Prove that if a sequence converges then $lim{x_n} = lim sup {x_n}$ or $lim{x_n} = lim inf {x_n}$
up vote
0
down vote
favorite
Given a convergent sequence ${x_n}$ prove that either:
$$
lim_{n toinfty}{x_n} = lim_{n to infty} sup {x_n}
$$
or
$$
lim_{n toinfty}{x_n} = lim_{n to infty} inf {x_n}
$$
I believe this problem has been solved several times here, but i couldn't find such a question (probably due to translation issues, since the original problem is in another other).
I've started with gathering what is given in the problem statement. So we have that a sequence is convergent, thus:
$$
lim_{ntoinfty}x_n = L iff { forallvarepsilon >0, exists Nin mathbb N:forall n> N implies |x_n-L|<varepsilon }
$$
Also we have that the sequence is bounded, so:
$$
m = inf{x_n} le x_nle sup{x_n} = M \
m le x_n le M
$$
Now using these facts I believe I should make some assumption (for example that $x_n$ doesn't reach any bound and proceed by contradiction), but i can't wrap my mind for several hours already.
I would appreciate if someone could show me how to prove this or point to an already answered question.
calculus limits epsilon-delta upper-lower-bounds
add a comment |
up vote
0
down vote
favorite
Given a convergent sequence ${x_n}$ prove that either:
$$
lim_{n toinfty}{x_n} = lim_{n to infty} sup {x_n}
$$
or
$$
lim_{n toinfty}{x_n} = lim_{n to infty} inf {x_n}
$$
I believe this problem has been solved several times here, but i couldn't find such a question (probably due to translation issues, since the original problem is in another other).
I've started with gathering what is given in the problem statement. So we have that a sequence is convergent, thus:
$$
lim_{ntoinfty}x_n = L iff { forallvarepsilon >0, exists Nin mathbb N:forall n> N implies |x_n-L|<varepsilon }
$$
Also we have that the sequence is bounded, so:
$$
m = inf{x_n} le x_nle sup{x_n} = M \
m le x_n le M
$$
Now using these facts I believe I should make some assumption (for example that $x_n$ doesn't reach any bound and proceed by contradiction), but i can't wrap my mind for several hours already.
I would appreciate if someone could show me how to prove this or point to an already answered question.
calculus limits epsilon-delta upper-lower-bounds
What exactly do you mean with $inf, sup$ and reaches? It is certainly wrong with the standard definition en.wikipedia.org/wiki/Infimum_and_supremum. Do you mean en.wikipedia.org/wiki/Limit_superior_and_limit_inferior?
– gammatester
Nov 26 at 16:24
@gammatester in case of this question "the sequence reaches an exact bound" would mean $lim_{nto infty}x_n = limsup{x_n}$ or $lim_{nto infty}x_n = liminf{x_n}$. I meant limit superior/inferior.
– roman
Nov 26 at 16:34
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Given a convergent sequence ${x_n}$ prove that either:
$$
lim_{n toinfty}{x_n} = lim_{n to infty} sup {x_n}
$$
or
$$
lim_{n toinfty}{x_n} = lim_{n to infty} inf {x_n}
$$
I believe this problem has been solved several times here, but i couldn't find such a question (probably due to translation issues, since the original problem is in another other).
I've started with gathering what is given in the problem statement. So we have that a sequence is convergent, thus:
$$
lim_{ntoinfty}x_n = L iff { forallvarepsilon >0, exists Nin mathbb N:forall n> N implies |x_n-L|<varepsilon }
$$
Also we have that the sequence is bounded, so:
$$
m = inf{x_n} le x_nle sup{x_n} = M \
m le x_n le M
$$
Now using these facts I believe I should make some assumption (for example that $x_n$ doesn't reach any bound and proceed by contradiction), but i can't wrap my mind for several hours already.
