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Let $(x_n)$ and $(y_n)$ be given sequences and define $(z_n)$ to be the “shuffled†sequence…
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Let $(x_n)$ and $(y_n)$ be given sequences and define $(z_n)$ to be the "shuffled" sequence $(x_1,y_1,x_2,y_2,x_3,y_3,..., x_n,y_n,...)$. Prove that $(z_n)$ is convergent if and only if $(x_n)$ and $(y_n)$ are convergent and $lim x_n=lim y_n$.
Can someone prove this theorem for me? I am lost and all I have so far is:
So we can write that $(z_2n-1)=(x_n)$ and $(z_2n)=(y_n)$ for $n in mathbb N.$
$(Longrightarrow)$ (Is this correct?)
Assume for $n ge N_1 $ $Rightarrow$ $|x_n-l|< epsilon$ and assume for $n ge N_2 $ $Rightarrow$ $|y_n-l|< epsilon$. Choose $M ge maxN_1, N_2$ and then for $n ge M$ for $2n$ and $2n+1$ is we get $|z_n -l| lt epsilon$.
($Longleftarrow$) I don't have anything for this direction.
Note that this question If $x_n to a$ and $x'_n to a$, then $x_1, x'_1, x_2, x'_2, ... to a$ only addresses the theorem in one direction so it isn't exactly a duplicate.
real-analysis sequences-and-series epsilon-delta
asked Jul 18 at 1:12
Red
1,747733
add a comment |Â
up vote
3
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favorite
Let $(x_n)$ and $(y_n)$ be given sequences and define $(z_n)$ to be the "shuffled" sequence $(x_1,y_1,x_2,y_2,x_3,y_3,..., x_n,y_n,...)$. Prove that $(z_n)$ is convergent if and only if $(x_n)$ and $(y_n)$ are convergent and $lim x_n=lim y_n$.
Can someone prove this theorem for me? I am lost and all I have so far is:
So we can write that $(z_2n-1)=(x_n)$ and $(z_2n)=(y_n)$ for $n in mathbb N.$
$(Longrightarrow)$ (Is this correct?)
Assume for $n ge N_1 $ $Rightarrow$ $|x_n-l|< epsilon$ and assume for $n ge N_2 $ $Rightarrow$ $|y_n-l|< epsilon$. Choose $M ge maxN_1, N_2$ and then for $n ge M$ for $2n$ and $2n+1$ is we get $|z_n -l| lt epsilon$.
($Longleftarrow$) I don't have anything for this direction.
Note that this question If $x_n to a$ and $x'_n to a$, then $x_1, x'_1, x_2, x'_2, ... to a$ only addresses the theorem in one direction so it isn't exactly a duplicate.
real-analysis sequences-and-series epsilon-delta
asked Jul 18 at 1:12
Red
1,747733
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Let $(x_n)$ and $(y_n)$ be given sequences and define $(z_n)$ to be the "shuffled" sequence $(x_1,y_1,x_2,y_2,x_3,y_3,..., x_n,y_n,...)$. Prove that $(z_n)$ is convergent if and only if $(x_n)$ and $(y_n)$ are convergent and $lim x_n=lim y_n$.
Can someone prove this theorem for me? I am lost and all I have so far is:
So we can write that $(z_2n-1)=(x_n)$ and $(z_2n)=(y_n)$ for $n in mathbb N.$
$(Longrightarrow)$ (Is this correct?)
Assume for $n ge N_1 $ $Rightarrow$ $|x_n-l|< epsilon$ and assume for $n ge N_2 $ $Rightarrow$ $|y_n-l|< epsilon$. Choose $M ge maxN_1, N_2$ and then for $n ge M$ for $2n$ and $2n+1$ is we get $|z_n -l| lt epsilon$.
($Longleftarrow$) I don't have anything for this direction.
Note that this question If $x_n to a$ and $x'_n to a$, then $x_1, x'_1, x_2, x'_2, ... to a$ only addresses the theorem in one direction so it isn't exactly a duplicate.
real-analysis sequences-and-series epsilon-delta
asked Jul 18 at 1:12
Red
1,747733
Let $(x_n)$ and $(y_n)$ be given sequences and define $(z_n)$ to be the "shuffled" sequence $(x_1,y_1,x_2,y_2,x_3,y_3,..., x_n,y_n,...)$. Prove that $(z_n)$ is convergent if and only if $(x_n)$ and $(y_n)$ are convergent and $lim x_n=lim y_n$.
Can someone prove this theorem for me? I am lost and all I have so far is:
So we can write that $(z_2n-1)=(x_n)$ and $(z_2n)=(y_n)$ for $n in mathbb N.$
$(Longrightarrow)$ (Is this correct?)
