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Convergence proof with Cauchy sequences
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Consider the sequence $left z_k right $ in $mathbbR.$ Let the sequence $$left x_n right = sum_k=1^nz_k   textand  left y_n right = sum_k=1^nleft | z_k right |.$$ My goal is to show that if $left y_n right $ converges, then $left x_n right $ converges, using the ideas of a Cauchy sequence and $textbfwithout$ any ideas from infinite series. ÂÂ
I know that a sequence $left a_n right $ in $mathbbR$ is a Cauchy sequence if$ $ $forall epsilon > 0, exists N in mathbbN$ such that whenever natural numbers $m,n > N$,$$left | a_m-a_n right |< epsilon.$$
From what I understand, this is saying that the distance between any two elements of $left a_n right $ get closer and closer to each other as $nrightarrow infty.$ It is sort of like the elements are "bundling up" near the supremum. ÂÂ
One thought I had initially was to consider one sequence as a subsequence of the other, but I do not think that would be true in this case. ÂÂ
Does anyone have any ideas? Thanks in advance!ÂÂ
real-analysis sequences-and-series cauchy-sequences
edited 40 mins ago
José Carlos Santos
123k17101186
asked 51 mins ago
StatGuy
3329
add a comment |Â
up vote
2
down vote
favorite
Consider the sequence $left z_k right $ in $mathbbR.$ Let the sequence $$left x_n right = sum_k=1^nz_k   textand  left y_n right = sum_k=1^nleft | z_k right |.$$ My goal is to show that if $left y_n right $ converges, then $left x_n right $ converges, using the ideas of a Cauchy sequence and $textbfwithout$ any ideas from infinite series. ÂÂ
I know that a sequence $left a_n right $ in $mathbbR$ is a Cauchy sequence if$ $ $forall epsilon > 0, exists N in mathbbN$ such that whenever natural numbers $m,n > N$,$$left | a_m-a_n right |< epsilon.$$
From what I understand, this is saying that the distance between any two elements of $left a_n right $ get closer and closer to each other as $nrightarrow infty.$ It is sort of like the elements are "bundling up" near the supremum. ÂÂ
One thought I had initially was to consider one sequence as a subsequence of the other, but I do not think that would be true in this case. ÂÂ
Does anyone have any ideas? Thanks in advance!ÂÂ
real-analysis sequences-and-series cauchy-sequences
edited 40 mins ago
José Carlos Santos
123k17101186
asked 51 mins ago
StatGuy
3329
The task should not be difficult. The sum of $|z_i |$ terms can be done as small as desired, hence the sum of $z_i$ terms (being less or equal) can also.
– AnyAD
43 mins ago
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Consider the sequence $left z_k right $ in $mathbbR.$ Let the sequence $$left x_n right = sum_k=1^nz_k   textand  left y_n right = sum_k=1^nleft | z_k right |.$$ My goal is to show that if $left y_n right $ converges, then $left x_n right $ converges, using the ideas of a Cauchy sequence and $textbfwithout$ any ideas from infinite series. ÂÂ
I know that a sequence $left a_n right $ in $mathbbR$ is a Cauchy sequence if$ $ $forall epsilon > 0, exists N in mathbbN$ such that whenever natural numbers $m,n > N$,$$left | a_m-a_n right |< epsilon.$$
From what I understand, this is saying that the distance between any two elements of $left a_n right $ get closer and closer to each other as $nrightarrow infty.$ It is sort of like the elements are "bundling up" near the supremum. ÂÂ
One thought I had initially was to consider one sequence as a subsequence of the other, but I do not think that would be true in this case. ÂÂ
Does anyone have any ideas? Thanks in advance!ÂÂ
real-analysis sequences-and-series cauchy-sequences
edited 40 mins ago
José Carlos Santos
123k17101186
asked 51 mins ago
StatGuy
3329
Consider the sequence $left z_k right $ in $mathbbR.$ Let the sequence $$left x_n right = sum_k=1^nz_k   textand  left y_n right = sum_k=1^nleft | z_k right |.$$ My goal is to show that if $left y_n right $ converges, then $left x_n right $ converges, using the ideas of a Cauchy sequence and $textbfwithout$ any ideas from infinite series. ÂÂ
