Limit comparison test; is this proof correct, and can it be improved?












1












$begingroup$



Suppose that $forall n: 0< a_n, b_n$ and that $lim_{n to infty} dfrac {a_n}{b_n}=c$ with $c not = 0$. Then



$sum_{n=0}^{infty}a_n$ converges $iff$ $sum_{n=0}^{infty}b_n$ converges




Suppose first that $displaystyle sum_{n=0}^{infty} b_n$ converges.



There exists an $N$ such that $forall (n ge N): left|dfrac {a_n}{b_n} -c right|<1$, so $forall (n ge N): |a_n-cb_n|<b_n$.



Therefore $displaystyle sum_{k=N}^{infty}|a_k-cb_k|< sum_{k=N}^{infty}b_k$. Adding $displaystyle sum_{k=N}^{infty}|cb_k|$ to both sides we get



$displaystylesum_{k=N}^{infty}a_k < sum_{k=N}^{infty}b_k+sum_{k=N}^{infty}cb_k$. This shows that $displaystylesum_{k=N}^{infty}a_k$ converges, so $displaystylesum_{k=1}^{infty}a_k$ converges.



Now for the other direction of the implcation; we can simply use that $lim_{n to infty} dfrac{b_n}{a_n}= dfrac 1c$ (which is OK since $c not = 0$) and do the same proof.






real-analysis sequences-and-series







share|cite|improve this question









$endgroup$












  • $begingroup$
    It seems your proof assumes that $lvert a_k - cb_k rvert + lvert cb_k rvert = lvert a_k rvert$, which I don't believe is necessarily always true.
    $endgroup$
    – John Omielan
    Dec 28 '18 at 18:33








  • 1




    $begingroup$
    $a_k leq (c+1)b_k$ for $kgeq N$, and now use summation. Everything is positive so there shouldn't be an issue
    $endgroup$
    – Jakobian
    Dec 28 '18 at 18:34
















1












$begingroup$



Suppose that $forall n: 0< a_n, b_n$ and that $lim_{n to infty} dfrac {a_n}{b_n}=c$ with $c not = 0$. Then



$sum_{n=0}^{infty}a_n$ converges $iff$ $sum_{n=0}^{infty}b_n$ converges




Suppose first that $displaystyle sum_{n=0}^{infty} b_n$ converges.



There exists an $N$ such that $forall (n ge N): left|dfrac {a_n}{b_n} -c right|<1$, so $forall (n ge N): |a_n-cb_n|<b_n$.



Therefore $displaystyle sum_{k=N}^{infty}|a_k-cb_k|< sum_{k=N}^{infty}b_k$. Adding $displaystyle sum_{k=N}^{infty}|cb_k|$ to both sides we get



$displaystylesum_{k=N}^{infty}a_k < sum_{k=N}^{infty}b_k+sum_{k=N}^{infty}cb_k$. This shows that $displaystylesum_{k=N}^{infty}a_k$ converges, so $displaystylesum_{k=1}^{infty}a_k$ converges.



Now for the other direction of the implcation; we can simply use that $lim_{n to infty} dfrac{b_n}{a_n}= dfrac 1c$ (which is OK since $c not = 0$) and do the same proof.






real-analysis sequences-and-series







share|cite|improve this question









$endgroup$












  • $begingroup$
    It seems your proof assumes that $lvert a_k - cb_k rvert + lvert cb_k rvert = lvert a_k rvert$, which I don't believe is necessarily always true.
    $endgroup$
    – John Omielan
    Dec 28 '18 at 18:33








  • 1




    $begingroup$
    $a_k leq (c+1)b_k$ for $kgeq N$, and now use summation. Everything is positive so there shouldn't be an issue
    $endgroup$
    – Jakobian
    Dec 28 '18 at 18:34














1












1








1





$begingroup$



Suppose that $forall n: 0< a_n, b_n$ and that $lim_{n to infty} dfrac {a_n}{b_n}=c$ with $c not = 0$. Then



$sum_{n=0}^{infty}a_n$ converges $iff$ $sum_{n=0}^{infty}b_n$ converges




Suppose first that $displaystyle sum_{n=0}^{infty} b_n$ converges.



