Given a sequence in R is it sufficient to prove limit d(xn,xn+1) is 0 to show that it is cauchy?
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I saw a proof in a fixed point theorem which showed that d(xn,xn+p) is 0 as n goes to infinity hence the sequence is cauchy.
using triangle inequality if
d(xn,xn+p) less than equal to d(xn,xn+1)+d(xn+1,xn+2)+....+d(xn+p-1,xn+p)
And since each of these terms tend to 0 as n tends to infinity then that implies d(xn,xn+p) tends to 0 .Therefore,sequence is cauchy.
Am i missing a condition here?
real-analysis
asked Jan 19 at 6:03
math123math123
153
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add a comment |
$begingroup$
I saw a proof in a fixed point theorem which showed that d(xn,xn+p) is 0 as n goes to infinity hence the sequence is cauchy.
using triangle inequality if
d(xn,xn+p) less than equal to d(xn,xn+1)+d(xn+1,xn+2)+....+d(xn+p-1,xn+p)
And since each of these terms tend to 0 as n tends to infinity then that implies d(xn,xn+p) tends to 0 .Therefore,sequence is cauchy.
Am i missing a condition here?
real-analysis
asked Jan 19 at 6:03
math123math123
153
$endgroup$
2
$begingroup$
What if $x_n=frac11+frac12+frac13+cdots+frac1n$ ?
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– bof
Jan 19 at 6:07
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Given a series $sum_{n=1}^infty a_n$ in $Bbb R$, is it sufficient to prove that $lim_{ntoinfty}a_n=0$ to show the series is convergent?
$endgroup$
– Lord Shark the Unknown
Jan 19 at 6:13
$begingroup$
One thing: $p$ could be related to $n$, like $p = n^2$. Then the sum is actually an infinite sum, and clearly the existence of such sum cannot be solely determined by the assumption $d(x_n, x_{n+1}) to 0$.
$endgroup$
– xbh
Jan 19 at 6:46
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No. See this.
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– David Mitra
Jan 19 at 6:48
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For an Euclidean metric d(xn,xn+1) = xn + 1 - xn = 1.
$endgroup$
– William Elliot
Jan 19 at 8:58
add a comment |
$begingroup$
I saw a proof in a fixed point theorem which showed that d(xn,xn+p) is 0 as n goes to infinity hence the sequence is cauchy.
using triangle inequality if
d(xn,xn+p) less than equal to d(xn,xn+1)+d(xn+1,xn+2)+....+d(xn+p-1,xn+p)
And since each of these terms tend to 0 as n tends to infinity then that implies d(xn,xn+p) tends to 0 .Therefore,sequence is cauchy.
Am i missing a condition here?
real-analysis
asked Jan 19 at 6:03
math123math123
153
$endgroup$
I saw a proof in a fixed point theorem which showed that d(xn,xn+p) is 0 as n goes to infinity hence the sequence is cauchy.
using triangle inequality if
d(xn,xn+p) less than equal to d(xn,xn+1)+d(xn+1,xn+2)+....+d(xn+p-1,xn+p)
And since each of these terms tend to 0 as n tends to infinity then that implies d(xn,xn+p) tends to 0 .Therefore,sequence is cauchy.
Am i missing a condition here?
real-analysis
real-analysis
asked Jan 19 at 6:03
math123math123
153
asked Jan 19 at 6:03
math123math123
153
edited Jan 19 at 7:48
math123
asked Jan 19 at 6:03
math123math123
153
asked Jan 19 at 6:03
math123math123
153
asked Jan 19 at 6:03
math123math123
153
153
2
$begingroup$
What if $x_n=frac11+frac12+frac13+cdots+frac1n$ ?
$endgroup$
– bof
Jan 19 at 6:07
$begingroup$
Given a series $sum_{n=1}^infty a_n$ in $Bbb R$, is it sufficient to prove that $lim_{ntoinfty}a_n=0$ to show the series is convergent?
$endgroup$
– Lord Shark the Unknown
Jan 19 at 6:13
$begingroup$
One thing: $p$ could be related to $n$, like $p = n^2$. Then the sum is actually an infinite sum, and clearly the existence of such sum cannot be solely determined by the assumption $d(x_n, x_{n+1}) to 0$.
$endgroup$
– xbh
Jan 19 at 6:46
$begingroup$
No. See this.
