Making sure if it is Cauchy
$begingroup$
In my real analysis exam I had a problem in which I proved that $|x_{n+1} - x_n|lt {a^n}$ for all natural numbers $n$ and for all positive number $alt 1$ then $(x_n)$ is a Cauchy sequence.
This was solved successfully but the question is if $|x_{n+1} - x_n|lt frac 1n$ does that mean $(x_n)$ is Cauchy? Well my answer was yes because I could write this in the form of the first one, but now I am somehow confused with what I have answered since $1/n$ is a sequence of $n$ so maybe the answer is not necessarily true... Can you please provide me with the correct answer for this question?
real-analysis cauchy-sequences
edited Nov 26 '18 at 5:41
Matt A Pelto
2,622621
asked Nov 26 '18 at 0:29
user7857462user7857462
784
$endgroup$
add a comment |
$begingroup$
In my real analysis exam I had a problem in which I proved that $|x_{n+1} - x_n|lt {a^n}$ for all natural numbers $n$ and for all positive number $alt 1$ then $(x_n)$ is a Cauchy sequence.
This was solved successfully but the question is if $|x_{n+1} - x_n|lt frac 1n$ does that mean $(x_n)$ is Cauchy? Well my answer was yes because I could write this in the form of the first one, but now I am somehow confused with what I have answered since $1/n$ is a sequence of $n$ so maybe the answer is not necessarily true... Can you please provide me with the correct answer for this question?
real-analysis cauchy-sequences
edited Nov 26 '18 at 5:41
Matt A Pelto
2,622621
asked Nov 26 '18 at 0:29
user7857462user7857462
784
$endgroup$
$begingroup$
Are you about the first question? I am confused the the "for all a" and even more by the "a<1" (which makes the upper bound on the differences rather large).
$endgroup$
– Dirk
Nov 26 '18 at 5:31
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Seems to be some sort of error. The proper correction should be either changing "$frac 1{a^n}$" to "$a^n$" or changing "$a<1$" to "$a>1$" (minding the radius of convergence for geometric series).
$endgroup$
– Matt A Pelto
Nov 26 '18 at 5:44
add a comment |
$begingroup$
In my real analysis exam I had a problem in which I proved that $|x_{n+1} - x_n|lt {a^n}$ for all natural numbers $n$ and for all positive number $alt 1$ then $(x_n)$ is a Cauchy sequence.
This was solved successfully but the question is if $|x_{n+1} - x_n|lt frac 1n$ does that mean $(x_n)$ is Cauchy? Well my answer was yes because I could write this in the form of the first one, but now I am somehow confused with what I have answered since $1/n$ is a sequence of $n$ so maybe the answer is not necessarily true... Can you please provide me with the correct answer for this question?
real-analysis cauchy-sequences
edited Nov 26 '18 at 5:41
Matt A Pelto
2,622621
asked Nov 26 '18 at 0:29
user7857462user7857462
784
$endgroup$
In my real analysis exam I had a problem in which I proved that $|x_{n+1} - x_n|lt {a^n}$ for all natural numbers $n$ and for all positive number $alt 1$ then $(x_n)$ is a Cauchy sequence.
This was solved successfully but the question is if $|x_{n+1} - x_n|lt frac 1n$ does that mean $(x_n)$ is Cauchy? Well my answer was yes because I could write this in the form of the first one, but now I am somehow confused with what I have answered since $1/n$ is a sequence of $n$ so maybe the answer is not necessarily true... Can you please provide me with the correct answer for this question?
real-analysis cauchy-sequences
real-analysis cauchy-sequences
edited Nov 26 '18 at 5:41
Matt A Pelto
2,622621
asked Nov 26 '18 at 0:29
user7857462user7857462
784
edited Nov 26 '18 at 5:41
Matt A Pelto
2,622621
asked Nov 26 '18 at 0:29
user7857462user7857462
784
edited Nov 26 '18 at 5:41
Matt A Pelto
2,622621
edited Nov 26 '18 at 5:41
Matt A Pelto
2,622621
edited Nov 26 '18 at 5:41
Matt A Pelto
2,622621
2,622621
asked Nov 26 '18 at 0:29
user7857462user7857462
784
asked Nov 26 '18 at 0:29
user7857462user7857462
784
asked Nov 26 '18 at 0:29
user7857462user7857462
784
784
$begingroup$
Are you about the first question? I am confused the the "for all a" and even more by the "a<1" (which makes the upper bound on the differences rather large).