There exists an $N$ such that $forall (n ge N): left|dfrac {a_n}{b_n} -c right|<1$, so $forall (n ge N): |a_n-cb_n|<b_n$.



Therefore $displaystyle sum_{k=N}^{infty}|a_k-cb_k|< sum_{k=N}^{infty}b_k$. Adding $displaystyle sum_{k=N}^{infty}|cb_k|$ to both sides we get



$displaystylesum_{k=N}^{infty}a_k < sum_{k=N}^{infty}b_k+sum_{k=N}^{infty}cb_k$. This shows that $displaystylesum_{k=N}^{infty}a_k$ converges, so $displaystylesum_{k=1}^{infty}a_k$ converges.



Now for the other direction of the implcation; we can simply use that $lim_{n to infty} dfrac{b_n}{a_n}= dfrac 1c$ (which is OK since $c not = 0$) and do the same proof.






real-analysis sequences-and-series







share|cite|improve this question









$endgroup$





Suppose that $forall n: 0< a_n, b_n$ and that $lim_{n to infty} dfrac {a_n}{b_n}=c$ with $c not = 0$. Then



$sum_{n=0}^{infty}a_n$ converges $iff$ $sum_{n=0}^{infty}b_n$ converges




Suppose first that $displaystyle sum_{n=0}^{infty} b_n$ converges.



There exists an $N$ such that $forall (n ge N): left|dfrac {a_n}{b_n} -c right|<1$, so $forall (n ge N): |a_n-cb_n|<b_n$.



Therefore $displaystyle sum_{k=N}^{infty}|a_k-cb_k|< sum_{k=N}^{infty}b_k$. Adding $displaystyle sum_{k=N}^{infty}|cb_k|$ to both sides we get



$displaystylesum_{k=N}^{infty}a_k < sum_{k=N}^{infty}b_k+sum_{k=N}^{infty}cb_k$. This shows that $displaystylesum_{k=N}^{infty}a_k$ converges, so $displaystylesum_{k=1}^{infty}a_k$ converges.



Now for the other direction of the implcation; we can simply use that $lim_{n to infty} dfrac{b_n}{a_n}= dfrac 1c$ (which is OK since $c not = 0$) and do the same proof.






real-analysis sequences-and-series




real-analysis sequences-and-series






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asked Dec 28 '18 at 18:24









OviOvi

12.3k1038112




12.3k1038112












  • $begingroup$
    It seems your proof assumes that $lvert a_k - cb_k rvert + lvert cb_k rvert = lvert a_k rvert$, which I don't believe is necessarily always true.
    $endgroup$
    – John Omielan
    Dec 28 '18 at 18:33








  • 1




    $begingroup$
    $a_k leq (c+1)b_k$ for $kgeq N$, and now use summation. Everything is positive so there shouldn't be an issue
    $endgroup$
    – Jakobian
    Dec 28 '18 at 18:34


















  • $begingroup$
    It seems your proof assumes that $lvert a_k - cb_k rvert + lvert cb_k rvert = lvert a_k rvert$, which I don't believe is necessarily always true.
    $endgroup$
    – John Omielan
    Dec 28 '18 at 18:33








  • 1




    $begingroup$
    $a_k leq (c+1)b_k$ for $kgeq N$, and now use summation. Everything is positive so there shouldn't be an issue
    $endgroup$
    – Jakobian
    Dec 28 '18 at 18:34
















$begingroup$
It seems your proof assumes that $lvert a_k - cb_k rvert + lvert cb_k rvert = lvert a_k rvert$, which I don't believe is necessarily always true.
$endgroup$
– John Omielan
Dec 28 '18 at 18:33






$begingroup$
It seems your proof assumes that $lvert a_k - cb_k rvert + lvert cb_k rvert = lvert a_k rvert$, which I don't believe is necessarily always true.
$endgroup$
– John Omielan
Dec 28 '18 at 18:33






1




1




$begingroup$
$a_k leq (c+1)b_k$ for $kgeq N$, and now use summation. Everything is positive so there shouldn't be an issue
$endgroup$
– Jakobian
Dec 28 '18 at 18:34




$begingroup$
$a_k leq (c+1)b_k$ for $kgeq N$, and now use summation. Everything is positive so there shouldn't be an issue
$endgroup$
– Jakobian
Dec 28 '18 at 18:34










3 Answers
3






active

oldest

votes


















1












$begingroup$


... Therefore $displaystyle sum_{k=N}^{infty}|a_k-cb_k|< sum_{k=N}^{infty}b_k$.




