Criteria for convergence in Rudin












0












$begingroup$



Theorem 3.22: $sum a_n $ converges if and only if for every $epsilon >0$ there is an integer $N$ such that $$ left|sum_{k=n}^m a_k right | leq epsilon $$ if $ m geq n geq N $



In particular, by taking $m=n$ becomes $$ left|a_n right | leq epsilon ~~~~~ (ngeq N)$$




However, the condition $ left|a_n right | leq epsilon $ is satisfied by $a_n = frac{1}{n}$. Thus by the backward implication of the theorem, $sum frac{1}{n} $ should converge. However, as we know it, it does not.



Which part of the theorem have I misunderstood?



Edit:



Would this be the correct interpretation of the theorem?
$$sum a_n~converges iff (forallepsilon>0)(exists N in mathbb{N})left( m geq n geq N implies left|sum_{k=n}^m a_k right | leq epsilon right) $$






real-analysis sequences-and-series convergence







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$endgroup$












  • $begingroup$
    You should write if $$m>nge N$$ not $le$.
    $endgroup$
    – hamam_Abdallah
    Dec 23 '18 at 21:32
















0












$begingroup$



Theorem 3.22: $sum a_n $ converges if and only if for every $epsilon >0$ there is an integer $N$ such that $$ left|sum_{k=n}^m a_k right | leq epsilon $$ if $ m geq n geq N $



In particular, by taking $m=n$ becomes $$ left|a_n right | leq epsilon ~~~~~ (ngeq N)$$




However, the condition $ left|a_n right | leq epsilon $ is satisfied by $a_n = frac{1}{n}$. Thus by the backward implication of the theorem, $sum frac{1}{n} $ should converge. However, as we know it, it does not.



Which part of the theorem have I misunderstood?



Edit:



Would this be the correct interpretation of the theorem?
$$sum a_n~converges iff (forallepsilon>0)(exists N in mathbb{N})left( m geq n geq N implies left|sum_{k=n}^m a_k right | leq epsilon right) $$






real-analysis sequences-and-series convergence







share|cite|improve this question











$endgroup$












  • $begingroup$
    You should write if $$m>nge N$$ not $le$.
    $endgroup$
    – hamam_Abdallah
    Dec 23 '18 at 21:32














0












0








0


1



$begingroup$



Theorem 3.22: $sum a_n $ converges if and only if for every $epsilon >0$ there is an integer $N$ such that $$ left|sum_{k=n}^m a_k right | leq epsilon $$ if $ m geq n geq N $



In particular, by taking $m=n$ becomes $$ left|a_n right | leq epsilon ~~~~~ (ngeq N)$$




However, the condition $ left|a_n right | leq epsilon $ is satisfied by $a_n = frac{1}{n}$. Thus by the backward implication of the theorem, $sum frac{1}{n} $ should converge. However, as we know it, it does not.



Which part of the theorem have I misunderstood?



Edit:



Would this be the correct interpretation of the theorem?
$$sum a_n~converges iff (forallepsilon>0)(exists N in mathbb{N})left( m geq n geq N implies left|sum_{k=n}^m a_k right | leq epsilon right) $$






real-analysis sequences-and-series convergence







share|cite|improve this question











$endgroup$





Theorem 3.22: $sum a_n $ converges if and only if for every $epsilon >0$ there is an integer $N$ such that $$ left|sum_{k=n}^m a_k right | leq epsilon $$ if $ m geq n geq N $



In particular, by taking $m=n$ becomes $$ left|a_n right | leq epsilon ~~~~~ (ngeq N)$$




However, the condition $ left|a_n right | leq epsilon $ is satisfied by $a_n = frac{1}{n}$. Thus by the backward implication of the theorem, $sum frac{1}{n} $ should converge. However, as we know it, it does not.



Which part of the theorem have I misunderstood?



Edit:



Would this be the correct interpretation of the theorem?
$$sum a_n~converges iff (forallepsilon>0)(exists N in mathbb{N})left( m geq n geq N implies left|sum_{k=n}^m a_k right | leq epsilon right) $$






real-analysis sequences-and-series convergence




real-analysis sequences-and-series convergence






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 23 '18 at 21:45







mathnoob123

















asked Dec 23 '18 at 21:16









mathnoob123mathnoob123

693417




693417












  • $begingroup$
    You should write if $$m>nge N$$ not $le$.
    $endgroup$
    – hamam_Abdallah
    Dec 23 '18 at 21:32


















  • $begingroup$
    You should write if $$m>nge N$$ not $le$.
    $endgroup$
    – hamam_Abdallah
    Dec 23 '18 at 21:32
















