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What can I say about the consecutive difference of a convergent sequence? [closed]
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Suppose sequence $x_rto 0$. Then what can I say about sequence $x_r+1-x_r$?does it converge to zero?
sequences-and-series convergence
asked Jul 16 at 19:24
Dave
41
closed as off-topic by Xander Henderson, José Carlos Santos, Mostafa Ayaz, GNU Supporter, Shailesh Jul 16 at 23:20
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." РXander Henderson, Jos̩ Carlos Santos, Mostafa Ayaz, GNU Supporter, Shailesh
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Suppose sequence $x_rto 0$. Then what can I say about sequence $x_r+1-x_r$?does it converge to zero?
sequences-and-series convergence
asked Jul 16 at 19:24
Dave
41
closed as off-topic by Xander Henderson, José Carlos Santos, Mostafa Ayaz, GNU Supporter, Shailesh Jul 16 at 23:20
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." РXander Henderson, Jos̩ Carlos Santos, Mostafa Ayaz, GNU Supporter, Shailesh
2
$x^r+1-x^r = x^r(x-1) xrightarrow 0 cdot (x-1) = 0$
– mechanodroid
Jul 16 at 19:29
1
Do you intend the superscript $r$ to be a power or have you perversely intended it as an index for your sequence (which is widely indicated with subscripts)?
– Eric Towers
Jul 16 at 19:30
That was the index of the sequence. Edited.
– Dave
Jul 16 at 19:33
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up vote
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up vote
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down vote
favorite
Suppose sequence $x_rto 0$. Then what can I say about sequence $x_r+1-x_r$?does it converge to zero?
sequences-and-series convergence
asked Jul 16 at 19:24
Dave
41
Suppose sequence $x_rto 0$. Then what can I say about sequence $x_r+1-x_r$?does it converge to zero?
sequences-and-series convergence
asked Jul 16 at 19:24
Dave
41
edited Jul 16 at 19:32
asked Jul 16 at 19:24
Dave
41
asked Jul 16 at 19:24
Dave
41
asked Jul 16 at 19:24
Dave
41
41
closed as off-topic by Xander Henderson, José Carlos Santos, Mostafa Ayaz, GNU Supporter, Shailesh Jul 16 at 23:20
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." РXander Henderson, Jos̩ Carlos Santos, Mostafa Ayaz, GNU Supporter, Shailesh
closed as off-topic by Xander Henderson, José Carlos Santos, Mostafa Ayaz, GNU Supporter, Shailesh Jul 16 at 23:20
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." РXander Henderson, Jos̩ Carlos Santos, Mostafa Ayaz, GNU Supporter, Shailesh
2
$x^r+1-x^r = x^r(x-1) xrightarrow 0 cdot (x-1) = 0$
– mechanodroid
Jul 16 at 19:29
1
Do you intend the superscript $r$ to be a power or have you perversely intended it as an index for your sequence (which is widely indicated with subscripts)?
– Eric Towers
Jul 16 at 19:30
That was the index of the sequence. Edited.
– Dave
Jul 16 at 19:33
add a comment |Â
2
$x^r+1-x^r = x^r(x-1) xrightarrow 0 cdot (x-1) = 0$
– mechanodroid
Jul 16 at 19:29
1
Do you intend the superscript $r$ to be a power or have you perversely intended it as an index for your sequence (which is widely indicated with subscripts)?
– Eric Towers
Jul 16 at 19:30
That was the index of the sequence. Edited.
– Dave
Jul 16 at 19:33
2
2
$x^r+1-x^r = x^r(x-1) xrightarrow 0 cdot (x-1) = 0$
– mechanodroid
Jul 16 at 19:29
$x^r+1-x^r = x^r(x-1) xrightarrow 0 cdot (x-1) = 0$
– mechanodroid
Jul 16 at 19:29
1
1
Do you intend the superscript $r$ to be a power or have you perversely intended it as an index for your sequence (which is widely indicated with subscripts)?
– Eric Towers
Jul 16 at 19:30
Do you intend the superscript $r$ to be a power or have you perversely intended it as an index for your sequence (which is widely indicated with subscripts)?
– Eric Towers
Jul 16 at 19:30
That was the index of the sequence. Edited.
– Dave
Jul 16 at 19:33
That was the index of the sequence. Edited.
– Dave
Jul 16 at 19:33
add a comment |Â
3 Answers
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up vote
5
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The fact that $x_rrightarrow0$ is irrelevant, so let's proceed
assuming $x_rrightarrow x$ (i.e., the sequence converges).
Then,
$$
lim_rrightarrowinftyleft x_r-x_r+1right =lim_rrightarrowinftyx_r-lim_rrightarrowinftyx_r+1=x-x=0.
$$
In the above, we used the fact that $x_r+1rightarrow x$ and that the limit
of sums is the sum of limits.
answered Jul 16 at 19:39
parsiad
16k32253
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up vote
1
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Hint: Convergent sequences are Cauchy, thus...
answered Jul 16 at 19:34
Saucy O'Path
2,716220
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up vote
1
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Given $epsilon>0$
$$x_rto x implies exists Nge 0 ; : $$
$$; forall rge N ;; |x_r-x|<fracepsilon2$$
but for $rge N$, we have
$$r+1>rge N$$ thus
for $rge N,$
$$|x_r+1-x_r|le |x_r+1-x|+|x_r-x|<epsilon$$
done.
answered Jul 16 at 19:36
Salahamam_ Fatima
33.6k21229
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
The fact that $x_rrightarrow0$ is irrelevant, so let's proceed
assuming $x_rrightarrow x$ (i.e., the sequence converges).