I would use "$le$" instead of "$<$" here.




Adding $displaystyle sum_{k=N}^{infty}|cb_k|$ to both sides we get $displaystylesum_{k=N}^{infty}a_k < sum_{k=N}^{infty}b_k+sum_{k=N}^{infty}cb_k$.




How? I assume that you use the triangle inequality
$$
a_k=(a_k-cb_k)+cb_kleq |a_k-cb_k|+cb_kleq b_k+cb_k.
$$




Now for the other direction of the implcation; we can simply use that $lim_{n to infty} dfrac{b_n}{a_n}= dfrac 1c$ (which is OK since $c not = 0$) and do the same proof.




You don't need to "do the same proof". You only need to apply what you have shown to
$$
tilde{a}_n=b_n,quad tilde{b}_n=a_n,quad tilde{c}=frac1c.
$$






share|cite|improve this answer











$endgroup$





















    1












    $begingroup$

    In your proof, you took $epsilon=1$.



    If we take $epsilon=frac c2$,



    For large $n$,



    $$-frac c2<frac{a_n}{b_n}-c<frac c2$$



    thus



    $$frac c2b_n<a_n<frac{3c}{2}b_n$$



    and comparison test.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      (+1) This is the typical way forward
      $endgroup$
      – Mark Viola
      Dec 28 '18 at 19:12










    • $begingroup$
      @MarkViola Thanks and Happy New Year.
      $endgroup$
      – hamam_Abdallah
      Dec 28 '18 at 19:13










    • $begingroup$
      You're welcome. My pleasure. And Happy Holidays to you also.
      $endgroup$
      – Mark Viola
      Dec 28 '18 at 23:38



















    0












    $begingroup$

    I think you should take what you have and write it "backwards" for the sake of clarity.



    Assume that $sum b_n$ converges. Do your limit definition thing. Now note that:
    $$
    sum a_k = sum|a_k-cb_k+cb_k|leq sum|a_k-cb_k|+sum|cb_k|leq sum b_k+sum cb_k=(1+c)sum b_k
    $$
    Thus, by direct comparison, $sum a_k$ converges.






    share|cite|improve this answer









    $endgroup$













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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

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      active

      oldest

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      active

      oldest

      votes









      1












      $begingroup$


      ... Therefore $displaystyle sum_{k=N}^{infty}|a_k-cb_k|< sum_{k=N}^{infty}b_k$.




      I would use "$le$" instead of "$<$" here.




      Adding $displaystyle sum_{k=N}^{infty}|cb_k|$ to both sides we get $displaystylesum_{k=N}^{infty}a_k < sum_{k=N}^{infty}b_k+sum_{k=N}^{infty}cb_k$.




      How? I assume that you use the triangle inequality
      $$
      a_k=(a_k-cb_k)+cb_kleq |a_k-cb_k|+cb_kleq b_k+cb_k.
      $$




      Now for the other direction of the implcation; we can simply use that $lim_{n to infty} dfrac{b_n}{a_n}= dfrac 1c$ (which is OK since $c not = 0$) and do the same proof.




      You don't need to "do the same proof". You only need to apply what you have shown to
      $$
      tilde{a}_n=b_n,quad tilde{b}_n=a_n,quad tilde{c}=frac1c.
      $$






      share|cite|improve this answer











      $endgroup$


















        1












        $begingroup$


        ... Therefore $displaystyle sum_{k=N}^{infty}|a_k-cb_k|< sum_{k=N}^{infty}b_k$.




        I would use "$le$" instead of "$<$" here.




