeze theorem.
Otherwise your way seems fine, from here we can use that
$$n+n^2 >n >frac1{epsilon}$$
and then take $N>frac1{epsilon}$.
answered Nov 27 '18 at 10:32
gimusigimusi
1
What we could use, if we don't need to use a direct approach, is that
$$frac1{n+n^2} le frac1n$$
and then invoke the squeeze theorem.
Otherwise your way seems fine, from here we can use that
$$n+n^2 >n >frac1{epsilon}$$
and then take $N>frac1{epsilon}$.
answered Nov 27 '18 at 10:32
gimusigimusi
1
edited Nov 27 '18 at 10:44
answered Nov 27 '18 at 10:32
gimusigimusi
1
answered Nov 27 '18 at 10:32
gimusigimusi
1
answered Nov 27 '18 at 10:32
gimusigimusi
1
1
im trying to use a direct approach to see is this is provable the same way if you prove 1/n , using the squeeze theorem will not be the same way as in most cases you wont use the squeeze theorem to prove 1/n converges to 0. im just wondering is it ok to leave it as n + n^2 or is that not possible ?
– Micheal smith
Nov 27 '18 at 10:38
sorry did not see the continued post you made before i typed that comment in
– Micheal smith
Nov 27 '18 at 10:38
@Michealsmith Yes indeed after editing I see better your work on that and you are almost done.
– gimusi
Nov 27 '18 at 10:40
add a comment |
im trying to use a direct approach to see is this is provable the same way if you prove 1/n , using the squeeze theorem will not be the same way as in most cases you wont use the squeeze theorem to prove 1/n converges to 0. im just wondering is it ok to leave it as n + n^2 or is that not possible ?
– Micheal smith
Nov 27 '18 at 10:38
sorry did not see the continued post you made before i typed that comment in
– Micheal smith
Nov 27 '18 at 10:38
@Michealsmith Yes indeed after editing I see better your work on that and you are almost done.
– gimusi
Nov 27 '18 at 10:40
im trying to use a direct approach to see is this is provable the same way if you prove 1/n , using the squeeze theorem will not be the same way as in most cases you wont use the squeeze theorem to prove 1/n converges to 0. im just wondering is it ok to leave it as n + n^2 or is that not possible ?
– Micheal smith
Nov 27 '18 at 10:38
im trying to use a direct approach to see is this is provable the same way if you prove 1/n , using the squeeze theorem will not be the same way as in most cases you wont use the squeeze theorem to prove 1/n converges to 0. im just wondering is it ok to leave it as n + n^2 or is that not possible ?
– Micheal smith
Nov 27 '18 at 10:38
sorry did not see the continued post you made before i typed that comment in
– Micheal smith
Nov 27 '18 at 10:38
sorry did not see the continued post you made before i typed that comment in
– Micheal smith
Nov 27 '18 at 10:38
@Michealsmith Yes indeed after editing I see better your work on that and you are almost done.
– gimusi
Nov 27 '18 at 10:40
@Michealsmith Yes indeed after editing I see better your work on that and you are almost done.
– gimusi
Nov 27 '18 at 10:40
add a comment |
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' only 1 n stays on that side '
– Micheal smith
Nov 27 '18 at 10:30