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












1















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







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  • 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.
    – John Omielan
    Dec 28 '18 at 18:33








  • 1




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
















1















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






















  • 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.
    – John Omielan
    Dec 28 '18 at 18:33








  • 1




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














1












1








1








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














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

12.4k1038111




12.4k1038111












  • 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.
    – John Omielan
    Dec 28 '18 at 18:33








  • 1




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


















  • 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.
    – John Omielan
    Dec 28 '18 at 18:33








  • 1




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
















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.
– John Omielan
Dec 28 '18 at 18:33






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.
– John Omielan
Dec 28 '18 at 18:33






1




1




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




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










3 Answers
3






active

oldest

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1















... 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































    1














    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





















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










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










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



















    0














    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





















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      3 Answers
      3






      active

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      3 Answers
      3






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      active

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      1















      ... 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




























        1















        ... 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


























          1












          1








          1







          ... 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















          ... 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









          user587192user587192

          1,747215




          1,747215























              1














              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





















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










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










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
















              1














              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





















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










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










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














              1












              1








              1






              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












              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

              38k21634




              38k21634












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










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










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


















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










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










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
















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




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












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




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












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




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











              0














              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


























                0














                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
























                  0












                  0








                  0






                  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












                  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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                      Can not write log (Is /dev/pts mounted?) - openpty in Ubuntu-on-Windows?

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                      0 1 What exactly does the error message beginning with E: mean? I assume it has to do with the file structure on Ubuntu-on-Windows, but what exactly? The following package was automatically installed and is no longer required: os-prober Use 'apt-get autoremove' to remove it. The following extra packages will be installed: libxslt1.1 The following NEW packages will be installed: libxslt1.1 xmlstarlet 0 upgraded, 2 newly installed, 0 to remove and 0 not upgraded. Need to get 435 kB of archives. After this operation, 1023 kB of additional disk space will be used. Get:1 http://archive.ubuntu.com/ubuntu/ trusty/main libxslt1.1 amd64 1.1.28-2build1 [145 kB] Get:2 http://archive.ubuntu.com/ubuntu/ trusty/universe xmlstarlet amd64 1.5.0-1 [290 kB] ...
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