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Does $limlimits_{mrightarrowinfty}sumlimits_{k=m}^infty d(a_k,a_{k+1})$ equal to zero for any sequence...
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Let $(M,d)$ be a metric space and $(a_n)subset M$ a sequence. Is it always true that if $r_m=sumlimits_{k=m}^infty d(a_k,a_{k+1})$ we have $limlimits_{mrightarrowinfty}r_m=0$?
$$r_m=sumlimits_{k=0}^infty d(a_k,a_{k+1})-sumlimits_{k=0}^{m-1}d(a_k,a_{k+1})xrightarrow[mrightarrowinfty]{}sumlimits_{k=0}^infty d(a_k,a_{k+1})-sumlimits_{k=0}^{infty}d(a_k,a_{k+1})=0$$
Is this always correct or does the sequence have to satisfy $sumlimits_{k=0}^infty d(a_k,a_{k+1})<infty$?
sequences-and-series metric-spaces
asked Jan 19 at 10:44
John CataldoJohn Cataldo
1,1881316
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Let $(M,d)$ be a metric space and $(a_n)subset M$ a sequence. Is it always true that if $r_m=sumlimits_{k=m}^infty d(a_k,a_{k+1})$ we have $limlimits_{mrightarrowinfty}r_m=0$?
$$r_m=sumlimits_{k=0}^infty d(a_k,a_{k+1})-sumlimits_{k=0}^{m-1}d(a_k,a_{k+1})xrightarrow[mrightarrowinfty]{}sumlimits_{k=0}^infty d(a_k,a_{k+1})-sumlimits_{k=0}^{infty}d(a_k,a_{k+1})=0$$
Is this always correct or does the sequence have to satisfy $sumlimits_{k=0}^infty d(a_k,a_{k+1})<infty$?
sequences-and-series metric-spaces
asked Jan 19 at 10:44
John CataldoJohn Cataldo
1,1881316
$endgroup$
$begingroup$
If the sum is infinite, then $r_m=infty$ for all $m$...
$endgroup$
– Mindlack
Jan 19 at 11:10
add a comment |
$begingroup$
Let $(M,d)$ be a metric space and $(a_n)subset M$ a sequence. Is it always true that if $r_m=sumlimits_{k=m}^infty d(a_k,a_{k+1})$ we have $limlimits_{mrightarrowinfty}r_m=0$?
$$r_m=sumlimits_{k=0}^infty d(a_k,a_{k+1})-sumlimits_{k=0}^{m-1}d(a_k,a_{k+1})xrightarrow[mrightarrowinfty]{}sumlimits_{k=0}^infty d(a_k,a_{k+1})-sumlimits_{k=0}^{infty}d(a_k,a_{k+1})=0$$
Is this always correct or does the sequence have to satisfy $sumlimits_{k=0}^infty d(a_k,a_{k+1})<infty$?
sequences-and-series metric-spaces
asked Jan 19 at 10:44
John CataldoJohn Cataldo
1,1881316
$endgroup$
Let $(M,d)$ be a metric space and $(a_n)subset M$ a sequence. Is it always true that if $r_m=sumlimits_{k=m}^infty d(a_k,a_{k+1})$ we have $limlimits_{mrightarrowinfty}r_m=0$?
$$r_m=sumlimits_{k=0}^infty d(a_k,a_{k+1})-sumlimits_{k=0}^{m-1}d(a_k,a_{k+1})xrightarrow[mrightarrowinfty]{}sumlimits_{k=0}^infty d(a_k,a_{k+1})-sumlimits_{k=0}^{infty}d(a_k,a_{k+1})=0$$
Is this always correct or does the sequence have to satisfy $sumlimits_{k=0}^infty d(a_k,a_{k+1})<infty$?
sequences-and-series metric-spaces
sequences-and-series metric-spaces
asked Jan 19 at 10:44
John CataldoJohn Cataldo
1,1881316
asked Jan 19 at 10:44
John CataldoJohn Cataldo
1,1881316
asked Jan 19 at 10:44
John CataldoJohn Cataldo
1,1881316
asked Jan 19 at 10:44
John CataldoJohn Cataldo
1,1881316
asked Jan 19 at 10:44
John CataldoJohn Cataldo
1,1881316
1,1881316
$begingroup$
If the sum is infinite, then $r_m=infty$ for all $m$...
$endgroup$
– Mindlack
Jan 19 at 11:10
add a comment |
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If the sum is infinite, then $r_m=infty$ for all $m$...
$endgroup$
– Mindlack
Jan 19 at 11:10
$begingroup$
If the sum is infinite, then $r_m=infty$ for all $m$...
$endgroup$
– Mindlack
Jan 19 at 11:10
$begingroup$
If the sum is infinite, then $r_m=infty$ for all $m$...
$endgroup$
– Mindlack
Jan 19 at 11:10
add a comment |
1 Answer
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If $M$ is real line with usual metric and $a_n=1+frac 1 2+cdots+frac 1 n$ then $r_m=infty$ for all $m$. If you assume that $sum d(a_n,a_{n+1})<infty$ then $r_m to 0$ as $m to infty$.
answered Jan 19 at 11:44
Kavi Rama MurthyKavi Rama Murthy
63.7k42463
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1 Answer
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active
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1 Answer
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active
oldest
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active
oldest
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active
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$begingroup$
If $M$ is real line with usual metric and $a_n=1+frac 1 2+cdots+frac 1 n$ then $r_m=infty$ for all $m$. If you assume that $sum d(a_n,a_{n+1})<infty$ then $r_m to 0$ as $m to infty$.
answered Jan 19 at 11:44
Kavi Rama MurthyKavi Rama Murthy
63.7k42463
$endgroup$
add a comment |
$begingroup$
If $M$ is real line with usual metric and $a_n=1+frac 1 2+cdots+frac 1 n$ then $r_m=infty$ for all $m$. If you assume that $sum d(a_n,a_{n+1})<infty$ then $r_m to 0$ as $m to infty$.
answered Jan 19 at 11:44
Kavi Rama MurthyKavi Rama Murthy
63.7k42463
$endgroup$
add a comment |
$begingroup$
If $M$ is real line with usual metric and $a_n=1+frac 1 2+cdots+frac 1 n$ then $r_m=infty$ for all $m$. If you assume that $sum d(a_n,a_{n+1})<infty$ then $r_m to 0$ as $m to infty$.
answered Jan 19 at 11:44
Kavi Rama MurthyKavi Rama Murthy
63.7k42463
$endgroup$
If $M$ is real line with usual metric and $a_n=1+frac 1 2+cdots+frac 1 n$ then $r_m=infty$ for all $m$. If you assume that $sum d(a_n,a_{n+1})<infty$ then $r_m to 0$ as $m to infty$.
answered Jan 19 at 11:44
Kavi Rama MurthyKavi Rama Murthy
63.7k42463
answered Jan 19 at 11:44
Kavi Rama MurthyKavi Rama Murthy
63.7k42463
answered Jan 19 at 11:44
Kavi Rama MurthyKavi Rama Murthy
63.7k42463
answered Jan 19 at 11:44
Kavi Rama MurthyKavi Rama Murthy
63.7k42463
63.7k42463
add a comment |
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If the sum is infinite, then $r_m=infty$ for all $m$...
$endgroup$
– Mindlack
Jan 19 at 11:10