$begingroup$
You should write if $$m>nge N$$ not $le$.
$endgroup$
– hamam_Abdallah
Dec 23 '18 at 21:32




$begingroup$
You should write if $$m>nge N$$ not $le$.
$endgroup$
– hamam_Abdallah
Dec 23 '18 at 21:32










3 Answers
3






active

oldest

votes


















2












$begingroup$

Yes, it is true that, for every $varepsilon>0$, we have $frac1n<varepsilon$ if $n$ is large enough. But you can't deduce from thath fact that$$sum_{k=n}^mfrac1k<varepsilon$$if $mgeqslant n$ and $m$ and $n$ are large enough.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Why you cant deduce?
    $endgroup$
    – Jimmy Sabater
    Dec 23 '18 at 21:24










  • $begingroup$
    How so? I am just using the particular backward implication of the theorem which states $$ left|a_n right | leq epsilon ~~~~~ (ngeq N) implies sum a_n ~converges$$. The antecedent is fulfilled by $a_n = frac{1}{n}$
    $endgroup$
    – mathnoob123
    Dec 23 '18 at 21:24








  • 1




    $begingroup$
    You aren't interpreting the theorem correctly. Rudin says that a series $sum_{n=0}^infty a_n$ converges if and only if a certain conditions holds. Then he asserts that if this condition holds, then $lim_{ntoinfty}a_n=0$. He never claims that this is an equivalence.
    $endgroup$
    – José Carlos Santos
    Dec 23 '18 at 21:26












  • $begingroup$
    That section of the theorem IS NOT part of the if and only if. Only the first section is
    $endgroup$
    – Sandeep Silwal
    Dec 23 '18 at 21:26










  • $begingroup$
    Oh of course. Thank you very much
    $endgroup$
    – mathnoob123
    Dec 23 '18 at 21:29



















0












$begingroup$

In other words,
let $$S_n=sum_{k=0}^na_k$$



the criteria is



$$sum a_ntext{ converges } iff (S_n) text{ is Cauchy} iff$$



$$forall epsilon>0 ;; exists Nin Bbb N :$$



$$m>nge N implies |S_m-S_n|<epsilon$$






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$


    If $mge nge N$ then $left|sum_{k=n}^ma_kright|leepsilon$




    means for all $m$,$n$ such that $mge nge N$. Taking $m=n$ is not equivalent to convergence.






    share|cite|improve this answer









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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2












      $begingroup$

      Yes, it is true that, for every $varepsilon>0$, we have $frac1n<varepsilon$ if $n$ is large enough. But you can't deduce from thath fact that$$sum_{k=n}^mfrac1k<varepsilon$$if $mgeqslant n$ and $m$ and $n$ are large enough.






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        Why you cant deduce?
        $endgroup$
        – Jimmy Sabater
        Dec 23 '18 at 21:24










      • $begingroup$
        How so? I am just using the particular backward implication of the theorem which states $$ left|a_n right | leq epsilon ~~~~~ (ngeq N) implies sum a_n ~converges$$. The antecedent is fulfilled by $a_n = frac{1}{n}$
        $endgroup$
        – mathnoob123
        Dec 23 '18 at 21:24








      • 1




        $begingroup$
        You aren't interpreting the theorem correctly. Rudin says that a series $sum_{n=0}^infty a_n$ converges if and only if a certain conditions holds. Then he asserts that if this condition holds, then $lim_{ntoinfty}a_n=0$. He never claims that this is an equivalence.
        $endgroup$
        – José Carlos Santos
        Dec 23 '18 at 21:26












      • $begingroup$
        That section of the theorem IS NOT part of the if and only if. Only the first section is
        $endgroup$
        – Sandeep Silwal
        Dec 23 '18 at 21:26










      • $begingroup$
        Oh of course. Thank you very much
        $endgroup$
        – mathnoob123
        Dec 23 '18 at 21:29
















      2












      $begingroup$

      Yes, it is true that, for every $varepsilon>0$, we have $frac1n<varepsilon$ if $n$ is large enough. But you can't deduce from thath fact that$$sum_{k=n}^mfrac1k<varepsilon$$if $mgeqslant n$ and $m$ and $n$ are large enough.