$endgroup$
– Dirk
Nov 26 '18 at 5:31
$begingroup$
Seems to be some sort of error. The proper correction should be either changing "$frac 1{a^n}$" to "$a^n$" or changing "$a<1$" to "$a>1$" (minding the radius of convergence for geometric series).
$endgroup$
– Matt A Pelto
Nov 26 '18 at 5:44
add a comment |
$begingroup$
Are you about the first question? I am confused the the "for all a" and even more by the "a<1" (which makes the upper bound on the differences rather large).
$endgroup$
– Dirk
Nov 26 '18 at 5:31
$begingroup$
Seems to be some sort of error. The proper correction should be either changing "$frac 1{a^n}$" to "$a^n$" or changing "$a<1$" to "$a>1$" (minding the radius of convergence for geometric series).
$endgroup$
– Matt A Pelto
Nov 26 '18 at 5:44
$begingroup$
Are you about the first question? I am confused the the "for all a" and even more by the "a<1" (which makes the upper bound on the differences rather large).
$endgroup$
– Dirk
Nov 26 '18 at 5:31
$begingroup$
Are you about the first question? I am confused the the "for all a" and even more by the "a<1" (which makes the upper bound on the differences rather large).
$endgroup$
– Dirk
Nov 26 '18 at 5:31
$begingroup$
Seems to be some sort of error. The proper correction should be either changing "$frac 1{a^n}$" to "$a^n$" or changing "$a<1$" to "$a>1$" (minding the radius of convergence for geometric series).
$endgroup$
– Matt A Pelto
Nov 26 '18 at 5:44
$begingroup$
Seems to be some sort of error. The proper correction should be either changing "$frac 1{a^n}$" to "$a^n$" or changing "$a<1$" to "$a>1$" (minding the radius of convergence for geometric series).
$endgroup$
– Matt A Pelto
Nov 26 '18 at 5:44
add a comment |
6 Answers
6
active
oldest
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$begingroup$
Consider the harmonic series: $sum_{n=1}^{infty}frac 1n$.
$mid a_{n+1}-a_nmid=frac1{n+1}ltfrac1n$.
But it diverges.
answered Nov 26 '18 at 0:37
Chris CusterChris Custer
14.2k3827
$endgroup$
$begingroup$
This is the difference between $1/n$ and $1/(n+1)$, not between two terms of the sequence of partial sums. The sequence $(1/n)_{nin mathbb{N}}$ is indeed Cauchy.
$endgroup$
– Michael Lee
Nov 26 '18 at 0:55
$begingroup$
Yes. I'm referring to the sequence of partial sums of the series. It's not Cauchy because it doesn't converge. I was addressing the OP's example. This shows that $mid a_{n+1}-a_nmid$ can be less than $frac1n$, but the sequence can still not be Cauchy.
$endgroup$
– Chris Custer
Nov 26 '18 at 1:08
$begingroup$
The difference between partial sums is $1/(n+1)$, not $1/n(n+1)$. You add $1/(n+1)$ to get from the $n$th partial sum to the $(n+1)$th.
$endgroup$
– Michael Lee
Nov 26 '18 at 1:08
$begingroup$
Oh yeah. My mistake.
$endgroup$
– Chris Custer
Nov 26 '18 at 1:10
add a comment |
$begingroup$
No, $lvert x_{n+1}-x_nrvert < 1/n$ does not imply that $(x_n)_{nin mathbb{N}}$ is Cauchy. Consider $x_n = sum_{k=1}^n 1/2k$, which does not converge.
answered Nov 26 '18 at 0:34
Michael LeeMichael Lee
4,8281930
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add a comment |
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Take $x_n=1+frac 1 2+cdots+frac 1 n$. This is not Cauchy because the harmonic series $1+frac 1 2+cdots$ is divergent.
