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If $(x_n)$ is monotone and contains a convergent subsequence $(x_{n_i}),$ then $(x_n)$ is convergent....
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This question already has an answer here:
Let ($x_n$) be a monotone sequence and contain a convergent subsequence. Prove that ($x_n$) is convergent.
3 answers
If $(x_n)$ is monotone and contains a convergent subsequence $(x_{n_i}),$ then $(x_n)$ is convergent.
Demonstrating an alternative proof, please provide feedback :) Thank you.
Suppose $(x_n)$ is monotone and contains a convergent subsequence $(x_{n_i}).$
Given that $(x_{n_i})$ is convergent, it is bounded above by some upper bound $b in mathbb{N},$ that is, $x_{n_i} leq b,$ $forall i in mathbb{N}.$
Suppose $(x_n)$ is divergent. Given its monotonicity, it follows that $(x_n)$ is unbounded, that is, $forall M in mathbb{N},$ $exists N in mathbb{N}$ such that $forall n geq N$ it follows that $x_n > M$ (at some point "$N$," the sequence passes the boundary $M$ for any boundary $M in mathbb{N}).$ If $(x_n)$ is bounded, it is necessarily the case that $(x_n),$ given that it is monotone increasing, IS convergent (you can prove this).
So, $exists N in mathbb{N}$ such that $forall n geq N$ it follows that $x_n > b.$
Given that $(x_{n_i})$ is bounded above by $b,$ this means that $forall i in mathbb{N},$ $n_i < N.$
Hence, as by definition of a subsequence it is the case that $n_1 < n_2 < cdots n_i,$ the subsequence $(x_{n_i})$ contains fewer than $N$ elements (at most $N - 1$ elements).
However, a subsequence is defined as a function whose domain is the natural numbers. Given that $(x_{n_i})$ contains fewer than $N$ elements, its domain is a finite, proper subset of the natural numbers, that is, its domain is not the natural numbers. Contradiction.
Therefore, $(x_n)$ is convergent.
real-analysis sequences-and-series
asked Jan 28 at 23:11
Rafael VergnaudRafael Vergnaud
357217
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marked as duplicate by Alexander Gruber♦ Jan 30 at 1:28
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
$begingroup$
This question already has an answer here:
Let ($x_n$) be a monotone sequence and contain a convergent subsequence. Prove that ($x_n$) is convergent.
3 answers
If $(x_n)$ is monotone and contains a convergent subsequence $(x_{n_i}),$ then $(x_n)$ is convergent.
Demonstrating an alternative proof, please provide feedback :) Thank you.
Suppose $(x_n)$ is monotone and contains a convergent subsequence $(x_{n_i}).$
Given that $(x_{n_i})$ is convergent, it is bounded above by some upper bound $b in mathbb{N},$ that is, $x_{n_i} leq b,$ $forall i in mathbb{N}.$
Suppose $(x_n)$ is divergent. Given its monotonicity, it follows that $(x_n)$ is unbounded, that is, $forall M in mathbb{N},$ $exists N in mathbb{N}$ such that $forall n geq N$ it follows that $x_n > M$ (at some point "$N$," the sequence passes the boundary $M$ for any boundary $M in mathbb{N}).$ If $(x_n)$ is bounded, it is necessarily the case that $(x_n),$ given that it is monotone increasing, IS convergent (you can prove this).
So, $exists N in mathbb{N}$ such that $forall n geq N$ it follows that $x_n > b.$
Given that $(x_{n_i})$ is bounded above by $b,$ this means that $forall i in mathbb{N},$ $n_i < N.$
Hence, as by definition of a subsequence it is the case that $n_1 < n_2 < cdots n_i,$ the subsequence $(x_{n_i})$ contains fewer than $N$ elements (at most $N - 1$ elements).
However, a subsequence is defined as a function whose domain is the natural numbers. Given that $(x_{n_i})$ contains fewer than $N$ elements, its domain is a finite, proper subset of the natural numbers, that is, its domain is not the natural numbers. Contradiction.
Therefore, $(x_n)$ is convergent.
real-analysis sequences-and-series
asked Jan 28 at 23:11
Rafael VergnaudRafael Vergnaud
357217
$endgroup$
marked as duplicate by Alexander Gruber♦ Jan 30 at 1:28
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
$begingroup$
Misfire in my brain. Fixed it.
$endgroup$
– Rafael Vergnaud
Jan 28 at 23:14
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Thanks for pointing that out... lol
$endgroup$
– Rafael Vergnaud
Jan 28 at 23:14
$begingroup$
I fixed it!!! haha. Please do not vote to close. I wanted feedback on my potential answer.
$endgroup$
– Rafael Vergnaud
Jan 28 at 23:14
1
$begingroup$
Ok. I'll change it.
$endgroup$
– Rafael Vergnaud
Jan 28 at 23:15
add a comment |
$begingroup$
This question already has an answer here:
Let ($x_n$) be a monotone sequence and contain a convergent subsequence. Prove that ($x_n$) is convergent.
3 answers
If $(x_n)$ is monotone and contains a convergent subsequence $(x_{n_i}),$ then $(x_n)$ is convergent.