I know that a sequence $left a_n right $ in $mathbbR$ is a Cauchy sequence if$ $ $forall epsilon > 0, exists N in mathbbN$ such that whenever natural numbers $m,n > N$,$$left | a_m-a_n right |< epsilon.$$
From what I understand, this is saying that the distance between any two elements of $left a_n right $ get closer and closer to each other as $nrightarrow infty.$ It is sort of like the elements are "bundling up" near the supremum. ÂÂ
One thought I had initially was to consider one sequence as a subsequence of the other, but I do not think that would be true in this case. ÂÂ
Does anyone have any ideas? Thanks in advance!ÂÂ
real-analysis sequences-and-series cauchy-sequences
real-analysis sequences-and-series cauchy-sequences
edited 40 mins ago
José Carlos Santos
123k17101186
asked 51 mins ago
StatGuy
3329
edited 40 mins ago
José Carlos Santos
123k17101186
asked 51 mins ago
StatGuy
3329
edited 40 mins ago
José Carlos Santos
123k17101186
edited 40 mins ago
José Carlos Santos
123k17101186
edited 40 mins ago
José Carlos Santos
123k17101186
123k17101186
asked 51 mins ago
StatGuy
3329
asked 51 mins ago
StatGuy
3329
asked 51 mins ago
StatGuy
3329
3329
The task should not be difficult. The sum of $|z_i |$ terms can be done as small as desired, hence the sum of $z_i$ terms (being less or equal) can also.
– AnyAD
43 mins ago
add a comment |Â
The task should not be difficult. The sum of $|z_i |$ terms can be done as small as desired, hence the sum of $z_i$ terms (being less or equal) can also.
– AnyAD
43 mins ago
The task should not be difficult. The sum of $|z_i |$ terms can be done as small as desired, hence the sum of $z_i$ terms (being less or equal) can also.
– AnyAD
43 mins ago
The task should not be difficult. The sum of $|z_i |$ terms can be done as small as desired, hence the sum of $z_i$ terms (being less or equal) can also.
– AnyAD
43 mins ago
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
2
down vote
Take any $epsilon>0$. You assume the sequence $y_n$ is a cauchy sequence so you know that:
$exists (n_0inmathbbN)forall(m,ngeq n_0)[|y_m-y_n|<epsilon]$
Well, so let's take $m>ngeq n_0$. Then using the triangle inequality we get:
$|x_m-x_n|=|sum_k=1^mz_k-sum_k=1^nz_k|=|sum_k=n+1^mz_k|leq sum_k=n+1^m|z_k|=sum_k=1^m|z_k|-sum_k=1^n|z_k|=y_m-y_n<epsilon$
So $x_n$ is also a cauchy sequence.
answered 41 mins ago
Mark
1,766110
To the proposer: The first sentence of this A is the definition of the Cauchy Criterion. A sequence that satisfies the Cauchy Criterion is called a Cauchy sequence. In $Bbb R$ or in $Bbb C,$ a sequence converges to a limit iff it is a Cauchy sequence. The result in your Q is very usful in determining that a series converges. More generally, if $M>0$ and $|w_k|leq M|z_k|$ for all but finitely many $k$ and if $sum_k=1^n|z_k|$ converges then $sum_k=1^nw_k$ satisfies the Cauchy Criterion, so it converges......+1
– DanielWainfleet
18 mins ago
add a comment |Â
up vote
1
down vote
For $m<n$ You have $$|x_n-x_m|=|sum_k=m+1^nz_k|leqsum_k=m+1^n|z_k|=|y_n-y_m|$$
by the triangle inequality. Can You take it from here?
answered 42 mins ago
Peter Melech
2,153711
add a comment |Â
up vote
1
down vote
Concerning your idea: no, it is not true in general.
Take $varepsilon>0$. Since $(y_n)_ninmathbb N$ converges, there is a natural $N$ such that$$m,ngeqslant Nimplieslvert y_m-y_nrvert<varepsilon.tag1$$Now, note that, in$(1)$, if $m=n$, then the inequality trivially holds. So, nothing is changed if we add to $(1)$ that $mneq n$. So, in fact, $(1)$ is equivalent to$$m>ngeqslant Nimplieslvert y_m-y_nrvert<varepsilon.$$But this is equivalent to$$m>ngeqslant Nimpliessum_k=n+1^mlvert z_krvert<varepsilon.$$Therefore, if $m,ninmathbb N$ and $m>ngeqslant N$, we have$$leftlvertsum_k=n+1^mz_krightrvertleqslantsum_k=n+1^mlvert z_krvert<varepsilon$$It follows that $(x_n)_ninmathbb N$ is a Cauchy sequence and therefore that it converges.