answered Nov 26 '18 at 0:34
Kavi Rama MurthyKavi Rama Murthy
69.9k53170
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add a comment |
$begingroup$
This kind of thing works only if you can show $|x_{n + 1} - x_{n}| < a_n$ where $sum_{k = 0}^infty a_k < infty$ because, if this condition holds,
begin{align}
|x_{n} - x_{n + m}| &= |x_{n} - x_{n + 1} + x_{n + 1} - x_{n + 2} + x_{n + 2} - cdots + x_{n + m - 1} - x_{n + m}| \
&le |x_{n} - x_{n + 1}| + cdots + |x_{n + m - 1} - x_{n + m}| \
&le a_n + a_{n + 1} + dots + a_{n + m - 1} \
&le sum_{k = n}^infty a_k
end{align}
Now convergence of $sum a_k$ to $A$ means that for any $varepsilon > 0$ there exists $N$ such that for all $n ge N$,
$$ left| A - sum_{k = 0}^{n - 1} a_k right| = sum_{k = n}^infty a_k < varepsilon $$
Comparing this with the above, we have for every $n ge N$ and every $m ge 0$,
$$ |x_n - x_{n + m}| < varepsilon $$
Which means the sequence $(x_n)$ is Cauchy.
If the bound on $|x_{n + 1} - x_n|$ does not converge as a series, you need to use a different trick.
answered Nov 26 '18 at 0:46
Trevor GunnTrevor Gunn
14.9k32047
$endgroup$
1
$begingroup$
+1 for the most informative response and using varepsilon...no effort has been spared here. I will however nitpick at the last sentence by noting that a different trick may or may not still work, depending on whether the sequence is indeed Cauchy which I presume you know but just add as clarification for people such as the question asker. An example where a different trick might apply is the sequence of partial sums of the alternating series $sum_{n=1}^infty frac{(-1)^{n+1}}{n}$.
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– Matt A Pelto
Nov 26 '18 at 1:28
add a comment |
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For each $n in mathbb{N}$, define $x_n:=1^{-1}+2^{-1}+cdots+n^{-1}$. Notice $|x_{n+1}-x_n|=frac{1}{n+1}$ but the sequence ${x_n}_{n=1}^infty$ is not Cauchy as its terms are just the partial sums of the harmonic series which is known to not converge and $mathbb{R}$ is complete.
answered Nov 26 '18 at 0:38
Matt A PeltoMatt A Pelto
2,622621
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This would imply that any series with a general term which tends to $0$ is convergent. This is false (except for $p$-adic numbers…).
answered Nov 26 '18 at 0:37
BernardBernard
123k741117
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6 Answers
6
active
oldest
votes
6 Answers
6
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Consider the harmonic series: $sum_{n=1}^{infty}frac 1n$.
$mid a_{n+1}-a_nmid=frac1{n+1}ltfrac1n$.
But it diverges.
answered Nov 26 '18 at 0:37
Chris CusterChris Custer
14.2k3827
$endgroup$
$begingroup$
This is the difference between $1/n$ and $1/(n+1)$, not between two terms of the sequence of partial sums. The sequence $(1/n)_{nin mathbb{N}}$ is indeed Cauchy.
$endgroup$
– Michael Lee
Nov 26 '18 at 0:55
$begingroup$
Yes. I'm referring to the sequence of partial sums of the series. It's not Cauchy because it doesn't converge. I was addressing the OP's example. This shows that $mid a_{n+1}-a_nmid$ can be less than $frac1n$, but the sequence can still not be Cauchy.
$endgroup$
– Chris Custer
Nov 26 '18 at 1:08
$begingroup$
The difference between partial sums is $1/(n+1)$, not $1/n(n+1)$. You add $1/(n+1)$ to get from the $n$th partial sum to the $(n+1)$th.
$endgroup$
– Michael Lee
Nov 26 '18 at 1:08
$begingroup$
Oh yeah. My mistake.
$endgroup$
– Chris Custer
Nov 26 '18 at 1:10
add a comment |
$begingroup$
Consider the harmonic series: $sum_{n=1}^{infty}frac 1n$.
$mid a_{n+1}-a_nmid=frac1{n+1}ltfrac1n$.
But it diverges.
answered Nov 26 '18 at 0:37
Chris CusterChris Custer
14.2k3827
$endgroup$
$begingroup$
This is the difference between $1/n$ and $1/(n+1)$, not between two terms of the sequence of partial sums. The sequence $(1/n)_{nin mathbb{N}}$ is indeed Cauchy.
$endgroup$
– Michael Lee
Nov 26 '18 at 0:55
$begingroup$
Yes. I'm referring to the sequence of partial sums of the series. It's not Cauchy because it doesn't converge. I was addressing the OP's example. This shows that $mid a_{n+1}-a_nmid$ can be less than $frac1n$, but the sequence can still not be Cauchy.
