Limits conceptual proof

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Show that if $a_n geq b_n$ Then $lim_n a_n ge lim_n b_n$ provided the limit exists.




I'm trying to derive a contradiction and it feels obvious but I can't come up with a clean proof?






sequences-and-series limits convergence proof-writing







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  • 1




    Since $a_nge b_n$, then $lim a_n ge lim b_n$ qed.
    – Jimmy R.
    2 hours ago







  • 1




    Try writing the definition of the $lim(a_n)=a$ and then add and subtract $b_n$ and add and subtract $b$ to get $|(a_n-b_n)+(b_n-b)+b-a|<varepsilon$ and then use triangle inequality. I think it might work..
    – Fareed AF
    2 hours ago










  • See related answer math.stackexchange.com/a/1270897/72031 (especially the second part of the answer).
    – Paramanand Singh
    57 mins ago














up vote
1
down vote

favorite













Show that if $a_n geq b_n$ Then $lim_n a_n ge lim_n b_n$ provided the limit exists.




I'm trying to derive a contradiction and it feels obvious but I can't come up with a clean proof?






sequences-and-series limits convergence proof-writing







share|cite|improve this question



















  • 1




    Since $a_nge b_n$, then $lim a_n ge lim b_n$ qed.
    – Jimmy R.
    2 hours ago







  • 1




    Try writing the definition of the $lim(a_n)=a$ and then add and subtract $b_n$ and add and subtract $b$ to get $|(a_n-b_n)+(b_n-b)+b-a|<varepsilon$ and then use triangle inequality. I think it might work..
    – Fareed AF
    2 hours ago










  • See related answer math.stackexchange.com/a/1270897/72031 (especially the second part of the answer).
    – Paramanand Singh
    57 mins ago












up vote
1
down vote

favorite









up vote
1
down vote

favorite












Show that if $a_n geq b_n$ Then $lim_n a_n ge lim_n b_n$ provided the limit exists.




I'm trying to derive a contradiction and it feels obvious but I can't come up with a clean proof?






sequences-and-series limits convergence proof-writing







share|cite|improve this question
















Show that if $a_n geq b_n$ Then $lim_n a_n ge lim_n b_n$ provided the limit exists.




I'm trying to derive a contradiction and it feels obvious but I can't come up with a clean proof?






sequences-and-series limits convergence proof-writing




sequences-and-series limits convergence proof-writing






share|cite|improve this question















share|cite|improve this question













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edited 1 hour ago









Parcly Taxel

40.1k137098




40.1k137098










asked 2 hours ago









Dis-integrating

948325




948325







  • 1




    Since $a_nge b_n$, then $lim a_n ge lim b_n$ qed.
    – Jimmy R.
    2 hours ago







  • 1




    Try writing the definition of the $lim(a_n)=a$ and then add and subtract $b_n$ and add and subtract $b$ to get $|(a_n-b_n)+(b_n-b)+b-a|<varepsilon$ and then use triangle inequality. I think it might work..
    – Fareed AF
    2 hours ago










  • See related answer math.stackexchange.com/a/1270897/72031 (especially the second part of the answer).
    – Paramanand Singh
    57 mins ago












  • 1




    Since $a_nge b_n$, then $lim a_n ge lim b_n$ qed.
    – Jimmy R.
    2 hours ago







  • 1




    Try writing the definition of the $lim(a_n)=a$ and then add and subtract $b_n$ and add and subtract $b$ to get $|(a_n-b_n)+(b_n-b)+b-a|<varepsilon$ and then use triangle inequality. I think it might work..
    – Fareed AF
    2 hours ago










  • See related answer math.stackexchange.com/a/1270897/72031 (especially the second part of the answer).
    – Paramanand Singh
    57 mins ago







1




1




Since $a_nge b_n$, then $lim a_n ge lim b_n$ qed.
– Jimmy R.
2 hours ago





Since $a_nge b_n$, then $lim a_n ge lim b_n$ qed.
– Jimmy R.
2 hours ago





1




1




Try writing the definition of the $lim(a_n)=a$ and then add and subtract $b_n$ and add and subtract $b$ to get $|(a_n-b_n)+(b_n-b)+b-a|<varepsilon$ and then use triangle inequality. I think it might work..
– Fareed AF
2 hours ago




Try writing the definition of the $lim(a_n)=a$ and then add and subtract $b_n$ and add and subtract $b$ to get $|(a_n-b_n)+(b_n-b)+b-a|<varepsilon$ and then use triangle inequality. I think it might work..
– Fareed AF
2 hours ago












See related answer math.stackexchange.com/a/1270897/72031 (especially the second part of the answer).
– Paramanand Singh
57 mins ago




See related answer math.stackexchange.com/a/1270897/72031 (especially the second part of the answer).
– Paramanand Singh
57 mins ago