Assume for $n ge N_1 $ $Rightarrow$ $|x_n-l|< epsilon$ and assume for $n ge N_2 $ $Rightarrow$ $|y_n-l|< epsilon$. Choose $M ge maxN_1, N_2$ and then for $n ge M$ for $2n$ and $2n+1$ is we get $|z_n -l| lt epsilon$.
($Longleftarrow$) I don't have anything for this direction.
Note that this question If $x_n to a$ and $x'_n to a$, then $x_1, x'_1, x_2, x'_2, ... to a$ only addresses the theorem in one direction so it isn't exactly a duplicate.
real-analysis sequences-and-series epsilon-delta
asked Jul 18 at 1:12
Red
1,747733
edited Jul 18 at 1:33
asked Jul 18 at 1:12
Red
1,747733
asked Jul 18 at 1:12
Red
1,747733
asked Jul 18 at 1:12
Red
1,747733
1,747733
add a comment |Â
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
1
down vote
accepted
Let $x_n_n=1^infty$ and $y_n_n=1^infty$ be two sequences of real numbers. We define the sequence $z_n_n=1^infty$ by setting $z_2n-1:=x_n$ and $z_2n:=y_n$ for each $n=1,2,ldots$.
$(Longleftarrow)$ Suppose $undersetn to inftylimz_n=L.$ Let $varepsilon>0$ be given. Since $z_n to L$ as $n to infty$, there is a positive integer $N$ so that
beginequation |z_n-L|<varepsilon text whenever n geq N.
endequation
Now notice that if $n geq N$, then $2n-1 geq N$ and $2n geq N$. Thus we have
beginaligned &|x_n-L|=|z_2n-1-L|<varepsilon text, and \& |y_n-L|=|z_2n-L|<varepsilon text whenever n geq N.
endaligned
Therefore, $undersetn to inftylimx_n=L=undersetn to inftylimy_n$.
$(Longrightarrow)$ Suppose $undersetn to inftylimx_n=L=undersetn to inftylimy_n$. Let $varepsilon>0$ be given. Since the sequences $x_n_n=1^infty$ and $y_n_n=1^infty$ both converge to the common limit $L$, there exist positive integers $N_1$ and $N_2$ so that
beginequation |x_n-L|<varepsilon text whenever n geq N_1
endequation
and
beginequation |y_n-L|<varepsilon text whenever n geq N_2.
endequation
Set $N=max2N_1, 2N_2$. Thus we have
beginequation |z_n-L|<varepsilon text whenever n geq N.
endequation
Therefore, $z_n_n=1^infty$ is a convergent sequence, by definition.
answered Jul 18 at 1:55
Matt A Pelto
1,709419
add a comment |Â
up vote
0
down vote
Sometimes it helps to prove a more general statement.
Prove that if a sequence converges then so does any subsequence and both limits are the same. It is not too hard to visualise why this is true.
In this case both $x_n$ and $y_n$ are subsequences of $z_n$ so the proof will be done.
answered Jul 18 at 1:18
Bernard W
1,333411
I see the intuition behind it but writing is whats frustrating,,, Also the book hasn't yet introduce the theorem which says subsequences converge to the same limit so I didn't want to "cheat" and simply evoke it. I got the feeling the book wanted me to write it out explicitly.
– Red
Jul 18 at 1:21
What I am proposing is that you prove the statement. Showing that a specific result is a simple consequence of a more general result is good mathematical practice.
– Bernard W
Jul 18 at 1:24
add a comment |Â
up vote
0
down vote
$implies$: If $(a_n)$ converges with limit $a$, every subsequence converges with limit $a$.
Proof: For alle $epsilon>0$ exists $N$ such that $|a_n-a|<epsilon$ for all $nge N$ and notably $|a_n_k-a|<epsilon$ for all $n_kge N$. In other words there exists $K$ such that $n_Kge N$ and $|a_n_k-a|<epsilon$ for all $kge K$. Thus all subsequences converge with the same limit as the convergent sequence.
$impliedby$: Hint: Select the largest of the two $N$'s that leads to convergence of the subsequences and see where that takes you.
answered Jul 18 at 2:02
bubububub
939
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Let $x_n_n=1^infty$ and $y_n_n=1^infty$ be two sequences of real numbers. We define the sequence $z_n_n=1^infty$ by setting $z_2n-1:=x_n$ and $z_2n:=y_n$ for each $n=1,2,ldots$.