$endgroup$
– Chris Custer
Nov 26 '18 at 1:08
$begingroup$
The difference between partial sums is $1/(n+1)$, not $1/n(n+1)$. You add $1/(n+1)$ to get from the $n$th partial sum to the $(n+1)$th.
$endgroup$
– Michael Lee
Nov 26 '18 at 1:08
$begingroup$
Oh yeah. My mistake.
$endgroup$
– Chris Custer
Nov 26 '18 at 1:10
add a comment |
$begingroup$
Consider the harmonic series: $sum_{n=1}^{infty}frac 1n$.
$mid a_{n+1}-a_nmid=frac1{n+1}ltfrac1n$.
But it diverges.
answered Nov 26 '18 at 0:37
Chris CusterChris Custer
14.2k3827
$endgroup$
Consider the harmonic series: $sum_{n=1}^{infty}frac 1n$.
$mid a_{n+1}-a_nmid=frac1{n+1}ltfrac1n$.
But it diverges.
answered Nov 26 '18 at 0:37
Chris CusterChris Custer
14.2k3827
edited Nov 26 '18 at 1:17
answered Nov 26 '18 at 0:37
Chris CusterChris Custer
14.2k3827
answered Nov 26 '18 at 0:37
Chris CusterChris Custer
14.2k3827
answered Nov 26 '18 at 0:37
Chris CusterChris Custer
14.2k3827
14.2k3827
$begingroup$
This is the difference between $1/n$ and $1/(n+1)$, not between two terms of the sequence of partial sums. The sequence $(1/n)_{nin mathbb{N}}$ is indeed Cauchy.
$endgroup$
– Michael Lee
Nov 26 '18 at 0:55
$begingroup$
Yes. I'm referring to the sequence of partial sums of the series. It's not Cauchy because it doesn't converge. I was addressing the OP's example. This shows that $mid a_{n+1}-a_nmid$ can be less than $frac1n$, but the sequence can still not be Cauchy.
$endgroup$
– Chris Custer
Nov 26 '18 at 1:08
$begingroup$
The difference between partial sums is $1/(n+1)$, not $1/n(n+1)$. You add $1/(n+1)$ to get from the $n$th partial sum to the $(n+1)$th.
$endgroup$
– Michael Lee
Nov 26 '18 at 1:08
$begingroup$
Oh yeah. My mistake.
$endgroup$
– Chris Custer
Nov 26 '18 at 1:10
add a comment |
$begingroup$
This is the difference between $1/n$ and $1/(n+1)$, not between two terms of the sequence of partial sums. The sequence $(1/n)_{nin mathbb{N}}$ is indeed Cauchy.
$endgroup$
– Michael Lee
Nov 26 '18 at 0:55
$begingroup$
Yes. I'm referring to the sequence of partial sums of the series. It's not Cauchy because it doesn't converge. I was addressing the OP's example. This shows that $mid a_{n+1}-a_nmid$ can be less than $frac1n$, but the sequence can still not be Cauchy.
$endgroup$
– Chris Custer
Nov 26 '18 at 1:08
$begingroup$
The difference between partial sums is $1/(n+1)$, not $1/n(n+1)$. You add $1/(n+1)$ to get from the $n$th partial sum to the $(n+1)$th.
$endgroup$
– Michael Lee
Nov 26 '18 at 1:08
$begingroup$
Oh yeah. My mistake.
$endgroup$
– Chris Custer
Nov 26 '18 at 1:10
$begingroup$
This is the difference between $1/n$ and $1/(n+1)$, not between two terms of the sequence of partial sums. The sequence $(1/n)_{nin mathbb{N}}$ is indeed Cauchy.
$endgroup$
– Michael Lee
Nov 26 '18 at 0:55
$begingroup$
This is the difference between $1/n$ and $1/(n+1)$, not between two terms of the sequence of partial sums. The sequence $(1/n)_{nin mathbb{N}}$ is indeed Cauchy.
$endgroup$
– Michael Lee
Nov 26 '18 at 0:55
$begingroup$
Yes. I'm referring to the sequence of partial sums of the series. It's not Cauchy because it doesn't converge. I was addressing the OP's example. This shows that $mid a_{n+1}-a_nmid$ can be less than $frac1n$, but the sequence can still not be Cauchy.