Demonstrating an alternative proof, please provide feedback :) Thank you.
Suppose $(x_n)$ is monotone and contains a convergent subsequence $(x_{n_i}).$
Given that $(x_{n_i})$ is convergent, it is bounded above by some upper bound $b in mathbb{N},$ that is, $x_{n_i} leq b,$ $forall i in mathbb{N}.$
Suppose $(x_n)$ is divergent. Given its monotonicity, it follows that $(x_n)$ is unbounded, that is, $forall M in mathbb{N},$ $exists N in mathbb{N}$ such that $forall n geq N$ it follows that $x_n > M$ (at some point "$N$," the sequence passes the boundary $M$ for any boundary $M in mathbb{N}).$ If $(x_n)$ is bounded, it is necessarily the case that $(x_n),$ given that it is monotone increasing, IS convergent (you can prove this).
So, $exists N in mathbb{N}$ such that $forall n geq N$ it follows that $x_n > b.$
Given that $(x_{n_i})$ is bounded above by $b,$ this means that $forall i in mathbb{N},$ $n_i < N.$
Hence, as by definition of a subsequence it is the case that $n_1 < n_2 < cdots n_i,$ the subsequence $(x_{n_i})$ contains fewer than $N$ elements (at most $N - 1$ elements).
However, a subsequence is defined as a function whose domain is the natural numbers. Given that $(x_{n_i})$ contains fewer than $N$ elements, its domain is a finite, proper subset of the natural numbers, that is, its domain is not the natural numbers. Contradiction.
Therefore, $(x_n)$ is convergent.
real-analysis sequences-and-series
asked Jan 28 at 23:11
Rafael VergnaudRafael Vergnaud
357217
$endgroup$
This question already has an answer here:
Let ($x_n$) be a monotone sequence and contain a convergent subsequence. Prove that ($x_n$) is convergent.
3 answers
If $(x_n)$ is monotone and contains a convergent subsequence $(x_{n_i}),$ then $(x_n)$ is convergent.
Demonstrating an alternative proof, please provide feedback :) Thank you.
Suppose $(x_n)$ is monotone and contains a convergent subsequence $(x_{n_i}).$
Given that $(x_{n_i})$ is convergent, it is bounded above by some upper bound $b in mathbb{N},$ that is, $x_{n_i} leq b,$ $forall i in mathbb{N}.$
Suppose $(x_n)$ is divergent. Given its monotonicity, it follows that $(x_n)$ is unbounded, that is, $forall M in mathbb{N},$ $exists N in mathbb{N}$ such that $forall n geq N$ it follows that $x_n > M$ (at some point "$N$," the sequence passes the boundary $M$ for any boundary $M in mathbb{N}).$ If $(x_n)$ is bounded, it is necessarily the case that $(x_n),$ given that it is monotone increasing, IS convergent (you can prove this).
So, $exists N in mathbb{N}$ such that $forall n geq N$ it follows that $x_n > b.$
Given that $(x_{n_i})$ is bounded above by $b,$ this means that $forall i in mathbb{N},$ $n_i < N.$
Hence, as by definition of a subsequence it is the case that $n_1 < n_2 < cdots n_i,$ the subsequence $(x_{n_i})$ contains fewer than $N$ elements (at most $N - 1$ elements).
However, a subsequence is defined as a function whose domain is the natural numbers. Given that $(x_{n_i})$ contains fewer than $N$ elements, its domain is a finite, proper subset of the natural numbers, that is, its domain is not the natural numbers. Contradiction.
Therefore, $(x_n)$ is convergent.
This question already has an answer here:
Let ($x_n$) be a monotone sequence and contain a convergent subsequence. Prove that ($x_n$) is convergent.
3 answers
real-analysis sequences-and-series
real-analysis sequences-and-series
asked Jan 28 at 23:11
Rafael VergnaudRafael Vergnaud
357217
asked Jan 28 at 23:11
Rafael VergnaudRafael Vergnaud
357217
edited Jan 28 at 23:17
Rafael Vergnaud
asked Jan 28 at 23:11
Rafael VergnaudRafael Vergnaud
357217
asked Jan 28 at 23:11
Rafael VergnaudRafael Vergnaud
357217
asked Jan 28 at 23:11
Rafael VergnaudRafael Vergnaud
357217
357217
marked as duplicate by Alexander Gruber♦ Jan 30 at 1:28
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Alexander Gruber♦ Jan 30 at 1:28
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
$begingroup$
Misfire in my brain. Fixed it.
$endgroup$
– Rafael Vergnaud
Jan 28 at 23:14
$begingroup$
Thanks for pointing that out... lol
$endgroup$
– Rafael Vergnaud
Jan 28 at 23:14
$begingroup$
I fixed it!!! haha. Please do not vote to close. I wanted feedback on my potential answer.
$endgroup$
– Rafael Vergnaud
Jan 28 at 23:14
1
$begingroup$
Ok. I'll change it.
$endgroup$
– Rafael Vergnaud
Jan 28 at 23:15
add a comment |
$begingroup$
Misfire in my brain. Fixed it.