I would appreciate if someone could show me how to prove this or point to an already answered question.
calculus limits epsilon-delta upper-lower-bounds
Given a convergent sequence ${x_n}$ prove that either:
$$
lim_{n toinfty}{x_n} = lim_{n to infty} sup {x_n}
$$
or
$$
lim_{n toinfty}{x_n} = lim_{n to infty} inf {x_n}
$$
I believe this problem has been solved several times here, but i couldn't find such a question (probably due to translation issues, since the original problem is in another other).
I've started with gathering what is given in the problem statement. So we have that a sequence is convergent, thus:
$$
lim_{ntoinfty}x_n = L iff { forallvarepsilon >0, exists Nin mathbb N:forall n> N implies |x_n-L|<varepsilon }
$$
Also we have that the sequence is bounded, so:
$$
m = inf{x_n} le x_nle sup{x_n} = M \
m le x_n le M
$$
Now using these facts I believe I should make some assumption (for example that $x_n$ doesn't reach any bound and proceed by contradiction), but i can't wrap my mind for several hours already.
I would appreciate if someone could show me how to prove this or point to an already answered question.
calculus limits epsilon-delta upper-lower-bounds
calculus limits epsilon-delta upper-lower-bounds
edited Nov 26 at 16:45
asked Nov 26 at 16:14
roman
1,25711017
1,25711017
What exactly do you mean with $inf, sup$ and reaches? It is certainly wrong with the standard definition en.wikipedia.org/wiki/Infimum_and_supremum. Do you mean en.wikipedia.org/wiki/Limit_superior_and_limit_inferior?
– gammatester
Nov 26 at 16:24
@gammatester in case of this question "the sequence reaches an exact bound" would mean $lim_{nto infty}x_n = limsup{x_n}$ or $lim_{nto infty}x_n = liminf{x_n}$. I meant limit superior/inferior.
– roman
Nov 26 at 16:34
add a comment |
What exactly do you mean with $inf, sup$ and reaches? It is certainly wrong with the standard definition en.wikipedia.org/wiki/Infimum_and_supremum. Do you mean en.wikipedia.org/wiki/Limit_superior_and_limit_inferior?
– gammatester
Nov 26 at 16:24
@gammatester in case of this question "the sequence reaches an exact bound" would mean $lim_{nto infty}x_n = limsup{x_n}$ or $lim_{nto infty}x_n = liminf{x_n}$. I meant limit superior/inferior.
– roman
Nov 26 at 16:34
What exactly do you mean with $inf, sup$ and reaches? It is certainly wrong with the standard definition en.wikipedia.org/wiki/Infimum_and_supremum. Do you mean en.wikipedia.org/wiki/Limit_superior_and_limit_inferior?
– gammatester
Nov 26 at 16:24
What exactly do you mean with $inf, sup$ and reaches? It is certainly wrong with the standard definition en.wikipedia.org/wiki/Infimum_and_supremum. Do you mean en.wikipedia.org/wiki/Limit_superior_and_limit_inferior?
– gammatester
Nov 26 at 16:24
@gammatester in case of this question "the sequence reaches an exact bound" would mean $lim_{nto infty}x_n = limsup{x_n}$ or $lim_{nto infty}x_n = liminf{x_n}$. I meant limit superior/inferior.
– roman
Nov 26 at 16:34
@gammatester in case of this question "the sequence reaches an exact bound" would mean $lim_{nto infty}x_n = limsup{x_n}$ or $lim_{nto infty}x_n = liminf{x_n}$. I meant limit superior/inferior.
– roman
Nov 26 at 16:34
add a comment |
1 Answer
1
active
oldest
votes
up vote
0
down vote
If $m = M$, then the sequence is constant, so the result holds. If not, then either $m$ or $M$ (maybe both, but it doesn't matter: pick either in that case) is not equal to $L$. Whichever it is (call that one $k$), there is some $N$ such that for all $n > N$, $|x_n - L| < frac{|L-k|}{2}$. Since $k$ is an exact bound for $(x_n)$, there must, for any $delta > 0$ be some $n$ such that $|k - x_n| < delta$. But for any $delta < frac{|L-k|}{2}$, this can't happen after the $N$th term, so must be in the first $N$ somewhere, so $k$ is an exact bound for the set of the first $N$ terms of $(x_n)$. But there are finitely many such, and every finite set achieves its exact bounds, so in particular, there is some $n < N$ such that $x_n = k$.