$endgroup$
– Chris Custer
Nov 26 '18 at 1:08
$begingroup$
Yes. I'm referring to the sequence of partial sums of the series. It's not Cauchy because it doesn't converge. I was addressing the OP's example. This shows that $mid a_{n+1}-a_nmid$ can be less than $frac1n$, but the sequence can still not be Cauchy.
$endgroup$
– Chris Custer
Nov 26 '18 at 1:08
$begingroup$
The difference between partial sums is $1/(n+1)$, not $1/n(n+1)$. You add $1/(n+1)$ to get from the $n$th partial sum to the $(n+1)$th.
$endgroup$
– Michael Lee
Nov 26 '18 at 1:08
$begingroup$
The difference between partial sums is $1/(n+1)$, not $1/n(n+1)$. You add $1/(n+1)$ to get from the $n$th partial sum to the $(n+1)$th.
$endgroup$
– Michael Lee
Nov 26 '18 at 1:08
$begingroup$
Oh yeah. My mistake.
$endgroup$
– Chris Custer
Nov 26 '18 at 1:10
$begingroup$
Oh yeah. My mistake.
$endgroup$
– Chris Custer
Nov 26 '18 at 1:10
add a comment |
$begingroup$
No, $lvert x_{n+1}-x_nrvert < 1/n$ does not imply that $(x_n)_{nin mathbb{N}}$ is Cauchy. Consider $x_n = sum_{k=1}^n 1/2k$, which does not converge.
answered Nov 26 '18 at 0:34
Michael LeeMichael Lee
4,8281930
$endgroup$
add a comment |
$begingroup$
No, $lvert x_{n+1}-x_nrvert < 1/n$ does not imply that $(x_n)_{nin mathbb{N}}$ is Cauchy. Consider $x_n = sum_{k=1}^n 1/2k$, which does not converge.
answered Nov 26 '18 at 0:34
Michael LeeMichael Lee
4,8281930
$endgroup$
add a comment |
$begingroup$
No, $lvert x_{n+1}-x_nrvert < 1/n$ does not imply that $(x_n)_{nin mathbb{N}}$ is Cauchy. Consider $x_n = sum_{k=1}^n 1/2k$, which does not converge.
answered Nov 26 '18 at 0:34
Michael LeeMichael Lee
4,8281930
$endgroup$
No, $lvert x_{n+1}-x_nrvert < 1/n$ does not imply that $(x_n)_{nin mathbb{N}}$ is Cauchy. Consider $x_n = sum_{k=1}^n 1/2k$, which does not converge.
answered Nov 26 '18 at 0:34
Michael LeeMichael Lee
4,8281930
answered Nov 26 '18 at 0:34
Michael LeeMichael Lee
4,8281930
answered Nov 26 '18 at 0:34
Michael LeeMichael Lee
4,8281930
answered Nov 26 '18 at 0:34
Michael LeeMichael Lee
4,8281930
4,8281930
add a comment |
add a comment |
$begingroup$
Take $x_n=1+frac 1 2+cdots+frac 1 n$. This is not Cauchy because the harmonic series $1+frac 1 2+cdots$ is divergent.
answered Nov 26 '18 at 0:34
Kavi Rama MurthyKavi Rama Murthy
69.9k53170
$endgroup$
add a comment |
$begingroup$
Take $x_n=1+frac 1 2+cdots+frac 1 n$. This is not Cauchy because the harmonic series $1+frac 1 2+cdots$ is divergent.
answered Nov 26 '18 at 0:34
Kavi Rama MurthyKavi Rama Murthy
69.9k53170
$endgroup$
add a comment |
$begingroup$
Take $x_n=1+frac 1 2+cdots+frac 1 n$. This is not Cauchy because the harmonic series $1+frac 1 2+cdots$ is divergent.
answered Nov 26 '18 at 0:34
Kavi Rama MurthyKavi Rama Murthy
69.9k53170
$endgroup$
Take $x_n=1+frac 1 2+cdots+frac 1 n$. This is not Cauchy because the harmonic series $1+frac 1 2+cdots$ is divergent.