$(Longleftarrow)$ Suppose $undersetn to inftylimz_n=L.$ Let $varepsilon>0$ be given. Since $z_n to L$ as $n to infty$, there is a positive integer $N$ so that
beginequation |z_n-L|<varepsilon text whenever n geq N.
endequation
Now notice that if $n geq N$, then $2n-1 geq N$ and $2n geq N$. Thus we have
beginaligned &|x_n-L|=|z_2n-1-L|<varepsilon text, and \& |y_n-L|=|z_2n-L|<varepsilon text whenever n geq N.
endaligned
Therefore, $undersetn to inftylimx_n=L=undersetn to inftylimy_n$.
$(Longrightarrow)$ Suppose $undersetn to inftylimx_n=L=undersetn to inftylimy_n$. Let $varepsilon>0$ be given. Since the sequences $x_n_n=1^infty$ and $y_n_n=1^infty$ both converge to the common limit $L$, there exist positive integers $N_1$ and $N_2$ so that
beginequation |x_n-L|<varepsilon text whenever n geq N_1
endequation
and
beginequation |y_n-L|<varepsilon text whenever n geq N_2.
endequation
Set $N=max2N_1, 2N_2$. Thus we have
beginequation |z_n-L|<varepsilon text whenever n geq N.
endequation
Therefore, $z_n_n=1^infty$ is a convergent sequence, by definition.
answered Jul 18 at 1:55
Matt A Pelto
1,709419
add a comment |Â
up vote
1
down vote
accepted
Let $x_n_n=1^infty$ and $y_n_n=1^infty$ be two sequences of real numbers. We define the sequence $z_n_n=1^infty$ by setting $z_2n-1:=x_n$ and $z_2n:=y_n$ for each $n=1,2,ldots$.
$(Longleftarrow)$ Suppose $undersetn to inftylimz_n=L.$ Let $varepsilon>0$ be given. Since $z_n to L$ as $n to infty$, there is a positive integer $N$ so that
beginequation |z_n-L|<varepsilon text whenever n geq N.
endequation
Now notice that if $n geq N$, then $2n-1 geq N$ and $2n geq N$. Thus we have
beginaligned &|x_n-L|=|z_2n-1-L|<varepsilon text, and \& |y_n-L|=|z_2n-L|<varepsilon text whenever n geq N.
endaligned
Therefore, $undersetn to inftylimx_n=L=undersetn to inftylimy_n$.
$(Longrightarrow)$ Suppose $undersetn to inftylimx_n=L=undersetn to inftylimy_n$. Let $varepsilon>0$ be given. Since the sequences $x_n_n=1^infty$ and $y_n_n=1^infty$ both converge to the common limit $L$, there exist positive integers $N_1$ and $N_2$ so that
beginequation |x_n-L|<varepsilon text whenever n geq N_1
endequation
and
beginequation |y_n-L|<varepsilon text whenever n geq N_2.
endequation
Set $N=max2N_1, 2N_2$. Thus we have
beginequation |z_n-L|<varepsilon text whenever n geq N.
endequation
Therefore, $z_n_n=1^infty$ is a convergent sequence, by definition.
answered Jul 18 at 1:55
Matt A Pelto
1,709419
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Let $x_n_n=1^infty$ and $y_n_n=1^infty$ be two sequences of real numbers. We define the sequence $z_n_n=1^infty$ by setting $z_2n-1:=x_n$ and $z_2n:=y_n$ for each $n=1,2,ldots$.
$(Longleftarrow)$ Suppose $undersetn to inftylimz_n=L.$ Let $varepsilon>0$ be given. Since $z_n to L$ as $n to infty$, there is a positive integer $N$ so that
beginequation |z_n-L|<varepsilon text whenever n geq N.
endequation
Now notice that if $n geq N$, then $2n-1 geq N$ and $2n geq N$. Thus we have
beginaligned &|x_n-L|=|z_2n-1-L|<varepsilon text, and \& |y_n-L|=|z_2n-L|<varepsilon text whenever n geq N.
endaligned
Therefore, $undersetn to inftylimx_n=L=undersetn to inftylimy_n$.
$(Longrightarrow)$ Suppose $undersetn to inftylimx_n=L=undersetn to inftylimy_n$. Let $varepsilon>0$ be given. Since the sequences $x_n_n=1^infty$ and $y_n_n=1^infty$ both converge to the common limit $L$, there exist positive integers $N_1$ and $N_2$ so that
beginequation |x_n-L|<varepsilon text whenever n geq N_1
endequation
and
beginequation |y_n-L|<varepsilon text whenever n geq N_2.
endequation
Set $N=max2N_1, 2N_2$. Thus we have
beginequation |z_n-L|<varepsilon text whenever n geq N.
endequation
Therefore, $z_n_n=1^infty$ is a convergent sequence, by definition.
answered Jul 18 at 1:55
Matt A Pelto
1,709419
Let $x_n_n=1^infty$ and $y_n_n=1^infty$ be two sequences of real numbers. We define the sequence $z_n_n=1^infty$ by setting $z_2n-1:=x_n$ and $z_2n:=y_n$ for each $n=1,2,ldots$.
