would the proof to show that $1/(n+n^2)$ converges to 0 be the same as for $1/n$












2












$begingroup$


I have started this by doing $mid1/(n+n^2) - 0mid< epsilon$ and I have fixed $epsilon > 0$ which then leads me to $n+n^2 > 1/epsilon$ however I'm not sure if I should leave it like that or make it so only 1 stays on that side. I carried on otherwise and left it like that and by the archimedian property chose a value $N$ where $N>1/epsilon.$ Then for any $n > N$ I have $1/(n+n^2) < 1/N < epsilon.$






real-analysis







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$endgroup$












  • $begingroup$
    ' only 1 n stays on that side '
    $endgroup$
    – Micheal smith
    Nov 27 '18 at 10:30
















2












$begingroup$


I have started this by doing $mid1/(n+n^2) - 0mid< epsilon$ and I have fixed $epsilon > 0$ which then leads me to $n+n^2 > 1/epsilon$ however I'm not sure if I should leave it like that or make it so only 1 stays on that side. I carried on otherwise and left it like that and by the archimedian property chose a value $N$ where $N>1/epsilon.$ Then for any $n > N$ I have $1/(n+n^2) < 1/N < epsilon.$






real-analysis







share|cite|improve this question











$endgroup$












  • $begingroup$
    ' only 1 n stays on that side '
    $endgroup$
    – Micheal smith
    Nov 27 '18 at 10:30














2












2








2





$begingroup$


I have started this by doing $mid1/(n+n^2) - 0mid< epsilon$ and I have fixed $epsilon > 0$ which then leads me to $n+n^2 > 1/epsilon$ however I'm not sure if I should leave it like that or make it so only 1 stays on that side. I carried on otherwise and left it like that and by the archimedian property chose a value $N$ where $N>1/epsilon.$ Then for any $n > N$ I have $1/(n+n^2) < 1/N < epsilon.$






real-analysis







share|cite|improve this question











$endgroup$




I have started this by doing $mid1/(n+n^2) - 0mid< epsilon$ and I have fixed $epsilon > 0$ which then leads me to $n+n^2 > 1/epsilon$ however I'm not sure if I should leave it like that or make it so only 1 stays on that side. I carried on otherwise and left it like that and by the archimedian property chose a value $N$ where $N>1/epsilon.$ Then for any $n > N$ I have $1/(n+n^2) < 1/N < epsilon.$






real-analysis




real-analysis






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share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 27 '18 at 10:37









user376343

3,3933826




3,3933826










asked Nov 27 '18 at 10:29









Micheal smithMicheal smith

112




112












  • $begingroup$
    ' only 1 n stays on that side '
    $endgroup$
    – Micheal smith
    Nov 27 '18 at 10:30


















  • $begingroup$
    ' only 1 n stays on that side '
    $endgroup$
    – Micheal smith
    Nov 27 '18 at 10:30
















$begingroup$
' only 1 n stays on that side '
$endgroup$
– Micheal smith
Nov 27 '18 at 10:30




$begingroup$
' only 1 n stays on that side '
$endgroup$
– Micheal smith
Nov 27 '18 at 10:30










2 Answers
2






active

oldest

votes


















2












$begingroup$

Yes, the proof that $frac1{n+n^2}$ converges can be very similar to the one that $n$ converges. In fact, it will be even simpler, since, in the original proof, you needed to find an $N$ such that $n>frac1epsilon$ for $n>N$, and in this proof, you need to find and $N'$ such that $n+n^2>frac1epsilon$ for $n>N'$. Clearly, since $n+n^2>n$, simply taking the same $N'$ as in the original proof would be sufficient, but even smaller values of $N'$ would also work (but are not required).



An alternative way of proving convergence is to use the fact that $$0leq frac{1}{n+n^2} leq frac 1n$$ and using the squeeze theorem.






share|cite|improve this answer









$endgroup$





















    2












    $begingroup$

    What we could use, if we don't need to use a direct approach, is that



    $$frac1{n+n^2} le frac1n$$



    and then invoke the squeeze theorem.