        Adding $displaystyle sum_{k=N}^{infty}|cb_k|$ to both sides we get $displaystylesum_{k=N}^{infty}a_k < sum_{k=N}^{infty}b_k+sum_{k=N}^{infty}cb_k$.




        How? I assume that you use the triangle inequality
        $$
        a_k=(a_k-cb_k)+cb_kleq |a_k-cb_k|+cb_kleq b_k+cb_k.
        $$




        Now for the other direction of the implcation; we can simply use that $lim_{n to infty} dfrac{b_n}{a_n}= dfrac 1c$ (which is OK since $c not = 0$) and do the same proof.




        You don't need to "do the same proof". You only need to apply what you have shown to
        $$
        tilde{a}_n=b_n,quad tilde{b}_n=a_n,quad tilde{c}=frac1c.
        $$






        share|cite|improve this answer











        $endgroup$
















          1












          1








          1





          $begingroup$


          ... Therefore $displaystyle sum_{k=N}^{infty}|a_k-cb_k|< sum_{k=N}^{infty}b_k$.




          I would use "$le$" instead of "$<$" here.




          Adding $displaystyle sum_{k=N}^{infty}|cb_k|$ to both sides we get $displaystylesum_{k=N}^{infty}a_k < sum_{k=N}^{infty}b_k+sum_{k=N}^{infty}cb_k$.




          How? I assume that you use the triangle inequality
          $$
          a_k=(a_k-cb_k)+cb_kleq |a_k-cb_k|+cb_kleq b_k+cb_k.
          $$




          Now for the other direction of the implcation; we can simply use that $lim_{n to infty} dfrac{b_n}{a_n}= dfrac 1c$ (which is OK since $c not = 0$) and do the same proof.




          You don't need to "do the same proof". You only need to apply what you have shown to
          $$
          tilde{a}_n=b_n,quad tilde{b}_n=a_n,quad tilde{c}=frac1c.
          $$






          share|cite|improve this answer











          $endgroup$




          ... Therefore $displaystyle sum_{k=N}^{infty}|a_k-cb_k|< sum_{k=N}^{infty}b_k$.




          I would use "$le$" instead of "$<$" here.




          Adding $displaystyle sum_{k=N}^{infty}|cb_k|$ to both sides we get $displaystylesum_{k=N}^{infty}a_k < sum_{k=N}^{infty}b_k+sum_{k=N}^{infty}cb_k$.




          How? I assume that you use the triangle inequality
          $$
          a_k=(a_k-cb_k)+cb_kleq |a_k-cb_k|+cb_kleq b_k+cb_k.
          $$




          Now for the other direction of the implcation; we can simply use that $lim_{n to infty} dfrac{b_n}{a_n}= dfrac 1c$ (which is OK since $c not = 0$) and do the same proof.




          You don't need to "do the same proof". You only need to apply what you have shown to
          $$
          tilde{a}_n=b_n,quad tilde{b}_n=a_n,quad tilde{c}=frac1c.
          $$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 28 '18 at 19:07

























          answered Dec 28 '18 at 18:51







          user587192






























              1












              $begingroup$

              In your proof, you took $epsilon=1$.



              If we take $epsilon=frac c2$,



              For large $n$,



              $$-frac c2<frac{a_n}{b_n}-c<frac c2$$



              thus



              $$frac c2b_n<a_n<frac{3c}{2}b_n$$



              and comparison test.






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                (+1) This is the typical way forward
                $endgroup$
                – Mark Viola
                Dec 28 '18 at 19:12










              • $begingroup$
                @MarkViola Thanks and Happy New Year.
                $endgroup$
                – hamam_Abdallah
                Dec 28 '18 at 19:13










              • $begingroup$
                You're welcome. My pleasure. And Happy Holidays to you also.
                $endgroup$
                – Mark Viola
                Dec 28 '18 at 23:38
















              1












              $begingroup$

              In your proof, you took $epsilon=1$.



              If we take $epsilon=frac c2$,



              For large $n$,



              $$-frac c2<frac{a_n}{b_n}-c<frac c2$$



              thus



              $$frac c2b_n<a_n<frac{3c}{2}b_n$$



              and comparison test.