answered Nov 26 '18 at 0:34
Kavi Rama MurthyKavi Rama Murthy
69.9k53170
answered Nov 26 '18 at 0:34
Kavi Rama MurthyKavi Rama Murthy
69.9k53170
answered Nov 26 '18 at 0:34
Kavi Rama MurthyKavi Rama Murthy
69.9k53170
answered Nov 26 '18 at 0:34
Kavi Rama MurthyKavi Rama Murthy
69.9k53170
69.9k53170
add a comment |
add a comment |
$begingroup$
This kind of thing works only if you can show $|x_{n + 1} - x_{n}| < a_n$ where $sum_{k = 0}^infty a_k < infty$ because, if this condition holds,
begin{align}
|x_{n} - x_{n + m}| &= |x_{n} - x_{n + 1} + x_{n + 1} - x_{n + 2} + x_{n + 2} - cdots + x_{n + m - 1} - x_{n + m}| \
&le |x_{n} - x_{n + 1}| + cdots + |x_{n + m - 1} - x_{n + m}| \
&le a_n + a_{n + 1} + dots + a_{n + m - 1} \
&le sum_{k = n}^infty a_k
end{align}
Now convergence of $sum a_k$ to $A$ means that for any $varepsilon > 0$ there exists $N$ such that for all $n ge N$,
$$ left| A - sum_{k = 0}^{n - 1} a_k right| = sum_{k = n}^infty a_k < varepsilon $$
Comparing this with the above, we have for every $n ge N$ and every $m ge 0$,
$$ |x_n - x_{n + m}| < varepsilon $$
Which means the sequence $(x_n)$ is Cauchy.
If the bound on $|x_{n + 1} - x_n|$ does not converge as a series, you need to use a different trick.
answered Nov 26 '18 at 0:46
Trevor GunnTrevor Gunn
14.9k32047
$endgroup$
1
$begingroup$
+1 for the most informative response and using varepsilon...no effort has been spared here. I will however nitpick at the last sentence by noting that a different trick may or may not still work, depending on whether the sequence is indeed Cauchy which I presume you know but just add as clarification for people such as the question asker. An example where a different trick might apply is the sequence of partial sums of the alternating series $sum_{n=1}^infty frac{(-1)^{n+1}}{n}$.
$endgroup$
– Matt A Pelto
Nov 26 '18 at 1:28
add a comment |
$begingroup$
This kind of thing works only if you can show $|x_{n + 1} - x_{n}| < a_n$ where $sum_{k = 0}^infty a_k < infty$ because, if this condition holds,
begin{align}
|x_{n} - x_{n + m}| &= |x_{n} - x_{n + 1} + x_{n + 1} - x_{n + 2} + x_{n + 2} - cdots + x_{n + m - 1} - x_{n + m}| \
&le |x_{n} - x_{n + 1}| + cdots + |x_{n + m - 1} - x_{n + m}| \
&le a_n + a_{n + 1} + dots + a_{n + m - 1} \
&le sum_{k = n}^infty a_k
end{align}
Now convergence of $sum a_k$ to $A$ means that for any $varepsilon > 0$ there exists $N$ such that for all $n ge N$,
$$ left| A - sum_{k = 0}^{n - 1} a_k right| = sum_{k = n}^infty a_k < varepsilon $$
Comparing this with the above, we have for every $n ge N$ and every $m ge 0$,
$$ |x_n - x_{n + m}| < varepsilon $$
Which means the sequence $(x_n)$ is Cauchy.
If the bound on $|x_{n + 1} - x_n|$ does not converge as a series, you need to use a different trick.
answered Nov 26 '18 at 0:46
Trevor GunnTrevor Gunn
14.9k32047
$endgroup$
1
$begingroup$
+1 for the most informative response and using varepsilon...no effort has been spared here. I will however nitpick at the last sentence by noting that a different trick may or may not still work, depending on whether the sequence is indeed Cauchy which I presume you know but just add as clarification for people such as the question asker. An example where a different trick might apply is the sequence of partial sums of the alternating series $sum_{n=1}^infty frac{(-1)^{n+1}}{n}$.
