Each sequence that is unbounded above / below has a subsequence that converges (improperly) to $+ infty /...
To Prove: Each sequence that is unbounded above / below has a subsequence
that converges (improperly) to $+ infty / −infty$. Conclude that each sequence
of real numbers has a properly or improperly convergent subsequence.
Whenever a sequence is unbounded, we have that it has an increasing/decreasing subsequence. Why is this the case? For any number $M$ you give me, there exists an $min mathbb{N}$ such that $a_m$>M. We choose the subsequence ${a_i, a_j, a_k dots }$ such that the index set is strictly increasing/decreasing and also the terms are strictly increasing/decreasing. Does this argument work? how can I formalise this better?
An example of my method applied would be the following sequence: ${0,1,2,1,0,3,4,5,4,3,6 dots dots }$ My method would pick the following subsequence ${0, 1,2,3,4,5,6 dots}$. This is the sequence $a_l=l$ where $l$ is now a new index.
Whenever a sequence is bounded it has a convergent subsequence by Bolzano Weierstrass (it has an accumulation point, hence the subsequence converges to this accumulation point.)
real-analysis sequences-and-series
asked Nov 25 at 6:06
Wesley Strik
1,534422
add a comment |
To Prove: Each sequence that is unbounded above / below has a subsequence
that converges (improperly) to $+ infty / −infty$. Conclude that each sequence
of real numbers has a properly or improperly convergent subsequence.
Whenever a sequence is unbounded, we have that it has an increasing/decreasing subsequence. Why is this the case? For any number $M$ you give me, there exists an $min mathbb{N}$ such that $a_m$>M. We choose the subsequence ${a_i, a_j, a_k dots }$ such that the index set is strictly increasing/decreasing and also the terms are strictly increasing/decreasing. Does this argument work? how can I formalise this better?
An example of my method applied would be the following sequence: ${0,1,2,1,0,3,4,5,4,3,6 dots dots }$ My method would pick the following subsequence ${0, 1,2,3,4,5,6 dots}$. This is the sequence $a_l=l$ where $l$ is now a new index.
Whenever a sequence is bounded it has a convergent subsequence by Bolzano Weierstrass (it has an accumulation point, hence the subsequence converges to this accumulation point.)
real-analysis sequences-and-series
asked Nov 25 at 6:06
Wesley Strik
1,534422
1
Simply build recursively a subsequence $(a_{n_j})$ such that $a_{n_0}=a_0$, say, and, for every $j$, $$a_{n_{j+1}}geqslant a_{n_j}+1$$
– Did
Nov 25 at 6:15
That's a nice way to put it.
– Wesley Strik
Nov 25 at 6:18
add a comment |
To Prove: Each sequence that is unbounded above / below has a subsequence
that converges (improperly) to $+ infty / −infty$. Conclude that each sequence
of real numbers has a properly or improperly convergent subsequence.
Whenever a sequence is unbounded, we have that it has an increasing/decreasing subsequence. Why is this the case? For any number $M$ you give me, there exists an $min mathbb{N}$ such that $a_m$>M. We choose the subsequence ${a_i, a_j, a_k dots }$ such that the index set is strictly increasing/decreasing and also the terms are strictly increasing/decreasing. Does this argument work? how can I formalise this better?
An example of my method applied would be the following sequence: ${0,1,2,1,0,3,4,5,4,3,6 dots dots }$ My method would pick the following subsequence ${0, 1,2,3,4,5,6 dots}$. This is the sequence $a_l=l$ where $l$ is now a new index.
Whenever a sequence is bounded it has a convergent subsequence by Bolzano Weierstrass (it has an accumulation point, hence the subsequence converges to this accumulation point.)
real-analysis sequences-and-series
asked Nov 25 at 6:06
Wesley Strik
1,534422
To Prove: Each sequence that is unbounded above / below has a subsequence
that converges (improperly) to $+ infty / −infty$. Conclude that each sequence
of real numbers has a properly or improperly convergent subsequence.