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                (+1) This is the typical way forward
                $endgroup$
                – Mark Viola
                Dec 28 '18 at 19:12










              • $begingroup$
                @MarkViola Thanks and Happy New Year.
                $endgroup$
                – hamam_Abdallah
                Dec 28 '18 at 19:13










              • $begingroup$
                You're welcome. My pleasure. And Happy Holidays to you also.
                $endgroup$
                – Mark Viola
                Dec 28 '18 at 23:38














              1












              1








              1





              $begingroup$

              In your proof, you took $epsilon=1$.



              If we take $epsilon=frac c2$,



              For large $n$,



              $$-frac c2<frac{a_n}{b_n}-c<frac c2$$



              thus



              $$frac c2b_n<a_n<frac{3c}{2}b_n$$



              and comparison test.






              share|cite|improve this answer









              $endgroup$



              In your proof, you took $epsilon=1$.



              If we take $epsilon=frac c2$,



              For large $n$,



              $$-frac c2<frac{a_n}{b_n}-c<frac c2$$



              thus



              $$frac c2b_n<a_n<frac{3c}{2}b_n$$



              and comparison test.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Dec 28 '18 at 19:07









              hamam_Abdallahhamam_Abdallah

              38.1k21634




              38.1k21634












              • $begingroup$
                (+1) This is the typical way forward
                $endgroup$
                – Mark Viola
                Dec 28 '18 at 19:12










              • $begingroup$
                @MarkViola Thanks and Happy New Year.
                $endgroup$
                – hamam_Abdallah
                Dec 28 '18 at 19:13










              • $begingroup$
                You're welcome. My pleasure. And Happy Holidays to you also.
                $endgroup$
                – Mark Viola
                Dec 28 '18 at 23:38


















              • $begingroup$
                (+1) This is the typical way forward
                $endgroup$
                – Mark Viola
                Dec 28 '18 at 19:12










              • $begingroup$
                @MarkViola Thanks and Happy New Year.
                $endgroup$
                – hamam_Abdallah
                Dec 28 '18 at 19:13










              • $begingroup$
                You're welcome. My pleasure. And Happy Holidays to you also.
                $endgroup$
                – Mark Viola
                Dec 28 '18 at 23:38
















              $begingroup$
              (+1) This is the typical way forward
              $endgroup$
              – Mark Viola
              Dec 28 '18 at 19:12




              $begingroup$
              (+1) This is the typical way forward
              $endgroup$
              – Mark Viola
              Dec 28 '18 at 19:12












              $begingroup$
              @MarkViola Thanks and Happy New Year.
              $endgroup$
              – hamam_Abdallah
              Dec 28 '18 at 19:13




              $begingroup$
              @MarkViola Thanks and Happy New Year.
              $endgroup$
              – hamam_Abdallah
              Dec 28 '18 at 19:13












              $begingroup$
              You're welcome. My pleasure. And Happy Holidays to you also.
              $endgroup$
              – Mark Viola
              Dec 28 '18 at 23:38




              $begingroup$
              You're welcome. My pleasure. And Happy Holidays to you also.
              $endgroup$
              – Mark Viola
              Dec 28 '18 at 23:38











              0












              $begingroup$

              I think you should take what you have and write it "backwards" for the sake of clarity.



              Assume that $sum b_n$ converges. Do your limit definition thing. Now note that:
              $$
              sum a_k = sum|a_k-cb_k+cb_k|leq sum|a_k-cb_k|+sum|cb_k|leq sum b_k+sum cb_k=(1+c)sum b_k
              $$
              Thus, by direct comparison, $sum a_k$ converges.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                I think you should take what you have and write it "backwards" for the sake of clarity.



                Assume that $sum b_n$ converges. Do your limit definition thing. Now note that:
                $$
                sum a_k = sum|a_k-cb_k+cb_k|leq sum|a_k-cb_k|+sum|cb_k|leq sum b_k+sum cb_k=(1+c)sum b_k
                $$
                Thus, by direct comparison, $sum a_k$ converges.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  I think you should take what you have and write it "backwards" for the sake of clarity.