    Otherwise your way seems fine, from here we can use that



    $$n+n^2 >n >frac1{epsilon}$$



    and then take $N>frac1{epsilon}$.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      im trying to use a direct approach to see is this is provable the same way if you prove 1/n , using the squeeze theorem will not be the same way as in most cases you wont use the squeeze theorem to prove 1/n converges to 0. im just wondering is it ok to leave it as n + n^2 or is that not possible ?
      $endgroup$
      – Micheal smith
      Nov 27 '18 at 10:38










    • $begingroup$
      sorry did not see the continued post you made before i typed that comment in
      $endgroup$
      – Micheal smith
      Nov 27 '18 at 10:38










    • $begingroup$
      @Michealsmith Yes indeed after editing I see better your work on that and you are almost done.
      $endgroup$
      – gimusi
      Nov 27 '18 at 10:40











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    2 Answers
    2






    active

    oldest

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    2 Answers
    2






    active

    oldest

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    active

    oldest

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    active

    oldest

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    2












    $begingroup$

    Yes, the proof that $frac1{n+n^2}$ converges can be very similar to the one that $n$ converges. In fact, it will be even simpler, since, in the original proof, you needed to find an $N$ such that $n>frac1epsilon$ for $n>N$, and in this proof, you need to find and $N'$ such that $n+n^2>frac1epsilon$ for $n>N'$. Clearly, since $n+n^2>n$, simply taking the same $N'$ as in the original proof would be sufficient, but even smaller values of $N'$ would also work (but are not required).



    An alternative way of proving convergence is to use the fact that $$0leq frac{1}{n+n^2} leq frac 1n$$ and using the squeeze theorem.






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      Yes, the proof that $frac1{n+n^2}$ converges can be very similar to the one that $n$ converges. In fact, it will be even simpler, since, in the original proof, you needed to find an $N$ such that $n>frac1epsilon$ for $n>N$, and in this proof, you need to find and $N'$ such that $n+n^2>frac1epsilon$ for $n>N'$. Clearly, since $n+n^2>n$, simply taking the same $N'$ as in the original proof would be sufficient, but even smaller values of $N'$ would also work (but are not required).



      An alternative way of proving convergence is to use the fact that $$0leq frac{1}{n+n^2} leq frac 1n$$ and using the squeeze theorem.






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        Yes, the proof that $frac1{n+n^2}$ converges can be very similar to the one that $n$ converges. In fact, it will be even simpler, since, in the original proof, you needed to find an $N$ such that $n>frac1epsilon$ for $n>N$, and in this proof, you need to find and $N'$ such that $n+n^2>frac1epsilon$ for $n>N'$. Clearly, since $n+n^2>n$, simply taking the same $N'$ as in the original proof would be sufficient, but even smaller values of $N'$ would also work (but are not required).



        An alternative way of proving convergence is to use the fact that $$0leq frac{1}{n+n^2} leq frac 1n$$ and using the squeeze theorem.






        share|cite|improve this answer









        $endgroup$



        Yes, the proof that $frac1{n+n^2}$ converges can be very similar to the one that $n$ converges. In fact, it will be even simpler, since, in the original proof, you needed to find an $N$ such that $n>frac1epsilon$ for $n>N$, and in this proof, you need to find and $N'$ such that $n+n^2>frac1epsilon$ for $n>N'$. Clearly, since $n+n^2>n$, simply taking the same $N'$ as in the original proof would be sufficient, but even smaller values of $N'$ would also work (but are not required).



        An alternative way of proving convergence is to use the fact that $$0leq frac{1}{n+n^2} leq frac 1n$$ and using the squeeze theorem.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 27 '18 at 10:36









        5xum5xum

        90.2k394161




        90.2k394161























            2












            $begingroup$

            What we could use, if we don't need to use a direct approach, is that



            $$frac1{n+n^2} le frac1n$$



            and then invoke the squeeze theorem.



            Otherwise your way seems fine, from here we can use that



            $$n+n^2 >n >frac1{epsilon}$$



            and then take $N>frac1{epsilon}$.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              im trying to use a direct approach to see is this is provable the same way if you prove 1/n , using the squeeze theorem will not be the same way as in most cases you wont use the squeeze theorem to prove 1/n converges to 0. im just wondering is it ok to leave it as n + n^2 or is that not possible ?
              $endgroup$
              – Micheal smith
              Nov 27 '18 at 10:38










            • $begingroup$
              sorry did not see the continued post you made before i typed that comment in
              $endgroup$
              – Micheal smith
              Nov 27 '18 at 10:38










            • $begingroup$
              @Michealsmith Yes indeed after editing I see better your work on that and you are almost done.
              $endgroup$
              – gimusi
              Nov 27 '18 at 10:40
















            2












            $begingroup$

            What we could use, if we don't need to use a direct approach, is that



            $$frac1{n+n^2} le frac1n$$



            and then invoke the squeeze theorem.