Whenever a sequence is unbounded, we have that it has an increasing/decreasing subsequence. Why is this the case? For any number $M$ you give me, there exists an $min mathbb{N}$ such that $a_m$>M. We choose the subsequence ${a_i, a_j, a_k dots }$ such that the index set is strictly increasing/decreasing and also the terms are strictly increasing/decreasing. Does this argument work? how can I formalise this better?
An example of my method applied would be the following sequence: ${0,1,2,1,0,3,4,5,4,3,6 dots dots }$ My method would pick the following subsequence ${0, 1,2,3,4,5,6 dots}$. This is the sequence $a_l=l$ where $l$ is now a new index.
Whenever a sequence is bounded it has a convergent subsequence by Bolzano Weierstrass (it has an accumulation point, hence the subsequence converges to this accumulation point.)
real-analysis sequences-and-series
real-analysis sequences-and-series
asked Nov 25 at 6:06
Wesley Strik
1,534422
asked Nov 25 at 6:06
Wesley Strik
1,534422
asked Nov 25 at 6:06
Wesley Strik
1,534422
asked Nov 25 at 6:06
Wesley Strik
1,534422
asked Nov 25 at 6:06
Wesley Strik
1,534422
1,534422
1
Simply build recursively a subsequence $(a_{n_j})$ such that $a_{n_0}=a_0$, say, and, for every $j$, $$a_{n_{j+1}}geqslant a_{n_j}+1$$
– Did
Nov 25 at 6:15
That's a nice way to put it.
– Wesley Strik
Nov 25 at 6:18
add a comment |
1
Simply build recursively a subsequence $(a_{n_j})$ such that $a_{n_0}=a_0$, say, and, for every $j$, $$a_{n_{j+1}}geqslant a_{n_j}+1$$
– Did
Nov 25 at 6:15
That's a nice way to put it.
– Wesley Strik
Nov 25 at 6:18
1
1
Simply build recursively a subsequence $(a_{n_j})$ such that $a_{n_0}=a_0$, say, and, for every $j$, $$a_{n_{j+1}}geqslant a_{n_j}+1$$
– Did
Nov 25 at 6:15
Simply build recursively a subsequence $(a_{n_j})$ such that $a_{n_0}=a_0$, say, and, for every $j$, $$a_{n_{j+1}}geqslant a_{n_j}+1$$
– Did
Nov 25 at 6:15
That's a nice way to put it.
– Wesley Strik
Nov 25 at 6:18
That's a nice way to put it.
– Wesley Strik
Nov 25 at 6:18
add a comment |
1 Answer
1
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We build a recursive subsequence defined by:
$$a_{n_0}=a_0 land n_{j+1} > n_j land a_{n_{j+1}} geq a_{n_j}+1 $$
We can do so because the set is unbounded, so we can always handpick the next term that is greater than our previous one, because there should never exist a bound. Suppose we would not be able to pick a greater term, we would get a contradiction since $a_{n_j}+1$ would be a bound for the entire sequence, which was supposed to be unbounded.
We also notice that the index set is strictly increasing and that the subsequence itself is increasing, the question is, is it increasing fast enough?
Notice that we take some arbitrary term for the sequence and handpick the next one that is at least greater than the current one plus $1$. This looks very familiar:
$$ a_{n+1} = a_{n} + 1$$
Is equivalent to linear growth, or also in a direct formula:
$$ a_n = n + a_0$$
We have thus desired that our sequence grows at least as fast as this sequence. However, we know that this sequence diverges. Because or minorant diverges, our original subsequence must also diverge. $square$
answered Nov 27 at 0:21
Wesley Strik
1,534422
add a comment |
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1 Answer
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We build a recursive subsequence defined by:
$$a_{n_0}=a_0 land n_{j+1} > n_j land a_{n_{j+1}} geq a_{n_j}+1 $$
We can do so because the set is unbounded, so we can always handpick the next term that is greater than our previous one, because there should never exist a bound. Suppose we would not be able to pick a greater term, we would get a contradiction since $a_{n_j}+1$ would be a bound for the entire sequence, which was supposed to be unbounded.
We also notice that the index set is strictly increasing and that the subsequence itself is increasing, the question is, is it increasing fast enough?