$(Longleftarrow)$ Suppose $undersetn to inftylimz_n=L.$ Let $varepsilon>0$ be given. Since $z_n to L$ as $n to infty$, there is a positive integer $N$ so that
beginequation |z_n-L|<varepsilon text whenever n geq N.
endequation
Now notice that if $n geq N$, then $2n-1 geq N$ and $2n geq N$. Thus we have
beginaligned &|x_n-L|=|z_2n-1-L|<varepsilon text, and \& |y_n-L|=|z_2n-L|<varepsilon text whenever n geq N.
endaligned
Therefore, $undersetn to inftylimx_n=L=undersetn to inftylimy_n$.
$(Longrightarrow)$ Suppose $undersetn to inftylimx_n=L=undersetn to inftylimy_n$. Let $varepsilon>0$ be given. Since the sequences $x_n_n=1^infty$ and $y_n_n=1^infty$ both converge to the common limit $L$, there exist positive integers $N_1$ and $N_2$ so that
beginequation |x_n-L|<varepsilon text whenever n geq N_1
endequation
and
beginequation |y_n-L|<varepsilon text whenever n geq N_2.
endequation
Set $N=max2N_1, 2N_2$. Thus we have
beginequation |z_n-L|<varepsilon text whenever n geq N.
endequation
Therefore, $z_n_n=1^infty$ is a convergent sequence, by definition.
answered Jul 18 at 1:55
Matt A Pelto
1,709419
edited Jul 18 at 6:40
answered Jul 18 at 1:55
Matt A Pelto
1,709419
answered Jul 18 at 1:55
Matt A Pelto
1,709419
answered Jul 18 at 1:55
Matt A Pelto
1,709419
1,709419
add a comment |Â
add a comment |Â
up vote
0
down vote
Sometimes it helps to prove a more general statement.
Prove that if a sequence converges then so does any subsequence and both limits are the same. It is not too hard to visualise why this is true.
In this case both $x_n$ and $y_n$ are subsequences of $z_n$ so the proof will be done.
answered Jul 18 at 1:18
Bernard W
1,333411
I see the intuition behind it but writing is whats frustrating,,, Also the book hasn't yet introduce the theorem which says subsequences converge to the same limit so I didn't want to "cheat" and simply evoke it. I got the feeling the book wanted me to write it out explicitly.
– Red
Jul 18 at 1:21
What I am proposing is that you prove the statement. Showing that a specific result is a simple consequence of a more general result is good mathematical practice.
– Bernard W
Jul 18 at 1:24
add a comment |Â
up vote
0
down vote
Sometimes it helps to prove a more general statement.
Prove that if a sequence converges then so does any subsequence and both limits are the same. It is not too hard to visualise why this is true.
In this case both $x_n$ and $y_n$ are subsequences of $z_n$ so the proof will be done.
answered Jul 18 at 1:18
Bernard W
1,333411
I see the intuition behind it but writing is whats frustrating,,, Also the book hasn't yet introduce the theorem which says subsequences converge to the same limit so I didn't want to "cheat" and simply evoke it. I got the feeling the book wanted me to write it out explicitly.
– Red
Jul 18 at 1:21
What I am proposing is that you prove the statement. Showing that a specific result is a simple consequence of a more general result is good mathematical practice.
– Bernard W
Jul 18 at 1:24
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Sometimes it helps to prove a more general statement.
Prove that if a sequence converges then so does any subsequence and both limits are the same. It is not too hard to visualise why this is true.
In this case both $x_n$ and $y_n$ are subsequences of $z_n$ so the proof will be done.
answered Jul 18 at 1:18
Bernard W
1,333411
Sometimes it helps to prove a more general statement.
Prove that if a sequence converges then so does any subsequence and both limits are the same. It is not too hard to visualise why this is true.
In this case both $x_n$ and $y_n$ are subsequences of $z_n$ so the proof will be done.
answered Jul 18 at 1:18
Bernard W
1,333411
answered Jul 18 at 1:18
Bernard W
1,333411
answered Jul 18 at 1:18
Bernard W
1,333411
answered Jul 18 at 1:18
Bernard W
1,333411
1,333411
I see the intuition behind it but writing is whats frustrating,,, Also the book hasn't yet introduce the theorem which says subsequences converge to the same limit so I didn't want to "cheat" and simply evoke it. I got the feeling the book wanted me to write it out explicitly.