                  Assume that $sum b_n$ converges. Do your limit definition thing. Now note that:
                  $$
                  sum a_k = sum|a_k-cb_k+cb_k|leq sum|a_k-cb_k|+sum|cb_k|leq sum b_k+sum cb_k=(1+c)sum b_k
                  $$
                  Thus, by direct comparison, $sum a_k$ converges.






                  share|cite|improve this answer









                  $endgroup$



                  I think you should take what you have and write it "backwards" for the sake of clarity.



                  Assume that $sum b_n$ converges. Do your limit definition thing. Now note that:
                  $$
                  sum a_k = sum|a_k-cb_k+cb_k|leq sum|a_k-cb_k|+sum|cb_k|leq sum b_k+sum cb_k=(1+c)sum b_k
                  $$
                  Thus, by direct comparison, $sum a_k$ converges.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 28 '18 at 18:36









                  DudeManDudeMan

                  1113




                  1113






























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                      Måne

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                      För jordens måne, se Månen. Några månar i solsystemet i skala till vår måne En måne , naturlig satellit eller drabant är en himlakropp som kretsar kring en planet eller en asteroid i ett solsystem. Ordet i bestämd form – månen – avser planeten jordens måne. Det finns inte i något fall belagt att någon måne har egna månar. De flesta stora månar i solsystemet är bundna i synkron rotation med en och samma sida vänd mot planeten. Ett exempel på en måne som inte beter sig på detta vis, är Hyperion som kaotiskt roterar kring Saturnus. En asteroidmåne är en måne som kretsar kring en asteroid. En måne kan ha bildats samtidigt som moderplaneten eller senare infångats av planetens gravitationskraft. I vårt eget solsystem känner vi i nuläget till 173 månar runt sex av de åtta planeterna [ 1 ] : Jorden – månen Mars – Phobos och Deimos Jupiter – Ganymedes (som är solsystemets största måne), Io, Europa, Callisto och 59 andra kända månar Saturnus – Titan (den enda måne...
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                      Storängen

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                      För stadsdelen i Mariehamn, se Storängen, Åland. För ett naturreservat i Norrtälje kommun, se Storängens naturreservat. Storängen från luften i september 1967, vy mot väst. Till vänster syns Järlasjön och Saltsjöbanans spår. Till höger löper Värmdövägen och längst ut till höger syns anläggningsarbetena för Värmdöleden. Storängen är ett villaområde i centrala Nacka i Stockholms län. Området är beläget strax sydost om Nacka Forum vid Saltsjöbanan mellan Lillängen och Saltsjö-Duvnäs. Storängens villasamhälle grundades 1904 av "Tjänstemännens Egnahemsförening vid Storängen" och de första villorna började byggas samma år. Redan 1909 var 107 tomter bebyggda av totalt runt 160. [ 1 ] År 2013 fanns drygt 150 villor med lika många familjer i området. Egnahemsföreningen existerar fortfarande och fungerar idag som intresseförening för de boende i Storängen. Storängen anses vara ett av Sveriges bäst bevarade villasamhällen från 1900-talets början, tack vare sina tidstypis...
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                      VLT Carioca

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                      VLT Carioca Informações Proprietário Prefeitura do Rio de Janeiro Local [[Rio de Janeiro, RJ]] País {{country data Brasil| country flag2| name = Brasil| variant = | size = }} Tipo de transporte Bonde (Veículo leve sobre trilhos) Número de linhas 2 (+1 em construção) Número de estações 26 (+3 em construção) Tráfego 80 mil passageiros/dia [ 1 ] (dezembro/2018) Website www.vltrio.com.br Funcionamento Início de funcionamento 5 de junho de 2016 (2 anos) Operadora(s) Concessionária do VLT Carioca S.A. Número de veículos 32 Alstom Citadis Comprimento dos veículos 44 m (144 ft) Dados técnicos Extensão do sistema 28 km (17,4 mi) [ 2 ] Frequência 5–10 minutos Bitola Bitola padrão/internacional 1 435 mm (4,71 ft) Eletrificação APS, 750V DC Velocidade média 15 km/h (9,32 mph) [ 2 ] Velocidade máxima 50 km/h (31,1 mph) ...
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