answered 41 mins ago
José Carlos Santos
123k17101186
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
Take any $epsilon>0$. You assume the sequence $y_n$ is a cauchy sequence so you know that:
$exists (n_0inmathbbN)forall(m,ngeq n_0)[|y_m-y_n|<epsilon]$
Well, so let's take $m>ngeq n_0$. Then using the triangle inequality we get:
$|x_m-x_n|=|sum_k=1^mz_k-sum_k=1^nz_k|=|sum_k=n+1^mz_k|leq sum_k=n+1^m|z_k|=sum_k=1^m|z_k|-sum_k=1^n|z_k|=y_m-y_n<epsilon$
So $x_n$ is also a cauchy sequence.
answered 41 mins ago
Mark
1,766110
To the proposer: The first sentence of this A is the definition of the Cauchy Criterion. A sequence that satisfies the Cauchy Criterion is called a Cauchy sequence. In $Bbb R$ or in $Bbb C,$ a sequence converges to a limit iff it is a Cauchy sequence. The result in your Q is very usful in determining that a series converges. More generally, if $M>0$ and $|w_k|leq M|z_k|$ for all but finitely many $k$ and if $sum_k=1^n|z_k|$ converges then $sum_k=1^nw_k$ satisfies the Cauchy Criterion, so it converges......+1
– DanielWainfleet
18 mins ago
add a comment |Â
up vote
2
down vote
Take any $epsilon>0$. You assume the sequence $y_n$ is a cauchy sequence so you know that:
$exists (n_0inmathbbN)forall(m,ngeq n_0)[|y_m-y_n|<epsilon]$
Well, so let's take $m>ngeq n_0$. Then using the triangle inequality we get:
$|x_m-x_n|=|sum_k=1^mz_k-sum_k=1^nz_k|=|sum_k=n+1^mz_k|leq sum_k=n+1^m|z_k|=sum_k=1^m|z_k|-sum_k=1^n|z_k|=y_m-y_n<epsilon$
So $x_n$ is also a cauchy sequence.
answered 41 mins ago
Mark
1,766110
To the proposer: The first sentence of this A is the definition of the Cauchy Criterion. A sequence that satisfies the Cauchy Criterion is called a Cauchy sequence. In $Bbb R$ or in $Bbb C,$ a sequence converges to a limit iff it is a Cauchy sequence. The result in your Q is very usful in determining that a series converges. More generally, if $M>0$ and $|w_k|leq M|z_k|$ for all but finitely many $k$ and if $sum_k=1^n|z_k|$ converges then $sum_k=1^nw_k$ satisfies the Cauchy Criterion, so it converges......+1
– DanielWainfleet
18 mins ago
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Take any $epsilon>0$. You assume the sequence $y_n$ is a cauchy sequence so you know that:
$exists (n_0inmathbbN)forall(m,ngeq n_0)[|y_m-y_n|<epsilon]$
Well, so let's take $m>ngeq n_0$. Then using the triangle inequality we get:
$|x_m-x_n|=|sum_k=1^mz_k-sum_k=1^nz_k|=|sum_k=n+1^mz_k|leq sum_k=n+1^m|z_k|=sum_k=1^m|z_k|-sum_k=1^n|z_k|=y_m-y_n<epsilon$
So $x_n$ is also a cauchy sequence.
answered 41 mins ago
Mark
1,766110
Take any $epsilon>0$. You assume the sequence $y_n$ is a cauchy sequence so you know that:
$exists (n_0inmathbbN)forall(m,ngeq n_0)[|y_m-y_n|<epsilon]$
Well, so let's take $m>ngeq n_0$. Then using the triangle inequality we get:
$|x_m-x_n|=|sum_k=1^mz_k-sum_k=1^nz_k|=|sum_k=n+1^mz_k|leq sum_k=n+1^m|z_k|=sum_k=1^m|z_k|-sum_k=1^n|z_k|=y_m-y_n<epsilon$
So $x_n$ is also a cauchy sequence.