– Red
Jul 18 at 1:21
What I am proposing is that you prove the statement. Showing that a specific result is a simple consequence of a more general result is good mathematical practice.
– Bernard W
Jul 18 at 1:24
add a comment |Â
I see the intuition behind it but writing is whats frustrating,,, Also the book hasn't yet introduce the theorem which says subsequences converge to the same limit so I didn't want to "cheat" and simply evoke it. I got the feeling the book wanted me to write it out explicitly.
– Red
Jul 18 at 1:21
What I am proposing is that you prove the statement. Showing that a specific result is a simple consequence of a more general result is good mathematical practice.
– Bernard W
Jul 18 at 1:24
I see the intuition behind it but writing is whats frustrating,,, Also the book hasn't yet introduce the theorem which says subsequences converge to the same limit so I didn't want to "cheat" and simply evoke it. I got the feeling the book wanted me to write it out explicitly.
– Red
Jul 18 at 1:21
I see the intuition behind it but writing is whats frustrating,,, Also the book hasn't yet introduce the theorem which says subsequences converge to the same limit so I didn't want to "cheat" and simply evoke it. I got the feeling the book wanted me to write it out explicitly.
– Red
Jul 18 at 1:21
What I am proposing is that you prove the statement. Showing that a specific result is a simple consequence of a more general result is good mathematical practice.
– Bernard W
Jul 18 at 1:24
What I am proposing is that you prove the statement. Showing that a specific result is a simple consequence of a more general result is good mathematical practice.
– Bernard W
Jul 18 at 1:24
add a comment |Â
up vote
0
down vote
$implies$: If $(a_n)$ converges with limit $a$, every subsequence converges with limit $a$.
Proof: For alle $epsilon>0$ exists $N$ such that $|a_n-a|<epsilon$ for all $nge N$ and notably $|a_n_k-a|<epsilon$ for all $n_kge N$. In other words there exists $K$ such that $n_Kge N$ and $|a_n_k-a|<epsilon$ for all $kge K$. Thus all subsequences converge with the same limit as the convergent sequence.
$impliedby$: Hint: Select the largest of the two $N$'s that leads to convergence of the subsequences and see where that takes you.
answered Jul 18 at 2:02
bubububub
939
add a comment |Â
up vote
0
down vote
$implies$: If $(a_n)$ converges with limit $a$, every subsequence converges with limit $a$.
Proof: For alle $epsilon>0$ exists $N$ such that $|a_n-a|<epsilon$ for all $nge N$ and notably $|a_n_k-a|<epsilon$ for all $n_kge N$. In other words there exists $K$ such that $n_Kge N$ and $|a_n_k-a|<epsilon$ for all $kge K$. Thus all subsequences converge with the same limit as the convergent sequence.
$impliedby$: Hint: Select the largest of the two $N$'s that leads to convergence of the subsequences and see where that takes you.
answered Jul 18 at 2:02
bubububub
939
add a comment |Â
up vote
0
down vote
up vote
0
down vote
$implies$: If $(a_n)$ converges with limit $a$, every subsequence converges with limit $a$.
Proof: For alle $epsilon>0$ exists $N$ such that $|a_n-a|<epsilon$ for all $nge N$ and notably $|a_n_k-a|<epsilon$ for all $n_kge N$. In other words there exists $K$ such that $n_Kge N$ and $|a_n_k-a|<epsilon$ for all $kge K$. Thus all subsequences converge with the same limit as the convergent sequence.
$impliedby$: Hint: Select the largest of the two $N$'s that leads to convergence of the subsequences and see where that takes you.
answered Jul 18 at 2:02
bubububub
939
$implies$: If $(a_n)$ converges with limit $a$, every subsequence converges with limit $a$.
Proof: For alle $epsilon>0$ exists $N$ such that $|a_n-a|<epsilon$ for all $nge N$ and notably $|a_n_k-a|<epsilon$ for all $n_kge N$. In other words there exists $K$ such that $n_Kge N$ and $|a_n_k-a|<epsilon$ for all $kge K$. Thus all subsequences converge with the same limit as the convergent sequence.
$impliedby$: Hint: Select the largest of the two $N$'s that leads to convergence of the subsequences and see where that takes you.
answered Jul 18 at 2:02
bubububub
939
answered Jul 18 at 2:02
bubububub
939
answered Jul 18 at 2:02
bubububub
939
answered Jul 18 at 2:02
bubububub
939
939
add a comment |Â
add a comment |Â
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