3 Answers
3






active

oldest

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up vote
2
down vote



accepted










Take $c_n=a_n-b_n geq 0$ and let $a=lim a_n$ and $b=lim b_n$. Then we can define $lim c_n$ as follows:



$$lim c_n=lim a_n-lim b_n=a-b$$



Claim: $a-b geq 0$



Suppose $a-b <0$, then $varepsilon =-(a-b)>0$. Since $c_n rightarrow a-b$, so $$(a-b)-varepsilon <c_n <(a-b)+varepsilon$$ for all $n geq K$. In paricular, $$c_K<(a-b)+varepsilon=(a-b)+(-(a-b))=0$$ contradiction to the fact $c_n geq 0$. Thus $a-b geq 0$ and so $lim a_n geq lim b_n$






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  • thanks, but somehow i always manage to complete problems myself just after asking somebody for help
    – Dis-integrating
    1 hour ago

















up vote
2
down vote













Here's a hint:



For convenience, define $a=lim_ntoinfty a_n$ and $b=lim_ntoinfty b_n$. Suppose that $a_ngeq b_n$ but $a<b$. Define $epsilon=fracb-a2$. For large enough $n$, we have that



$$|a_n-a|<epsilon$$



Similarly, for large enough $n$, we have



$$|b_n-b|<epsilon$$



Using these inequalities, we can derive a contradiction. What is $a_n$ smaller than? What is $b_n$ larger than?






share|cite|improve this answer








New contributor




Josh B. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
























    up vote
    0
    down vote













    So looking online, I think that the first problem from



    http://www.stat.uchicago.edu/~lekheng/courses/104s09/math104s09-hw4sol.pdf is a pretty good resource.



    proof. Let's first define $lim a_n = a$ and $lim b_n = b$. Now for the sake of contradiction, let's suppose that $lim a_n < lim b_n$, or equivalently, $a < b$. We can set $epsilon = fracb-a2>0$.



    By the definition of limit, there exists an $N_1$ such that for all $n>N_1$, we have that



    $(1)$ $|a_n - a| < epsilon$ $iff$ $a-epsilon < a_n < a + epsilon$



    Likewise, there exists an $N_2$ such that for all $n>N_2$, we have that



    $(2)$ $|b_n - b| < epsilon$ $iff$ $b-epsilon < b_n < b+ epsilon$.



    We can let $N = maxN_1, N_2$. From $(1)$ and $(2)$, we get that for $n> N$



    $b_n > b - epsilon = b - fracb-a2 = a + fracb-a2 = a + epsilon > a_n$



    which contradicts the hypothesis given. Thus if $a_n ge b_n$, it follows that $lim a_n ge lim b_n$






    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








      up vote
      2
      down vote



      accepted










      Take $c_n=a_n-b_n geq 0$ and let $a=lim a_n$ and $b=lim b_n$. Then we can define $lim c_n$ as follows:



      $$lim c_n=lim a_n-lim b_n=a-b$$



      Claim: $a-b geq 0$



      Suppose $a-b <0$, then $varepsilon =-(a-b)>0$. Since $c_n rightarrow a-b$, so $$(a-b)-varepsilon <c_n <(a-b)+varepsilon$$ for all $n geq K$. In paricular, $$c_K<(a-b)+varepsilon=(a-b)+(-(a-b))=0$$ contradiction to the fact $c_n geq 0$. Thus $a-b geq 0$ and so $lim a_n geq lim b_n$






      share|cite|improve this answer






















      • thanks, but somehow i always manage to complete problems myself just after asking somebody for help
        – Dis-integrating
        1 hour ago














      up vote
      2
      down vote



      accepted










      Take $c_n=a_n-b_n geq 0$ and let $a=lim a_n$ and $b=lim b_n$. Then we can define $lim c_n$ as follows:



      $$lim c_n=lim a_n-lim b_n=a-b$$



      Claim: $a-b geq 0$



      Suppose $a-b <0$, then $varepsilon =-(a-b)>0$. Since $c_n rightarrow a-b$, so $$(a-b)-varepsilon <c_n <(a-b)+varepsilon$$ for all $n geq K$. In paricular, $$c_K<(a-b)+varepsilon=(a-b)+(-(a-b))=0$$ contradiction to the fact $c_n geq 0$. Thus $a-b geq 0$ and so $lim a_n geq lim b_n$






      share|cite|improve this answer






















      • thanks, but somehow i always manage to complete problems myself just after asking somebody for help
        – Dis-integrating
        1 hour ago












      up vote
      2
      down vote



      accepted







      up vote
      2
      down vote



      accepted






      Take $c_n=a_n-b_n geq 0$ and let $a=lim a_n$ and $b=lim b_n$. Then we can define $lim c_n$ as follows:



      $$lim c_n=lim a_n-lim b_n=a-b$$



      Claim: $a-b geq 0$



      Suppose $a-b <0$, then $varepsilon =-(a-b)>0$. Since $c_n rightarrow a-b$, so $$(a-b)-varepsilon <c_n <(a-b)+varepsilon$$ for all $n geq K$. In paricular, $$c_K<(a-b)+varepsilon=(a-b)+(-(a-b))=0$$ contradiction to the fact $c_n geq 0$. Thus $a-b geq 0$ and so $lim a_n geq lim b_n$






      share|cite|improve this answer














      Take $c_n=a_n-b_n geq 0$ and let $a=lim a_n$ and $b=lim b_n$. Then we can define $lim c_n$ as follows:



      $$lim c_n=lim a_n-lim b_n=a-b$$



      Claim: $a-b geq 0$



      Suppose $a-b <0$, then $varepsilon =-(a-b)>0$. Since $c_n rightarrow a-b$, so $$(a-b)-varepsilon <c_n <(a-b)+varepsilon$$ for all $n geq K$. In paricular, $$c_K<(a-b)+varepsilon=(a-b)+(-(a-b))=0$$ contradiction to the fact $c_n geq 0$. Thus $a-b geq 0$ and so $lim a_n geq lim b_n$







      share|cite|improve this answer














      share|cite|improve this answer



      share|cite|improve this answer








      edited 20 mins ago









      K.M

      518312




      518312










      answered 2 hours ago









      Chinnapparaj R

      4,338725




      4,338725











      • thanks, but somehow i always manage to complete problems myself just after asking somebody for help
        – Dis-integrating
        1 hour ago
















      • thanks, but somehow i always manage to complete problems myself just after asking somebody for help
        – Dis-integrating
        1 hour ago















      thanks, but somehow i always manage to complete problems myself just after asking somebody for help
      – Dis-integrating
      1 hour ago




      thanks, but somehow i always manage to complete problems myself just after asking somebody for help
      – Dis-integrating
      1 hour ago










      up vote
      2
      down vote













      Here's a hint:



      For convenience, define $a=lim_ntoinfty a_n$ and $b=lim_ntoinfty b_n$. Suppose that $a_ngeq b_n$ but $a<b$. Define $epsilon=fracb-a2$. For large enough $n$, we have that



      $$|a_n-a|<epsilon$$



      Similarly, for large enough $n$, we have



      $$|b_n-b|<epsilon$$



      Using these inequalities, we can derive a contradiction. What is $a_n$ smaller than? What is $b_n$ larger than?






      share|cite|improve this answer








      New contributor




      Josh B. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.





















        up vote
        2
        down vote













        Here's a hint:



        For convenience, define $a=lim_ntoinfty a_n$ and $b=lim_ntoinfty b_n$. Suppose that $a_ngeq b_n$ but $a<b$. Define $epsilon=fracb-a2$. For large enough $n$, we have that



        $$|a_n-a|<epsilon$$



        Similarly, for large enough $n$, we have



        $$|b_n-b|<epsilon$$



        Using these inequalities, we can derive a contradiction. What is $a_n$ smaller than? What is $b_n$ larger than?






        share|cite|improve this answer








        New contributor




        Josh B. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
        Check out our Code of Conduct.



















          up vote
          2
          down vote










          up vote
          2
          down vote









          Here's a hint:



          For convenience, define $a=lim_ntoinfty a_n$ and $b=lim_ntoinfty b_n$. Suppose that $a_ngeq b_n$ but $a<b$. Define $epsilon=fracb-a2$. For large enough $n$, we have that



          $$|a_n-a|<epsilon$$



          Similarly, for large enough $n$, we have



          $$|b_n-b|<epsilon$$



          Using these inequalities, we can derive a contradiction. What is $a_n$ smaller than? What is $b_n$ larger than?






          share|cite|improve this answer








          New contributor




          Josh B. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.









          Here's a hint:



          For convenience, define $a=lim_ntoinfty a_n$ and $b=lim_ntoinfty b_n$. Suppose that $a_ngeq b_n$ but $a<b$. Define $epsilon=fracb-a2$. For large enough $n$, we have that



          $$|a_n-a|<epsilon$$



          Similarly, for large enough $n$, we have



          $$|b_n-b|<epsilon$$



          Using these inequalities, we can derive a contradiction. What is $a_n$ smaller than? What is $b_n$ larger than?







          share|cite|improve this answer








          New contributor




          Josh B. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.









          share|cite|improve this answer



          share|cite|improve this answer






          New contributor




          Josh B. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.









          answered 2 hours ago









          Josh B.

          1864




          1864




          New contributor




          Josh B. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.





          New contributor





          Josh B. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.