$endgroup$
– Matt A Pelto
Nov 26 '18 at 1:28
add a comment |
$begingroup$
This kind of thing works only if you can show $|x_{n + 1} - x_{n}| < a_n$ where $sum_{k = 0}^infty a_k < infty$ because, if this condition holds,
begin{align}
|x_{n} - x_{n + m}| &= |x_{n} - x_{n + 1} + x_{n + 1} - x_{n + 2} + x_{n + 2} - cdots + x_{n + m - 1} - x_{n + m}| \
&le |x_{n} - x_{n + 1}| + cdots + |x_{n + m - 1} - x_{n + m}| \
&le a_n + a_{n + 1} + dots + a_{n + m - 1} \
&le sum_{k = n}^infty a_k
end{align}
Now convergence of $sum a_k$ to $A$ means that for any $varepsilon > 0$ there exists $N$ such that for all $n ge N$,
$$ left| A - sum_{k = 0}^{n - 1} a_k right| = sum_{k = n}^infty a_k < varepsilon $$
Comparing this with the above, we have for every $n ge N$ and every $m ge 0$,
$$ |x_n - x_{n + m}| < varepsilon $$
Which means the sequence $(x_n)$ is Cauchy.
If the bound on $|x_{n + 1} - x_n|$ does not converge as a series, you need to use a different trick.
answered Nov 26 '18 at 0:46
Trevor GunnTrevor Gunn
14.9k32047
$endgroup$
This kind of thing works only if you can show $|x_{n + 1} - x_{n}| < a_n$ where $sum_{k = 0}^infty a_k < infty$ because, if this condition holds,
begin{align}
|x_{n} - x_{n + m}| &= |x_{n} - x_{n + 1} + x_{n + 1} - x_{n + 2} + x_{n + 2} - cdots + x_{n + m - 1} - x_{n + m}| \
&le |x_{n} - x_{n + 1}| + cdots + |x_{n + m - 1} - x_{n + m}| \
&le a_n + a_{n + 1} + dots + a_{n + m - 1} \
&le sum_{k = n}^infty a_k
end{align}
Now convergence of $sum a_k$ to $A$ means that for any $varepsilon > 0$ there exists $N$ such that for all $n ge N$,
$$ left| A - sum_{k = 0}^{n - 1} a_k right| = sum_{k = n}^infty a_k < varepsilon $$
Comparing this with the above, we have for every $n ge N$ and every $m ge 0$,
$$ |x_n - x_{n + m}| < varepsilon $$
Which means the sequence $(x_n)$ is Cauchy.
If the bound on $|x_{n + 1} - x_n|$ does not converge as a series, you need to use a different trick.
answered Nov 26 '18 at 0:46
Trevor GunnTrevor Gunn
14.9k32047
answered Nov 26 '18 at 0:46
Trevor GunnTrevor Gunn
14.9k32047
answered Nov 26 '18 at 0:46
Trevor GunnTrevor Gunn
14.9k32047
answered Nov 26 '18 at 0:46
Trevor GunnTrevor Gunn
14.9k32047
14.9k32047
1
$begingroup$
+1 for the most informative response and using varepsilon...no effort has been spared here. I will however nitpick at the last sentence by noting that a different trick may or may not still work, depending on whether the sequence is indeed Cauchy which I presume you know but just add as clarification for people such as the question asker. An example where a different trick might apply is the sequence of partial sums of the alternating series $sum_{n=1}^infty frac{(-1)^{n+1}}{n}$.
$endgroup$
– Matt A Pelto
Nov 26 '18 at 1:28
add a comment |
1
$begingroup$
+1 for the most informative response and using varepsilon...no effort has been spared here. I will however nitpick at the last sentence by noting that a different trick may or may not still work, depending on whether the sequence is indeed Cauchy which I presume you know but just add as clarification for people such as the question asker. An example where a different trick might apply is the sequence of partial sums of the alternating series $sum_{n=1}^infty frac{(-1)^{n+1}}{n}$.
$endgroup$
– Matt A Pelto
Nov 26 '18 at 1:28
1
1
$begingroup$
+1 for the most informative response and using varepsilon...no effort has been spared here. I will however nitpick at the last sentence by noting that a different trick may or may not still work, depending on whether the sequence is indeed Cauchy which I presume you know but just add as clarification for people such as the question asker. An example where a different trick might apply is the sequence of partial sums of the alternating series $sum_{n=1}^infty frac{(-1)^{n+1}}{n}$.
$endgroup$
– Matt A Pelto
Nov 26 '18 at 1:28
$begingroup$
+1 for the most informative response and using varepsilon...no effort has been spared here. I will however nitpick at the last sentence by noting that a different trick may or may not still work, depending on whether the sequence is indeed Cauchy which I presume you know but just add as clarification for people such as the question asker. An example where a different trick might apply is the sequence of partial sums of the alternating series $sum_{n=1}^infty frac{(-1)^{n+1}}{n}$.