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        Why you cant deduce?
        $endgroup$
        – Jimmy Sabater
        Dec 23 '18 at 21:24










      • $begingroup$
        How so? I am just using the particular backward implication of the theorem which states $$ left|a_n right | leq epsilon ~~~~~ (ngeq N) implies sum a_n ~converges$$. The antecedent is fulfilled by $a_n = frac{1}{n}$
        $endgroup$
        – mathnoob123
        Dec 23 '18 at 21:24








      • 1




        $begingroup$
        You aren't interpreting the theorem correctly. Rudin says that a series $sum_{n=0}^infty a_n$ converges if and only if a certain conditions holds. Then he asserts that if this condition holds, then $lim_{ntoinfty}a_n=0$. He never claims that this is an equivalence.
        $endgroup$
        – José Carlos Santos
        Dec 23 '18 at 21:26












      • $begingroup$
        That section of the theorem IS NOT part of the if and only if. Only the first section is
        $endgroup$
        – Sandeep Silwal
        Dec 23 '18 at 21:26










      • $begingroup$
        Oh of course. Thank you very much
        $endgroup$
        – mathnoob123
        Dec 23 '18 at 21:29














      2












      2








      2





      $begingroup$

      Yes, it is true that, for every $varepsilon>0$, we have $frac1n<varepsilon$ if $n$ is large enough. But you can't deduce from thath fact that$$sum_{k=n}^mfrac1k<varepsilon$$if $mgeqslant n$ and $m$ and $n$ are large enough.






      share|cite|improve this answer









      $endgroup$



      Yes, it is true that, for every $varepsilon>0$, we have $frac1n<varepsilon$ if $n$ is large enough. But you can't deduce from thath fact that$$sum_{k=n}^mfrac1k<varepsilon$$if $mgeqslant n$ and $m$ and $n$ are large enough.







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered Dec 23 '18 at 21:22









      José Carlos SantosJosé Carlos Santos

      163k22130233




      163k22130233












      • $begingroup$
        Why you cant deduce?
        $endgroup$
        – Jimmy Sabater
        Dec 23 '18 at 21:24










      • $begingroup$
        How so? I am just using the particular backward implication of the theorem which states $$ left|a_n right | leq epsilon ~~~~~ (ngeq N) implies sum a_n ~converges$$. The antecedent is fulfilled by $a_n = frac{1}{n}$
        $endgroup$
        – mathnoob123
        Dec 23 '18 at 21:24








      • 1




        $begingroup$
        You aren't interpreting the theorem correctly. Rudin says that a series $sum_{n=0}^infty a_n$ converges if and only if a certain conditions holds. Then he asserts that if this condition holds, then $lim_{ntoinfty}a_n=0$. He never claims that this is an equivalence.
        $endgroup$
        – José Carlos Santos
        Dec 23 '18 at 21:26












      • $begingroup$
        That section of the theorem IS NOT part of the if and only if. Only the first section is
        $endgroup$
        – Sandeep Silwal
        Dec 23 '18 at 21:26










      • $begingroup$
        Oh of course. Thank you very much
        $endgroup$
        – mathnoob123
        Dec 23 '18 at 21:29


















      • $begingroup$
        Why you cant deduce?
        $endgroup$
        – Jimmy Sabater
        Dec 23 '18 at 21:24










      • $begingroup$
        How so? I am just using the particular backward implication of the theorem which states $$ left|a_n right | leq epsilon ~~~~~ (ngeq N) implies sum a_n ~converges$$. The antecedent is fulfilled by $a_n = frac{1}{n}$
        $endgroup$
        – mathnoob123
        Dec 23 '18 at 21:24








      • 1




        $begingroup$
        You aren't interpreting the theorem correctly. Rudin says that a series $sum_{n=0}^infty a_n$ converges if and only if a certain conditions holds. Then he asserts that if this condition holds, then $lim_{ntoinfty}a_n=0$. He never claims that this is an equivalence.
        $endgroup$
        – José Carlos Santos
        Dec 23 '18 at 21:26












      • $begingroup$
        That section of the theorem IS NOT part of the if and only if. Only the first section is
        $endgroup$
        – Sandeep Silwal
        Dec 23 '18 at 21:26










      • $begingroup$
        Oh of course. Thank you very much
        $endgroup$
        – mathnoob123
        Dec 23 '18 at 21:29
















      $begingroup$
      Why you cant deduce?
      $endgroup$
      – Jimmy Sabater
      Dec 23 '18 at 21:24




      $begingroup$
      Why you cant deduce?
      $endgroup$
      – Jimmy Sabater
      Dec 23 '18 at 21:24












      $begingroup$
      How so? I am just using the particular backward implication of the theorem which states $$ left|a_n right | leq epsilon ~~~~~ (ngeq N) implies sum a_n ~converges$$. The antecedent is fulfilled by $a_n = frac{1}{n}$
      $endgroup$
      – mathnoob123
      Dec 23 '18 at 21:24






