Prove that if a sequence converges then $lim{x_n} = lim sup {x_n}$ or $lim{x_n} = lim inf {x_n}$
$begingroup$
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
asked Nov 26 '18 at 16:14
romanroman
2,02621222
$endgroup$
add a comment |
$begingroup$
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
asked Nov 26 '18 at 16:14
romanroman
2,02621222
$endgroup$
$begingroup$
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?
$endgroup$
– gammatester
Nov 26 '18 at 16:24
$begingroup$
@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.
$endgroup$
– roman
Nov 26 '18 at 16:34
add a comment |
$begingroup$
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
asked Nov 26 '18 at 16:14
romanroman
2,02621222
$endgroup$
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
asked Nov 26 '18 at 16:14
romanroman
2,02621222
asked Nov 26 '18 at 16:14
romanroman
2,02621222
edited Nov 26 '18 at 16:45
roman
asked Nov 26 '18 at 16:14
romanroman
2,02621222
asked Nov 26 '18 at 16:14
romanroman
2,02621222
asked Nov 26 '18 at 16:14
romanroman
2,02621222
2,02621222
$begingroup$
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?
$endgroup$
– gammatester
Nov 26 '18 at 16:24
$begingroup$
@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.
$endgroup$
– roman
Nov 26 '18 at 16:34
add a comment |
$begingroup$
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?
$endgroup$
– gammatester
Nov 26 '18 at 16:24
$begingroup$
@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.
$endgroup$
– roman
Nov 26 '18 at 16:34
$begingroup$
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?
$endgroup$
– gammatester
Nov 26 '18 at 16:24
$begingroup$
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?
$endgroup$
– gammatester
Nov 26 '18 at 16:24
$begingroup$
@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.
$endgroup$
– roman
Nov 26 '18 at 16:34
$begingroup$
@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.
$endgroup$
– roman
Nov 26 '18 at 16:34
add a comment |
1 Answer
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$begingroup$
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 '18 at 16:22
user3482749user3482749
4,057818
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add a comment |
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1 Answer
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$begingroup$
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 '18 at 16:22
user3482749user3482749
4,057818
$endgroup$
add a comment |
$begingroup$
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 '18 at 16:22
user3482749user3482749
4,057818
$endgroup$
add a comment |
$begingroup$
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 '18 at 16:22
user3482749user3482749
4,057818
$endgroup$
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 '18 at 16:22
user3482749user3482749
4,057818
answered Nov 26 '18 at 16:22
user3482749user3482749
4,057818
answered Nov 26 '18 at 16:22
user3482749user3482749
4,057818
answered Nov 26 '18 at 16:22
user3482749user3482749
4,057818
4,057818
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$begingroup$
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?
$endgroup$
– gammatester
Nov 26 '18 at 16:24
$begingroup$
@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.
$endgroup$
– roman
Nov 26 '18 at 16:34