$endgroup$
– Matt A Pelto
Nov 26 '18 at 1:28
add a comment |
$begingroup$
For each $n in mathbb{N}$, define $x_n:=1^{-1}+2^{-1}+cdots+n^{-1}$. Notice $|x_{n+1}-x_n|=frac{1}{n+1}$ but the sequence ${x_n}_{n=1}^infty$ is not Cauchy as its terms are just the partial sums of the harmonic series which is known to not converge and $mathbb{R}$ is complete.
answered Nov 26 '18 at 0:38
Matt A PeltoMatt A Pelto
2,622621
$endgroup$
add a comment |
$begingroup$
For each $n in mathbb{N}$, define $x_n:=1^{-1}+2^{-1}+cdots+n^{-1}$. Notice $|x_{n+1}-x_n|=frac{1}{n+1}$ but the sequence ${x_n}_{n=1}^infty$ is not Cauchy as its terms are just the partial sums of the harmonic series which is known to not converge and $mathbb{R}$ is complete.
answered Nov 26 '18 at 0:38
Matt A PeltoMatt A Pelto
2,622621
$endgroup$
add a comment |
$begingroup$
For each $n in mathbb{N}$, define $x_n:=1^{-1}+2^{-1}+cdots+n^{-1}$. Notice $|x_{n+1}-x_n|=frac{1}{n+1}$ but the sequence ${x_n}_{n=1}^infty$ is not Cauchy as its terms are just the partial sums of the harmonic series which is known to not converge and $mathbb{R}$ is complete.
answered Nov 26 '18 at 0:38
Matt A PeltoMatt A Pelto
2,622621
$endgroup$
For each $n in mathbb{N}$, define $x_n:=1^{-1}+2^{-1}+cdots+n^{-1}$. Notice $|x_{n+1}-x_n|=frac{1}{n+1}$ but the sequence ${x_n}_{n=1}^infty$ is not Cauchy as its terms are just the partial sums of the harmonic series which is known to not converge and $mathbb{R}$ is complete.
answered Nov 26 '18 at 0:38
Matt A PeltoMatt A Pelto
2,622621
edited Nov 30 '18 at 3:28
answered Nov 26 '18 at 0:38
Matt A PeltoMatt A Pelto
2,622621
answered Nov 26 '18 at 0:38
Matt A PeltoMatt A Pelto
2,622621
answered Nov 26 '18 at 0:38
Matt A PeltoMatt A Pelto
2,622621
2,622621
add a comment |
add a comment |
$begingroup$
This would imply that any series with a general term which tends to $0$ is convergent. This is false (except for $p$-adic numbers…).
answered Nov 26 '18 at 0:37
BernardBernard
123k741117
$endgroup$
add a comment |
$begingroup$
This would imply that any series with a general term which tends to $0$ is convergent. This is false (except for $p$-adic numbers…).
answered Nov 26 '18 at 0:37
BernardBernard
123k741117
$endgroup$
add a comment |
$begingroup$
This would imply that any series with a general term which tends to $0$ is convergent. This is false (except for $p$-adic numbers…).
answered Nov 26 '18 at 0:37
BernardBernard
123k741117
$endgroup$
This would imply that any series with a general term which tends to $0$ is convergent. This is false (except for $p$-adic numbers…).
answered Nov 26 '18 at 0:37
BernardBernard
123k741117
answered Nov 26 '18 at 0:37
BernardBernard
123k741117
answered Nov 26 '18 at 0:37
BernardBernard
123k741117
answered Nov 26 '18 at 0:37
BernardBernard
123k741117
123k741117
add a comment |
add a comment |
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$begingroup$
Are you about the first question? I am confused the the "for all a" and even more by the "a<1" (which makes the upper bound on the differences rather large).
$endgroup$
– Dirk
Nov 26 '18 at 5:31
$begingroup$
Seems to be some sort of error. The proper correction should be either changing "$frac 1{a^n}$" to "$a^n$" or changing "$a<1$" to "$a>1$" (minding the radius of convergence for geometric series).
$endgroup$
– Matt A Pelto
Nov 26 '18 at 5:44