            Otherwise your way seems fine, from here we can use that



            $$n+n^2 >n >frac1{epsilon}$$



            and then take $N>frac1{epsilon}$.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              im trying to use a direct approach to see is this is provable the same way if you prove 1/n , using the squeeze theorem will not be the same way as in most cases you wont use the squeeze theorem to prove 1/n converges to 0. im just wondering is it ok to leave it as n + n^2 or is that not possible ?
              $endgroup$
              – Micheal smith
              Nov 27 '18 at 10:38










            • $begingroup$
              sorry did not see the continued post you made before i typed that comment in
              $endgroup$
              – Micheal smith
              Nov 27 '18 at 10:38










            • $begingroup$
              @Michealsmith Yes indeed after editing I see better your work on that and you are almost done.
              $endgroup$
              – gimusi
              Nov 27 '18 at 10:40














            2












            2








            2





            $begingroup$

            What we could use, if we don't need to use a direct approach, is that



            $$frac1{n+n^2} le frac1n$$



            and then invoke the squeeze theorem.



            Otherwise your way seems fine, from here we can use that



            $$n+n^2 >n >frac1{epsilon}$$



            and then take $N>frac1{epsilon}$.






            share|cite|improve this answer











            $endgroup$



            What we could use, if we don't need to use a direct approach, is that



            $$frac1{n+n^2} le frac1n$$



            and then invoke the squeeze theorem.



            Otherwise your way seems fine, from here we can use that



            $$n+n^2 >n >frac1{epsilon}$$



            and then take $N>frac1{epsilon}$.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Nov 27 '18 at 10:44

























            answered Nov 27 '18 at 10:32









            gimusigimusi

            92.8k84494




            92.8k84494












            • $begingroup$
              im trying to use a direct approach to see is this is provable the same way if you prove 1/n , using the squeeze theorem will not be the same way as in most cases you wont use the squeeze theorem to prove 1/n converges to 0. im just wondering is it ok to leave it as n + n^2 or is that not possible ?
              $endgroup$
              – Micheal smith
              Nov 27 '18 at 10:38










            • $begingroup$
              sorry did not see the continued post you made before i typed that comment in
              $endgroup$
              – Micheal smith
              Nov 27 '18 at 10:38










            • $begingroup$
              @Michealsmith Yes indeed after editing I see better your work on that and you are almost done.
              $endgroup$
              – gimusi
              Nov 27 '18 at 10:40


















            • $begingroup$
              im trying to use a direct approach to see is this is provable the same way if you prove 1/n , using the squeeze theorem will not be the same way as in most cases you wont use the squeeze theorem to prove 1/n converges to 0. im just wondering is it ok to leave it as n + n^2 or is that not possible ?
              $endgroup$
              – Micheal smith
              Nov 27 '18 at 10:38










            • $begingroup$
              sorry did not see the continued post you made before i typed that comment in
              $endgroup$
              – Micheal smith
              Nov 27 '18 at 10:38










            • $begingroup$
              @Michealsmith Yes indeed after editing I see better your work on that and you are almost done.
              $endgroup$
              – gimusi
              Nov 27 '18 at 10:40
















            $begingroup$
            im trying to use a direct approach to see is this is provable the same way if you prove 1/n , using the squeeze theorem will not be the same way as in most cases you wont use the squeeze theorem to prove 1/n converges to 0. im just wondering is it ok to leave it as n + n^2 or is that not possible ?
            $endgroup$
            – Micheal smith
            Nov 27 '18 at 10:38




            $begingroup$
            im trying to use a direct approach to see is this is provable the same way if you prove 1/n , using the squeeze theorem will not be the same way as in most cases you wont use the squeeze theorem to prove 1/n converges to 0. im just wondering is it ok to leave it as n + n^2 or is that not possible ?
            $endgroup$
            – Micheal smith
            Nov 27 '18 at 10:38












            $begingroup$
            sorry did not see the continued post you made before i typed that comment in
            $endgroup$
            – Micheal smith
            Nov 27 '18 at 10:38




            $begingroup$
            sorry did not see the continued post you made before i typed that comment in
            $endgroup$
            – Micheal smith
            Nov 27 '18 at 10:38












            $begingroup$
            @Michealsmith Yes indeed after editing I see better your work on that and you are almost done.
            $endgroup$
            – gimusi
            Nov 27 '18 at 10:40




            $begingroup$
            @Michealsmith Yes indeed after editing I see better your work on that and you are almost done.
            $endgroup$
            – gimusi
            Nov 27 '18 at 10:40


















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