Then,
$$
lim_rrightarrowinftyleft x_r-x_r+1right =lim_rrightarrowinftyx_r-lim_rrightarrowinftyx_r+1=x-x=0.
$$
In the above, we used the fact that $x_r+1rightarrow x$ and that the limit
of sums is the sum of limits.
answered Jul 16 at 19:39
parsiad
16k32253
add a comment |Â
up vote
5
down vote
The fact that $x_rrightarrow0$ is irrelevant, so let's proceed
assuming $x_rrightarrow x$ (i.e., the sequence converges).
Then,
$$
lim_rrightarrowinftyleft x_r-x_r+1right =lim_rrightarrowinftyx_r-lim_rrightarrowinftyx_r+1=x-x=0.
$$
In the above, we used the fact that $x_r+1rightarrow x$ and that the limit
of sums is the sum of limits.
answered Jul 16 at 19:39
parsiad
16k32253
add a comment |Â
up vote
5
down vote
up vote
5
down vote
The fact that $x_rrightarrow0$ is irrelevant, so let's proceed
assuming $x_rrightarrow x$ (i.e., the sequence converges).
Then,
$$
lim_rrightarrowinftyleft x_r-x_r+1right =lim_rrightarrowinftyx_r-lim_rrightarrowinftyx_r+1=x-x=0.
$$
In the above, we used the fact that $x_r+1rightarrow x$ and that the limit
of sums is the sum of limits.
answered Jul 16 at 19:39
parsiad
16k32253
The fact that $x_rrightarrow0$ is irrelevant, so let's proceed
assuming $x_rrightarrow x$ (i.e., the sequence converges).
Then,
$$
lim_rrightarrowinftyleft x_r-x_r+1right =lim_rrightarrowinftyx_r-lim_rrightarrowinftyx_r+1=x-x=0.
$$
In the above, we used the fact that $x_r+1rightarrow x$ and that the limit
of sums is the sum of limits.
answered Jul 16 at 19:39
parsiad
16k32253
answered Jul 16 at 19:39
parsiad
16k32253
answered Jul 16 at 19:39
parsiad
16k32253
answered Jul 16 at 19:39
parsiad
16k32253
16k32253
add a comment |Â
add a comment |Â
up vote
1
down vote
Hint: Convergent sequences are Cauchy, thus...
answered Jul 16 at 19:34
Saucy O'Path
2,716220
add a comment |Â
up vote
1
down vote
Hint: Convergent sequences are Cauchy, thus...
answered Jul 16 at 19:34
Saucy O'Path
2,716220
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Hint: Convergent sequences are Cauchy, thus...
answered Jul 16 at 19:34
Saucy O'Path
2,716220
Hint: Convergent sequences are Cauchy, thus...
answered Jul 16 at 19:34
Saucy O'Path
2,716220
answered Jul 16 at 19:34
Saucy O'Path
2,716220
answered Jul 16 at 19:34
Saucy O'Path
2,716220
answered Jul 16 at 19:34
Saucy O'Path
2,716220
2,716220
add a comment |Â
add a comment |Â
up vote
1
down vote
Given $epsilon>0$
$$x_rto x implies exists Nge 0 ; : $$
$$; forall rge N ;; |x_r-x|<fracepsilon2$$
but for $rge N$, we have
$$r+1>rge N$$ thus
for $rge N,$
$$|x_r+1-x_r|le |x_r+1-x|+|x_r-x|<epsilon$$
done.
answered Jul 16 at 19:36
Salahamam_ Fatima
33.6k21229
add a comment |Â
up vote
1
down vote
Given $epsilon>0$
$$x_rto x implies exists Nge 0 ; : $$
$$; forall rge N ;; |x_r-x|<fracepsilon2$$
but for $rge N$, we have
$$r+1>rge N$$ thus
for $rge N,$
$$|x_r+1-x_r|le |x_r+1-x|+|x_r-x|<epsilon$$
done.
answered Jul 16 at 19:36
Salahamam_ Fatima
33.6k21229
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Given $epsilon>0$
$$x_rto x implies exists Nge 0 ; : $$
$$; forall rge N ;; |x_r-x|<fracepsilon2$$
but for $rge N$, we have
$$r+1>rge N$$ thus
for $rge N,$
$$|x_r+1-x_r|le |x_r+1-x|+|x_r-x|<epsilon$$
done.
answered Jul 16 at 19:36
Salahamam_ Fatima
33.6k21229
Given $epsilon>0$
$$x_rto x implies exists Nge 0 ; : $$
$$; forall rge N ;; |x_r-x|<fracepsilon2$$
but for $rge N$, we have
$$r+1>rge N$$ thus
for $rge N,$
$$|x_r+1-x_r|le |x_r+1-x|+|x_r-x|<epsilon$$
done.
answered Jul 16 at 19:36
Salahamam_ Fatima
33.6k21229
edited Jul 16 at 19:46
answered Jul 16 at 19:36
Salahamam_ Fatima
33.6k21229
answered Jul 16 at 19:36
Salahamam_ Fatima
33.6k21229
answered Jul 16 at 19:36
Salahamam_ Fatima
33.6k21229
33.6k21229
add a comment |Â
add a comment |Â
2
$x^r+1-x^r = x^r(x-1) xrightarrow 0 cdot (x-1) = 0$
– mechanodroid
Jul 16 at 19:29
1
Do you intend the superscript $r$ to be a power or have you perversely intended it as an index for your sequence (which is widely indicated with subscripts)?
– Eric Towers
Jul 16 at 19:30
That was the index of the sequence. Edited.
– Dave
Jul 16 at 19:33