          Josh B. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.




















              up vote
              0
              down vote













              So looking online, I think that the first problem from



              http://www.stat.uchicago.edu/~lekheng/courses/104s09/math104s09-hw4sol.pdf is a pretty good resource.



              proof. Let's first define $lim a_n = a$ and $lim b_n = b$. Now for the sake of contradiction, let's suppose that $lim a_n < lim b_n$, or equivalently, $a < b$. We can set $epsilon = fracb-a2>0$.



              By the definition of limit, there exists an $N_1$ such that for all $n>N_1$, we have that



              $(1)$ $|a_n - a| < epsilon$ $iff$ $a-epsilon < a_n < a + epsilon$



              Likewise, there exists an $N_2$ such that for all $n>N_2$, we have that



              $(2)$ $|b_n - b| < epsilon$ $iff$ $b-epsilon < b_n < b+ epsilon$.



              We can let $N = maxN_1, N_2$. From $(1)$ and $(2)$, we get that for $n> N$



              $b_n > b - epsilon = b - fracb-a2 = a + fracb-a2 = a + epsilon > a_n$



              which contradicts the hypothesis given. Thus if $a_n ge b_n$, it follows that $lim a_n ge lim b_n$






              share|cite|improve this answer
























                up vote
                0
                down vote













                So looking online, I think that the first problem from



                http://www.stat.uchicago.edu/~lekheng/courses/104s09/math104s09-hw4sol.pdf is a pretty good resource.



                proof. Let's first define $lim a_n = a$ and $lim b_n = b$. Now for the sake of contradiction, let's suppose that $lim a_n < lim b_n$, or equivalently, $a < b$. We can set $epsilon = fracb-a2>0$.



                By the definition of limit, there exists an $N_1$ such that for all $n>N_1$, we have that



                $(1)$ $|a_n - a| < epsilon$ $iff$ $a-epsilon < a_n < a + epsilon$



                Likewise, there exists an $N_2$ such that for all $n>N_2$, we have that



                $(2)$ $|b_n - b| < epsilon$ $iff$ $b-epsilon < b_n < b+ epsilon$.



                We can let $N = maxN_1, N_2$. From $(1)$ and $(2)$, we get that for $n> N$



                $b_n > b - epsilon = b - fracb-a2 = a + fracb-a2 = a + epsilon > a_n$



                which contradicts the hypothesis given. Thus if $a_n ge b_n$, it follows that $lim a_n ge lim b_n$






                share|cite|improve this answer






















                  up vote
                  0
                  down vote










                  up vote
                  0
                  down vote









                  So looking online, I think that the first problem from



                  http://www.stat.uchicago.edu/~lekheng/courses/104s09/math104s09-hw4sol.pdf is a pretty good resource.



                  proof. Let's first define $lim a_n = a$ and $lim b_n = b$. Now for the sake of contradiction, let's suppose that $lim a_n < lim b_n$, or equivalently, $a < b$. We can set $epsilon = fracb-a2>0$.



                  By the definition of limit, there exists an $N_1$ such that for all $n>N_1$, we have that



                  $(1)$ $|a_n - a| < epsilon$ $iff$ $a-epsilon < a_n < a + epsilon$



                  Likewise, there exists an $N_2$ such that for all $n>N_2$, we have that



                  $(2)$ $|b_n - b| < epsilon$ $iff$ $b-epsilon < b_n < b+ epsilon$.



                  We can let $N = maxN_1, N_2$. From $(1)$ and $(2)$, we get that for $n> N$



                  $b_n > b - epsilon = b - fracb-a2 = a + fracb-a2 = a + epsilon > a_n$



                  which contradicts the hypothesis given. Thus if $a_n ge b_n$, it follows that $lim a_n ge lim b_n$






                  share|cite|improve this answer












                  So looking online, I think that the first problem from



                  http://www.stat.uchicago.edu/~lekheng/courses/104s09/math104s09-hw4sol.pdf is a pretty good resource.



                  proof. Let's first define $lim a_n = a$ and $lim b_n = b$. Now for the sake of contradiction, let's suppose that $lim a_n < lim b_n$, or equivalently, $a < b$. We can set $epsilon = fracb-a2>0$.



                  By the definition of limit, there exists an $N_1$ such that for all $n>N_1$, we have that



                  $(1)$ $|a_n - a| < epsilon$ $iff$ $a-epsilon < a_n < a + epsilon$



                  Likewise, there exists an $N_2$ such that for all $n>N_2$, we have that



                  $(2)$ $|b_n - b| < epsilon$ $iff$ $b-epsilon < b_n < b+ epsilon$.



                  We can let $N = maxN_1, N_2$. From $(1)$ and $(2)$, we get that for $n> N$



                  $b_n > b - epsilon = b - fracb-a2 = a + fracb-a2 = a + epsilon > a_n$



                  which contradicts the hypothesis given. Thus if $a_n ge b_n$, it follows that $lim a_n ge lim b_n$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 41 mins ago









                  K.M

                  518312




                  518312



























                       

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