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
If $m = M$, then the sequence is constant, so the result holds. If not, then either $m$ or $M$ (maybe both, but it doesn't matter: pick either in that case) is not equal to $L$. Whichever it is (call that one $k$), there is some $N$ such that for all $n > N$, $|x_n - L| < frac{|L-k|}{2}$. Since $k$ is an exact bound for $(x_n)$, there must, for any $delta > 0$ be some $n$ such that $|k - x_n| < delta$. But for any $delta < frac{|L-k|}{2}$, this can't happen after the $N$th term, so must be in the first $N$ somewhere, so $k$ is an exact bound for the set of the first $N$ terms of $(x_n)$. But there are finitely many such, and every finite set achieves its exact bounds, so in particular, there is some $n < N$ such that $x_n = k$.
add a comment |
up vote
0
down vote
If $m = M$, then the sequence is constant, so the result holds. If not, then either $m$ or $M$ (maybe both, but it doesn't matter: pick either in that case) is not equal to $L$. Whichever it is (call that one $k$), there is some $N$ such that for all $n > N$, $|x_n - L| < frac{|L-k|}{2}$. Since $k$ is an exact bound for $(x_n)$, there must, for any $delta > 0$ be some $n$ such that $|k - x_n| < delta$. But for any $delta < frac{|L-k|}{2}$, this can't happen after the $N$th term, so must be in the first $N$ somewhere, so $k$ is an exact bound for the set of the first $N$ terms of $(x_n)$. But there are finitely many such, and every finite set achieves its exact bounds, so in particular, there is some $n < N$ such that $x_n = k$.
add a comment |
up vote
0
down vote
up vote
0
down vote
If $m = M$, then the sequence is constant, so the result holds. If not, then either $m$ or $M$ (maybe both, but it doesn't matter: pick either in that case) is not equal to $L$. Whichever it is (call that one $k$), there is some $N$ such that for all $n > N$, $|x_n - L| < frac{|L-k|}{2}$. Since $k$ is an exact bound for $(x_n)$, there must, for any $delta > 0$ be some $n$ such that $|k - x_n| < delta$. But for any $delta < frac{|L-k|}{2}$, this can't happen after the $N$th term, so must be in the first $N$ somewhere, so $k$ is an exact bound for the set of the first $N$ terms of $(x_n)$. But there are finitely many such, and every finite set achieves its exact bounds, so in particular, there is some $n < N$ such that $x_n = k$.
If $m = M$, then the sequence is constant, so the result holds. If not, then either $m$ or $M$ (maybe both, but it doesn't matter: pick either in that case) is not equal to $L$. Whichever it is (call that one $k$), there is some $N$ such that for all $n > N$, $|x_n - L| < frac{|L-k|}{2}$. Since $k$ is an exact bound for $(x_n)$, there must, for any $delta > 0$ be some $n$ such that $|k - x_n| < delta$. But for any $delta < frac{|L-k|}{2}$, this can't happen after the $N$th term, so must be in the first $N$ somewhere, so $k$ is an exact bound for the set of the first $N$ terms of $(x_n)$. But there are finitely many such, and every finite set achieves its exact bounds, so in particular, there is some $n < N$ such that $x_n = k$.
answered Nov 26 at 16:22
user3482749
2,086414
2,086414
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3014519%2fprove-that-if-a-sequence-converges-then-limx-n-lim-sup-x-n-or-limx%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
What exactly do you mean with $inf, sup$ and reaches? It is certainly wrong with the standard definition en.wikipedia.org/wiki/Infimum_and_supremum. Do you mean en.wikipedia.org/wiki/Limit_superior_and_limit_inferior?
– gammatester
Nov 26 at 16:24
@gammatester in case of this question "the sequence reaches an exact bound" would mean $lim_{nto infty}x_n = limsup{x_n}$ or $lim_{nto infty}x_n = liminf{x_n}$. I meant limit superior/inferior.
– roman
Nov 26 at 16:34