$endgroup$
– David Mitra
Jan 19 at 6:48
$begingroup$
For an Euclidean metric d(xn,xn+1) = xn + 1 - xn = 1.
$endgroup$
– William Elliot
Jan 19 at 8:58
add a comment |
2
$begingroup$
What if $x_n=frac11+frac12+frac13+cdots+frac1n$ ?
$endgroup$
– bof
Jan 19 at 6:07
$begingroup$
Given a series $sum_{n=1}^infty a_n$ in $Bbb R$, is it sufficient to prove that $lim_{ntoinfty}a_n=0$ to show the series is convergent?
$endgroup$
– Lord Shark the Unknown
Jan 19 at 6:13
$begingroup$
One thing: $p$ could be related to $n$, like $p = n^2$. Then the sum is actually an infinite sum, and clearly the existence of such sum cannot be solely determined by the assumption $d(x_n, x_{n+1}) to 0$.
$endgroup$
– xbh
Jan 19 at 6:46
$begingroup$
No. See this.
$endgroup$
– David Mitra
Jan 19 at 6:48
$begingroup$
For an Euclidean metric d(xn,xn+1) = xn + 1 - xn = 1.
$endgroup$
– William Elliot
Jan 19 at 8:58
2
2
$begingroup$
What if $x_n=frac11+frac12+frac13+cdots+frac1n$ ?
$endgroup$
– bof
Jan 19 at 6:07
$begingroup$
What if $x_n=frac11+frac12+frac13+cdots+frac1n$ ?
$endgroup$
– bof
Jan 19 at 6:07
$begingroup$
Given a series $sum_{n=1}^infty a_n$ in $Bbb R$, is it sufficient to prove that $lim_{ntoinfty}a_n=0$ to show the series is convergent?
$endgroup$
– Lord Shark the Unknown
Jan 19 at 6:13
$begingroup$
Given a series $sum_{n=1}^infty a_n$ in $Bbb R$, is it sufficient to prove that $lim_{ntoinfty}a_n=0$ to show the series is convergent?
$endgroup$
– Lord Shark the Unknown
Jan 19 at 6:13
$begingroup$
One thing: $p$ could be related to $n$, like $p = n^2$. Then the sum is actually an infinite sum, and clearly the existence of such sum cannot be solely determined by the assumption $d(x_n, x_{n+1}) to 0$.
$endgroup$
– xbh
Jan 19 at 6:46
$begingroup$
One thing: $p$ could be related to $n$, like $p = n^2$. Then the sum is actually an infinite sum, and clearly the existence of such sum cannot be solely determined by the assumption $d(x_n, x_{n+1}) to 0$.
$endgroup$
– xbh
Jan 19 at 6:46
$begingroup$
No. See this.
$endgroup$
– David Mitra
Jan 19 at 6:48
$begingroup$
No. See this.
$endgroup$
– David Mitra
Jan 19 at 6:48
$begingroup$
For an Euclidean metric d(xn,xn+1) = xn + 1 - xn = 1.
$endgroup$
– William Elliot
Jan 19 at 8:58
$begingroup$
For an Euclidean metric d(xn,xn+1) = xn + 1 - xn = 1.
$endgroup$
– William Elliot
Jan 19 at 8:58
add a comment |
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2
$begingroup$
What if $x_n=frac11+frac12+frac13+cdots+frac1n$ ?
$endgroup$
– bof
Jan 19 at 6:07
$begingroup$
Given a series $sum_{n=1}^infty a_n$ in $Bbb R$, is it sufficient to prove that $lim_{ntoinfty}a_n=0$ to show the series is convergent?
$endgroup$
– Lord Shark the Unknown
Jan 19 at 6:13
$begingroup$
One thing: $p$ could be related to $n$, like $p = n^2$. Then the sum is actually an infinite sum, and clearly the existence of such sum cannot be solely determined by the assumption $d(x_n, x_{n+1}) to 0$.
$endgroup$
– xbh
Jan 19 at 6:46
$begingroup$
No. See this.
$endgroup$
– David Mitra
Jan 19 at 6:48
$begingroup$
For an Euclidean metric d(xn,xn+1) = xn + 1 - xn = 1.
$endgroup$
– William Elliot
Jan 19 at 8:58