How to check convergence of sequence in complete metric space.












1












$begingroup$



Let ${x_n}$ and ${y_n}$ be two sequences in a complete metric space $(X,d)$ such that





  1. $d(x_n,x_{n+1})lefrac{1}{n^2}$


  2. $d(y_n,y_{n+1})le frac{1}{n}$ , for all $nin mathbb{N}.$


Then which sequence would converge? Justify.




Now since the space is complete so every Cauchy sequence is convergent. Let $X=mathbb{R}$ and $d$, the usual metric. So I take the sequence of partial sums of Harmonic series ${H_n}_{nge1}$, then



$$d(H_n,H_{n+1})=|H_{n+1}-H_n|=1/(n+1)le1/n$$
But $H_n$ is not a Cauchy sequence so it does not converge in $X$.



So 2 need not converge.



Is my reasoning correct? Also I have a feeling that 1 would always converge, but I haven't be able to prove it. Can anyone help me with that? Thanks.










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  • $begingroup$
    for large $M < N,$ how big is $sum_{j =M+1}^N ; frac{1}{j^2} ; ; ? $
    $endgroup$
    – Will Jagy
    Jan 30 at 2:17
















1












$begingroup$



Let ${x_n}$ and ${y_n}$ be two sequences in a complete metric space $(X,d)$ such that





  1. $d(x_n,x_{n+1})lefrac{1}{n^2}$


  2. $d(y_n,y_{n+1})le frac{1}{n}$ , for all $nin mathbb{N}.$


Then which sequence would converge? Justify.




Now since the space is complete so every Cauchy sequence is convergent. Let $X=mathbb{R}$ and $d$, the usual metric. So I take the sequence of partial sums of Harmonic series ${H_n}_{nge1}$, then



$$d(H_n,H_{n+1})=|H_{n+1}-H_n|=1/(n+1)le1/n$$
But $H_n$ is not a Cauchy sequence so it does not converge in $X$.



So 2 need not converge.



Is my reasoning correct? Also I have a feeling that 1 would always converge, but I haven't be able to prove it. Can anyone help me with that? Thanks.










share|cite|improve this question









$endgroup$












  • $begingroup$
    for large $M < N,$ how big is $sum_{j =M+1}^N ; frac{1}{j^2} ; ; ? $
    $endgroup$
    – Will Jagy
    Jan 30 at 2:17














1












1








1


1



$begingroup$



Let ${x_n}$ and ${y_n}$ be two sequences in a complete metric space $(X,d)$ such that





  1. $d(x_n,x_{n+1})lefrac{1}{n^2}$


  2. $d(y_n,y_{n+1})le frac{1}{n}$ , for all $nin mathbb{N}.$


Then which sequence would converge? Justify.




Now since the space is complete so every Cauchy sequence is convergent. Let $X=mathbb{R}$ and $d$, the usual metric. So I take the sequence of partial sums of Harmonic series ${H_n}_{nge1}$, then



$$d(H_n,H_{n+1})=|H_{n+1}-H_n|=1/(n+1)le1/n$$
But $H_n$ is not a Cauchy sequence so it does not converge in $X$.



So 2 need not converge.



Is my reasoning correct? Also I have a feeling that 1 would always converge, but I haven't be able to prove it. Can anyone help me with that? Thanks.










share|cite|improve this question









$endgroup$





Let ${x_n}$ and ${y_n}$ be two sequences in a complete metric space $(X,d)$ such that





  1. $d(x_n,x_{n+1})lefrac{1}{n^2}$


  2. $d(y_n,y_{n+1})le frac{1}{n}$ , for all $nin mathbb{N}.$


Then which sequence would converge? Justify.




Now since the space is complete so every Cauchy sequence is convergent. Let $X=mathbb{R}$ and $d$, the usual metric. So I take the sequence of partial sums of Harmonic series ${H_n}_{nge1}$, then



$$d(H_n,H_{n+1})=|H_{n+1}-H_n|=1/(n+1)le1/n$$
But $H_n$ is not a Cauchy sequence so it does not converge in $X$.



So 2 need not converge.



Is my reasoning correct? Also I have a feeling that 1 would always converge, but I haven't be able to prove it. Can anyone help me with that? Thanks.







sequences-and-series convergence metric-spaces cauchy-sequences






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asked Jan 30 at 2:02









Kushal BhuyanKushal Bhuyan

5,08021246




5,08021246












  • $begingroup$
    for large $M < N,$ how big is $sum_{j =M+1}^N ; frac{1}{j^2} ; ; ? $
    $endgroup$
    – Will Jagy
    Jan 30 at 2:17


















  • $begingroup$
    for large $M < N,$ how big is $sum_{j =M+1}^N ; frac{1}{j^2} ; ; ? $
    $endgroup$
    – Will Jagy
    Jan 30 at 2:17
















$begingroup$
for large $M < N,$ how big is $sum_{j =M+1}^N ; frac{1}{j^2} ; ; ? $
$endgroup$
– Will Jagy
Jan 30 at 2:17




$begingroup$
for large $M < N,$ how big is $sum_{j =M+1}^N ; frac{1}{j^2} ; ; ? $
$endgroup$
– Will Jagy
Jan 30 at 2:17










1 Answer
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2












$begingroup$

Your reasoning is good.



For 2, use the triangle inequality to show that, for $m < n$,
$$d(x_n, x_m) le sum_{k = m}^{n - 1} d(x_k, x_{k+1}).$$
Cauchiness follows from the convergence of $sum frac{1}{n^2}$.






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    1 Answer
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    1 Answer
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    2












    $begingroup$

    Your reasoning is good.



    For 2, use the triangle inequality to show that, for $m < n$,
    $$d(x_n, x_m) le sum_{k = m}^{n - 1} d(x_k, x_{k+1}).$$
    Cauchiness follows from the convergence of $sum frac{1}{n^2}$.






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      Your reasoning is good.



      For 2, use the triangle inequality to show that, for $m < n$,
      $$d(x_n, x_m) le sum_{k = m}^{n - 1} d(x_k, x_{k+1}).$$
      Cauchiness follows from the convergence of $sum frac{1}{n^2}$.






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        Your reasoning is good.



        For 2, use the triangle inequality to show that, for $m < n$,
        $$d(x_n, x_m) le sum_{k = m}^{n - 1} d(x_k, x_{k+1}).$$
        Cauchiness follows from the convergence of $sum frac{1}{n^2}$.






        share|cite|improve this answer









        $endgroup$



        Your reasoning is good.



        For 2, use the triangle inequality to show that, for $m < n$,
        $$d(x_n, x_m) le sum_{k = m}^{n - 1} d(x_k, x_{k+1}).$$
        Cauchiness follows from the convergence of $sum frac{1}{n^2}$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 30 at 2:19









        Theo BenditTheo Bendit

        20.1k12354




        20.1k12354






























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