$endgroup$
– Rafael Vergnaud
Jan 28 at 23:14
$begingroup$
Thanks for pointing that out... lol
$endgroup$
– Rafael Vergnaud
Jan 28 at 23:14
$begingroup$
I fixed it!!! haha. Please do not vote to close. I wanted feedback on my potential answer.
$endgroup$
– Rafael Vergnaud
Jan 28 at 23:14
1
$begingroup$
Ok. I'll change it.
$endgroup$
– Rafael Vergnaud
Jan 28 at 23:15
$begingroup$
Misfire in my brain. Fixed it.
$endgroup$
– Rafael Vergnaud
Jan 28 at 23:14
$begingroup$
Misfire in my brain. Fixed it.
$endgroup$
– Rafael Vergnaud
Jan 28 at 23:14
$begingroup$
Thanks for pointing that out... lol
$endgroup$
– Rafael Vergnaud
Jan 28 at 23:14
$begingroup$
Thanks for pointing that out... lol
$endgroup$
– Rafael Vergnaud
Jan 28 at 23:14
$begingroup$
I fixed it!!! haha. Please do not vote to close. I wanted feedback on my potential answer.
$endgroup$
– Rafael Vergnaud
Jan 28 at 23:14
$begingroup$
I fixed it!!! haha. Please do not vote to close. I wanted feedback on my potential answer.
$endgroup$
– Rafael Vergnaud
Jan 28 at 23:14
1
1
$begingroup$
Ok. I'll change it.
$endgroup$
– Rafael Vergnaud
Jan 28 at 23:15
$begingroup$
Ok. I'll change it.
$endgroup$
– Rafael Vergnaud
Jan 28 at 23:15
add a comment |
1 Answer
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If $x_n$ did not converge then, by monotonicity, $x_n$ would be unbounded, in particular $x_n - 1 > x = lim x_{n_i}$ for all $n geq n_0,$ for some $n_0.$ But there exists a firs $i_0$ such that $n_i geq n_0$ for all $i geq i_0,$ and so $x_{n_i} geq x + 1$ for all $i geq i_0,$ a contradiction. Q.E.D.
answered Jan 28 at 23:28
Will M.Will M.
2,865315
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add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If $x_n$ did not converge then, by monotonicity, $x_n$ would be unbounded, in particular $x_n - 1 > x = lim x_{n_i}$ for all $n geq n_0,$ for some $n_0.$ But there exists a firs $i_0$ such that $n_i geq n_0$ for all $i geq i_0,$ and so $x_{n_i} geq x + 1$ for all $i geq i_0,$ a contradiction. Q.E.D.
answered Jan 28 at 23:28
Will M.Will M.
2,865315
$endgroup$
add a comment |
$begingroup$
If $x_n$ did not converge then, by monotonicity, $x_n$ would be unbounded, in particular $x_n - 1 > x = lim x_{n_i}$ for all $n geq n_0,$ for some $n_0.$ But there exists a firs $i_0$ such that $n_i geq n_0$ for all $i geq i_0,$ and so $x_{n_i} geq x + 1$ for all $i geq i_0,$ a contradiction. Q.E.D.
answered Jan 28 at 23:28
Will M.Will M.
2,865315
$endgroup$
add a comment |
$begingroup$
If $x_n$ did not converge then, by monotonicity, $x_n$ would be unbounded, in particular $x_n - 1 > x = lim x_{n_i}$ for all $n geq n_0,$ for some $n_0.$ But there exists a firs $i_0$ such that $n_i geq n_0$ for all $i geq i_0,$ and so $x_{n_i} geq x + 1$ for all $i geq i_0,$ a contradiction. Q.E.D.
answered Jan 28 at 23:28
Will M.Will M.
2,865315
$endgroup$
If $x_n$ did not converge then, by monotonicity, $x_n$ would be unbounded, in particular $x_n - 1 > x = lim x_{n_i}$ for all $n geq n_0,$ for some $n_0.$ But there exists a firs $i_0$ such that $n_i geq n_0$ for all $i geq i_0,$ and so $x_{n_i} geq x + 1$ for all $i geq i_0,$ a contradiction. Q.E.D.
answered Jan 28 at 23:28
Will M.Will M.
2,865315
answered Jan 28 at 23:28
Will M.Will M.
2,865315
answered Jan 28 at 23:28
Will M.Will M.
2,865315
answered Jan 28 at 23:28
Will M.Will M.
2,865315
2,865315
add a comment |
add a comment |
$begingroup$
Misfire in my brain. Fixed it.
$endgroup$
– Rafael Vergnaud
Jan 28 at 23:14
$begingroup$
Thanks for pointing that out... lol
$endgroup$
– Rafael Vergnaud
Jan 28 at 23:14
$begingroup$
I fixed it!!! haha. Please do not vote to close. I wanted feedback on my potential answer.
$endgroup$
– Rafael Vergnaud
Jan 28 at 23:14
1
$begingroup$
Ok. I'll change it.
$endgroup$
– Rafael Vergnaud
Jan 28 at 23:15