answered 41 mins ago
Mark
1,766110
answered 41 mins ago
Mark
1,766110
answered 41 mins ago
Mark
1,766110
answered 41 mins ago
Mark
1,766110
1,766110
To the proposer: The first sentence of this A is the definition of the Cauchy Criterion. A sequence that satisfies the Cauchy Criterion is called a Cauchy sequence. In $Bbb R$ or in $Bbb C,$ a sequence converges to a limit iff it is a Cauchy sequence. The result in your Q is very usful in determining that a series converges. More generally, if $M>0$ and $|w_k|leq M|z_k|$ for all but finitely many $k$ and if $sum_k=1^n|z_k|$ converges then $sum_k=1^nw_k$ satisfies the Cauchy Criterion, so it converges......+1
– DanielWainfleet
18 mins ago
add a comment |Â
To the proposer: The first sentence of this A is the definition of the Cauchy Criterion. A sequence that satisfies the Cauchy Criterion is called a Cauchy sequence. In $Bbb R$ or in $Bbb C,$ a sequence converges to a limit iff it is a Cauchy sequence. The result in your Q is very usful in determining that a series converges. More generally, if $M>0$ and $|w_k|leq M|z_k|$ for all but finitely many $k$ and if $sum_k=1^n|z_k|$ converges then $sum_k=1^nw_k$ satisfies the Cauchy Criterion, so it converges......+1
– DanielWainfleet
18 mins ago
To the proposer: The first sentence of this A is the definition of the Cauchy Criterion. A sequence that satisfies the Cauchy Criterion is called a Cauchy sequence. In $Bbb R$ or in $Bbb C,$ a sequence converges to a limit iff it is a Cauchy sequence. The result in your Q is very usful in determining that a series converges. More generally, if $M>0$ and $|w_k|leq M|z_k|$ for all but finitely many $k$ and if $sum_k=1^n|z_k|$ converges then $sum_k=1^nw_k$ satisfies the Cauchy Criterion, so it converges......+1
– DanielWainfleet
18 mins ago
To the proposer: The first sentence of this A is the definition of the Cauchy Criterion. A sequence that satisfies the Cauchy Criterion is called a Cauchy sequence. In $Bbb R$ or in $Bbb C,$ a sequence converges to a limit iff it is a Cauchy sequence. The result in your Q is very usful in determining that a series converges. More generally, if $M>0$ and $|w_k|leq M|z_k|$ for all but finitely many $k$ and if $sum_k=1^n|z_k|$ converges then $sum_k=1^nw_k$ satisfies the Cauchy Criterion, so it converges......+1
– DanielWainfleet
18 mins ago
add a comment |Â
up vote
1
down vote
For $m<n$ You have $$|x_n-x_m|=|sum_k=m+1^nz_k|leqsum_k=m+1^n|z_k|=|y_n-y_m|$$
by the triangle inequality. Can You take it from here?
answered 42 mins ago
Peter Melech
2,153711
add a comment |Â
up vote
1
down vote
For $m<n$ You have $$|x_n-x_m|=|sum_k=m+1^nz_k|leqsum_k=m+1^n|z_k|=|y_n-y_m|$$
by the triangle inequality. Can You take it from here?
answered 42 mins ago
Peter Melech
2,153711
add a comment |Â
up vote
1
down vote
up vote
1
down vote
For $m<n$ You have $$|x_n-x_m|=|sum_k=m+1^nz_k|leqsum_k=m+1^n|z_k|=|y_n-y_m|$$
by the triangle inequality. Can You take it from here?
answered 42 mins ago
Peter Melech
2,153711
For $m<n$ You have $$|x_n-x_m|=|sum_k=m+1^nz_k|leqsum_k=m+1^n|z_k|=|y_n-y_m|$$
by the triangle inequality. Can You take it from here?
answered 42 mins ago
Peter Melech
2,153711
answered 42 mins ago
Peter Melech
2,153711
answered 42 mins ago
Peter Melech
2,153711
answered 42 mins ago
Peter Melech
2,153711
2,153711
add a comment |Â
add a comment |Â
up vote
1
down vote
Concerning your idea: no, it is not true in general.