Notice that we take some arbitrary term for the sequence and handpick the next one that is at least greater than the current one plus $1$. This looks very familiar:
$$ a_{n+1} = a_{n} + 1$$
Is equivalent to linear growth, or also in a direct formula:
$$ a_n = n + a_0$$
We have thus desired that our sequence grows at least as fast as this sequence. However, we know that this sequence diverges. Because or minorant diverges, our original subsequence must also diverge. $square$
answered Nov 27 at 0:21
Wesley Strik
1,534422
add a comment |
We build a recursive subsequence defined by:
$$a_{n_0}=a_0 land n_{j+1} > n_j land a_{n_{j+1}} geq a_{n_j}+1 $$
We can do so because the set is unbounded, so we can always handpick the next term that is greater than our previous one, because there should never exist a bound. Suppose we would not be able to pick a greater term, we would get a contradiction since $a_{n_j}+1$ would be a bound for the entire sequence, which was supposed to be unbounded.
We also notice that the index set is strictly increasing and that the subsequence itself is increasing, the question is, is it increasing fast enough?
Notice that we take some arbitrary term for the sequence and handpick the next one that is at least greater than the current one plus $1$. This looks very familiar:
$$ a_{n+1} = a_{n} + 1$$
Is equivalent to linear growth, or also in a direct formula:
$$ a_n = n + a_0$$
We have thus desired that our sequence grows at least as fast as this sequence. However, we know that this sequence diverges. Because or minorant diverges, our original subsequence must also diverge. $square$
answered Nov 27 at 0:21
Wesley Strik
1,534422
add a comment |
We build a recursive subsequence defined by:
$$a_{n_0}=a_0 land n_{j+1} > n_j land a_{n_{j+1}} geq a_{n_j}+1 $$
We can do so because the set is unbounded, so we can always handpick the next term that is greater than our previous one, because there should never exist a bound. Suppose we would not be able to pick a greater term, we would get a contradiction since $a_{n_j}+1$ would be a bound for the entire sequence, which was supposed to be unbounded.
We also notice that the index set is strictly increasing and that the subsequence itself is increasing, the question is, is it increasing fast enough?
Notice that we take some arbitrary term for the sequence and handpick the next one that is at least greater than the current one plus $1$. This looks very familiar:
$$ a_{n+1} = a_{n} + 1$$
Is equivalent to linear growth, or also in a direct formula:
$$ a_n = n + a_0$$
We have thus desired that our sequence grows at least as fast as this sequence. However, we know that this sequence diverges. Because or minorant diverges, our original subsequence must also diverge. $square$
answered Nov 27 at 0:21
Wesley Strik
1,534422
We build a recursive subsequence defined by:
$$a_{n_0}=a_0 land n_{j+1} > n_j land a_{n_{j+1}} geq a_{n_j}+1 $$
We can do so because the set is unbounded, so we can always handpick the next term that is greater than our previous one, because there should never exist a bound. Suppose we would not be able to pick a greater term, we would get a contradiction since $a_{n_j}+1$ would be a bound for the entire sequence, which was supposed to be unbounded.
We also notice that the index set is strictly increasing and that the subsequence itself is increasing, the question is, is it increasing fast enough?
Notice that we take some arbitrary term for the sequence and handpick the next one that is at least greater than the current one plus $1$. This looks very familiar:
$$ a_{n+1} = a_{n} + 1$$
Is equivalent to linear growth, or also in a direct formula:
$$ a_n = n + a_0$$
We have thus desired that our sequence grows at least as fast as this sequence. However, we know that this sequence diverges. Because or minorant diverges, our original subsequence must also diverge. $square$
answered Nov 27 at 0:21
Wesley Strik
1,534422
answered Nov 27 at 0:21
Wesley Strik
1,534422
answered Nov 27 at 0:21
Wesley Strik
1,534422
answered Nov 27 at 0:21
Wesley Strik
1,534422
1,534422
add a comment |
add a comment |
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Simply build recursively a subsequence $(a_{n_j})$ such that $a_{n_0}=a_0$, say, and, for every $j$, $$a_{n_{j+1}}geqslant a_{n_j}+1$$
– Did
Nov 25 at 6:15
That's a nice way to put it.
– Wesley Strik
Nov 25 at 6:18