      $begingroup$
      How so? I am just using the particular backward implication of the theorem which states $$ left|a_n right | leq epsilon ~~~~~ (ngeq N) implies sum a_n ~converges$$. The antecedent is fulfilled by $a_n = frac{1}{n}$
      $endgroup$
      – mathnoob123
      Dec 23 '18 at 21:24






      1




      1




      $begingroup$
      You aren't interpreting the theorem correctly. Rudin says that a series $sum_{n=0}^infty a_n$ converges if and only if a certain conditions holds. Then he asserts that if this condition holds, then $lim_{ntoinfty}a_n=0$. He never claims that this is an equivalence.
      $endgroup$
      – José Carlos Santos
      Dec 23 '18 at 21:26






      $begingroup$
      You aren't interpreting the theorem correctly. Rudin says that a series $sum_{n=0}^infty a_n$ converges if and only if a certain conditions holds. Then he asserts that if this condition holds, then $lim_{ntoinfty}a_n=0$. He never claims that this is an equivalence.
      $endgroup$
      – José Carlos Santos
      Dec 23 '18 at 21:26














      $begingroup$
      That section of the theorem IS NOT part of the if and only if. Only the first section is
      $endgroup$
      – Sandeep Silwal
      Dec 23 '18 at 21:26




      $begingroup$
      That section of the theorem IS NOT part of the if and only if. Only the first section is
      $endgroup$
      – Sandeep Silwal
      Dec 23 '18 at 21:26












      $begingroup$
      Oh of course. Thank you very much
      $endgroup$
      – mathnoob123
      Dec 23 '18 at 21:29




      $begingroup$
      Oh of course. Thank you very much
      $endgroup$
      – mathnoob123
      Dec 23 '18 at 21:29











      0












      $begingroup$

      In other words,
      let $$S_n=sum_{k=0}^na_k$$



      the criteria is



      $$sum a_ntext{ converges } iff (S_n) text{ is Cauchy} iff$$



      $$forall epsilon>0 ;; exists Nin Bbb N :$$



      $$m>nge N implies |S_m-S_n|<epsilon$$






      share|cite|improve this answer









      $endgroup$


















        0












        $begingroup$

        In other words,
        let $$S_n=sum_{k=0}^na_k$$



        the criteria is



        $$sum a_ntext{ converges } iff (S_n) text{ is Cauchy} iff$$



        $$forall epsilon>0 ;; exists Nin Bbb N :$$



        $$m>nge N implies |S_m-S_n|<epsilon$$






        share|cite|improve this answer









        $endgroup$
















          0












          0








          0





          $begingroup$

          In other words,
          let $$S_n=sum_{k=0}^na_k$$



          the criteria is



          $$sum a_ntext{ converges } iff (S_n) text{ is Cauchy} iff$$



          $$forall epsilon>0 ;; exists Nin Bbb N :$$



          $$m>nge N implies |S_m-S_n|<epsilon$$






          share|cite|improve this answer









          $endgroup$



          In other words,
          let $$S_n=sum_{k=0}^na_k$$



          the criteria is



          $$sum a_ntext{ converges } iff (S_n) text{ is Cauchy} iff$$



          $$forall epsilon>0 ;; exists Nin Bbb N :$$



          $$m>nge N implies |S_m-S_n|<epsilon$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 23 '18 at 21:28









          hamam_Abdallahhamam_Abdallah

          38.1k21634




          38.1k21634























              0












              $begingroup$


              If $mge nge N$ then $left|sum_{k=n}^ma_kright|leepsilon$




              means for all $m$,$n$ such that $mge nge N$. Taking $m=n$ is not equivalent to convergence.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$


                If $mge nge N$ then $left|sum_{k=n}^ma_kright|leepsilon$




                means for all $m$,$n$ such that $mge nge N$. Taking $m=n$ is not equivalent to convergence.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$


                  If $mge nge N$ then $left|sum_{k=n}^ma_kright|leepsilon$




                  means for all $m$,$n$ such that $mge nge N$. Taking $m=n$ is not equivalent to convergence.






                  share|cite|improve this answer









                  $endgroup$




                  If $mge nge N$ then $left|sum_{k=n}^ma_kright|leepsilon$




                  means for all $m$,$n$ such that $mge nge N$. Taking $m=n$ is not equivalent to convergence.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 23 '18 at 21:29









                  A.Γ.A.Γ.

                  22.8k32656




                  22.8k32656






























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