Take $varepsilon>0$. Since $(y_n)_ninmathbb N$ converges, there is a natural $N$ such that$$m,ngeqslant Nimplieslvert y_m-y_nrvert<varepsilon.tag1$$Now, note that, in$(1)$, if $m=n$, then the inequality trivially holds. So, nothing is changed if we add to $(1)$ that $mneq n$. So, in fact, $(1)$ is equivalent to$$m>ngeqslant Nimplieslvert y_m-y_nrvert<varepsilon.$$But this is equivalent to$$m>ngeqslant Nimpliessum_k=n+1^mlvert z_krvert<varepsilon.$$Therefore, if $m,ninmathbb N$ and $m>ngeqslant N$, we have$$leftlvertsum_k=n+1^mz_krightrvertleqslantsum_k=n+1^mlvert z_krvert<varepsilon$$It follows that $(x_n)_ninmathbb N$ is a Cauchy sequence and therefore that it converges.
answered 41 mins ago
José Carlos Santos
123k17101186
add a comment |Â
up vote
1
down vote
Concerning your idea: no, it is not true in general.
Take $varepsilon>0$. Since $(y_n)_ninmathbb N$ converges, there is a natural $N$ such that$$m,ngeqslant Nimplieslvert y_m-y_nrvert<varepsilon.tag1$$Now, note that, in$(1)$, if $m=n$, then the inequality trivially holds. So, nothing is changed if we add to $(1)$ that $mneq n$. So, in fact, $(1)$ is equivalent to$$m>ngeqslant Nimplieslvert y_m-y_nrvert<varepsilon.$$But this is equivalent to$$m>ngeqslant Nimpliessum_k=n+1^mlvert z_krvert<varepsilon.$$Therefore, if $m,ninmathbb N$ and $m>ngeqslant N$, we have$$leftlvertsum_k=n+1^mz_krightrvertleqslantsum_k=n+1^mlvert z_krvert<varepsilon$$It follows that $(x_n)_ninmathbb N$ is a Cauchy sequence and therefore that it converges.
answered 41 mins ago
José Carlos Santos
123k17101186
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Concerning your idea: no, it is not true in general.
Take $varepsilon>0$. Since $(y_n)_ninmathbb N$ converges, there is a natural $N$ such that$$m,ngeqslant Nimplieslvert y_m-y_nrvert<varepsilon.tag1$$Now, note that, in$(1)$, if $m=n$, then the inequality trivially holds. So, nothing is changed if we add to $(1)$ that $mneq n$. So, in fact, $(1)$ is equivalent to$$m>ngeqslant Nimplieslvert y_m-y_nrvert<varepsilon.$$But this is equivalent to$$m>ngeqslant Nimpliessum_k=n+1^mlvert z_krvert<varepsilon.$$Therefore, if $m,ninmathbb N$ and $m>ngeqslant N$, we have$$leftlvertsum_k=n+1^mz_krightrvertleqslantsum_k=n+1^mlvert z_krvert<varepsilon$$It follows that $(x_n)_ninmathbb N$ is a Cauchy sequence and therefore that it converges.
answered 41 mins ago
José Carlos Santos
123k17101186
Concerning your idea: no, it is not true in general.
Take $varepsilon>0$. Since $(y_n)_ninmathbb N$ converges, there is a natural $N$ such that$$m,ngeqslant Nimplieslvert y_m-y_nrvert<varepsilon.tag1$$Now, note that, in$(1)$, if $m=n$, then the inequality trivially holds. So, nothing is changed if we add to $(1)$ that $mneq n$. So, in fact, $(1)$ is equivalent to$$m>ngeqslant Nimplieslvert y_m-y_nrvert<varepsilon.$$But this is equivalent to$$m>ngeqslant Nimpliessum_k=n+1^mlvert z_krvert<varepsilon.$$Therefore, if $m,ninmathbb N$ and $m>ngeqslant N$, we have$$leftlvertsum_k=n+1^mz_krightrvertleqslantsum_k=n+1^mlvert z_krvert<varepsilon$$It follows that $(x_n)_ninmathbb N$ is a Cauchy sequence and therefore that it converges.
answered 41 mins ago
José Carlos Santos
123k17101186
answered 41 mins ago
José Carlos Santos
123k17101186
answered 41 mins ago
José Carlos Santos
123k17101186
answered 41 mins ago
José Carlos Santos
123k17101186
123k17101186
add a comment |Â
add a comment |Â
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The task should not be difficult. The sum of $|z_i |$ terms can be done as small as desired, hence the sum of $z_i$ terms (being less or equal) can also.
– AnyAD
43 mins ago