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Proving $sum_limitsn=1^inftyfrac1(n+z)^2$ converges uniformly



The 2019 Stack Overflow Developer Survey Results Are InShow that $sum_k=0^infty frac1k^2-z$ converges uniformly by Weierstrass M Test$sumlimits_n=1^inftyfracsin(nx)(n^4+x^4)^frac13$ How to major $sumlimits_n=1^inftysin(nx)$?Study the convergence of $sum_limitsn=1^inftyfrac(-1)^n(x+n)^p$Convergence domain of $sum_limitsn=1^infty(-1)^n+1frac1n^x$?Absolute convergence of $sum_limitsn=1^infty(-1)^n+1e^-nsin x$$sum_limitsn=1^inftyfrac1n!x^n$ and $sum_limitsn=1^inftyfrac1(2n-1)x^n$ different convergence domains?Find the convergence domain of the series $sum_limitsn=1^inftyfrac(-1)^n-1n3^n(x-5)^n$Taking the derivative of $sum_limitsn=1^inftyarctan(fracxn^2)$Studying the convergence of $sum_limitsn=1^inftyfraccos(in)2^n$Studying the convergence of $sum_limitsn=1^inftyfraccosh(ifracpin)}{n^log n$










1












$begingroup$



Using the Weierstrass test show that the series $sum_limitsn=1^inftyfrac1(n+z)^2$ converge uniformly on $E=Re(z)geqslant 1$.




This solution was given to me but I am not understanding certain steps:



$|frac1(n+z)^2|leqslantfrac1^2leqslant frac1leqslantfrac1^2 $



Questions:



Which inequality was used here $frac1^2leqslant frac1$? How can the modulus of $|z|$ be subtracted? What is backing up that step?



Thanks in advance!






sequences-and-series complex-analysis proof-explanation







share|cite|improve this question









$endgroup$
















    1












    $begingroup$



    Using the Weierstrass test show that the series $sum_limitsn=1^inftyfrac1(n+z)^2$ converge uniformly on $E=Re(z)geqslant 1$.




    This solution was given to me but I am not understanding certain steps:



    $|frac1(n+z)^2|leqslantfrac1^2leqslant frac1leqslantfrac1^2 $



    Questions:



    Which inequality was used here $frac1^2leqslant frac1$? How can the modulus of $|z|$ be subtracted? What is backing up that step?



    Thanks in advance!






    sequences-and-series complex-analysis proof-explanation







    share|cite|improve this question









    $endgroup$














      1












      1








      1





      $begingroup$



      Using the Weierstrass test show that the series $sum_limitsn=1^inftyfrac1(n+z)^2$ converge uniformly on $E=Re(z)geqslant 1$.




      This solution was given to me but I am not understanding certain steps:



      $|frac1(n+z)^2|leqslantfrac1^2leqslant frac1leqslantfrac1^2 $



      Questions:



      Which inequality was used here $frac1^2leqslant frac1$? How can the modulus of $|z|$ be subtracted? What is backing up that step?



      Thanks in advance!






      sequences-and-series complex-analysis proof-explanation







      share|cite|improve this question









      $endgroup$





      Using the Weierstrass test show that the series $sum_limitsn=1^inftyfrac1(n+z)^2$ converge uniformly on $E=Re(z)geqslant 1$.




      This solution was given to me but I am not understanding certain steps:



      $|frac1(n+z)^2|leqslantfrac1^2leqslant frac1leqslantfrac1^2 $



      Questions:



      Which inequality was used here $frac1^2leqslant frac1$? How can the modulus of $|z|$ be subtracted? What is backing up that step?



      Thanks in advance!






      sequences-and-series complex-analysis proof-explanation




      sequences-and-series complex-analysis proof-explanation






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Mar 23 at 16:37









      Pedro GomesPedro Gomes

      2,0062821




      2,0062821




















          4 Answers
          4






          active

          oldest

          votes


















          2












          $begingroup$

          The triangle inequality gives
          $$|n+z|ge n-|-z|=n-|z|$$
          and
          $$|n+z|ge |z|-|-n|=-(n-|z|).$$
          In both cases
          $$|n+z|ge|n-|z||$$ and
          so
          $$|n+z|^2ge|n-|z||^2.$$
          Then
          $$frac1^2lefrac1$$
          provided that $|z|ne n$.






          share|cite|improve this answer









          $endgroup$




















            4












            $begingroup$

            HINT:



            Rather than using the proof based on the triangle inequality, $|x+y|ge ||x|-|y||$, simply write



            $$|n+z|^2=n^2+|z|^2+2ntextRe(z)ge n^2+1+2n=(n+1)^2$$






            share|cite|improve this answer









            $endgroup$




















              3












              $begingroup$

              What was used is the reverse triangle inequality $|z-w|geq ||z|-|w||$.






              share|cite|improve this answer









              $endgroup$




















                2












                $begingroup$

                If $mathrmRe,zge 1$, then
                $$
                |z+n|ge |mathrmRe,z+n|=mathrmRe,z+nge n+1
                $$
                and hence
                $$
                left|frac1(z+n)^2right|le frac1(n+1)^2.
                $$
                Comparison Test implies that $sum frac1(z+n)^2$ converges ABSOLUTELY, and since that right hand side does not depend on $z$, then it converges uniformly on $z$, for $mathrmRe,zge 1$.



                To understand this, observe that if
                $$
                s_n(z)=sum_k=1^nfrac1(z+k)^2
                $$
                then
                $$
                |s_m(z)-s_n(z)|le sum_k=n+1^mfrac1z+kle sum_k=n+1^mfrac1(k+1)^2
                $$
                and the right hand side becomes less that $varepsilon$, for $m,nge N$, when $N$ is sufficiently large, and this $N$ clearly depends only on $varepsilon$ and not on $z$.






                share|cite|improve this answer









                $endgroup$













                  Your Answer





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                  4 Answers
                  4






                  active

                  oldest

                  votes








                  4 Answers
                  4






                  active

                  oldest

                  votes









                  active

                  oldest

                  votes






                  active

                  oldest

                  votes









                  2












                  $begingroup$

                  The triangle inequality gives
                  $$|n+z|ge n-|-z|=n-|z|$$
                  and
                  $$|n+z|ge |z|-|-n|=-(n-|z|).$$
                  In both cases
                  $$|n+z|ge|n-|z||$$ and
                  so
                  $$|n+z|^2ge|n-|z||^2.$$
                  Then
                  $$frac1^2lefrac1$$
                  provided that $|z|ne n$.






                  share|cite|improve this answer









                  $endgroup$

















                    2












                    $begingroup$

                    The triangle inequality gives
                    $$|n+z|ge n-|-z|=n-|z|$$
                    and
                    $$|n+z|ge |z|-|-n|=-(n-|z|).$$
                    In both cases
                    $$|n+z|ge|n-|z||$$ and
                    so
                    $$|n+z|^2ge|n-|z||^2.$$
                    Then
                    $$frac1^2lefrac1$$
                    provided that $|z|ne n$.






                    share|cite|improve this answer









                    $endgroup$















                      2












                      2








                      2





                      $begingroup$

                      The triangle inequality gives
                      $$|n+z|ge n-|-z|=n-|z|$$
                      and
                      $$|n+z|ge |z|-|-n|=-(n-|z|).$$
                      In both cases
                      $$|n+z|ge|n-|z||$$ and
                      so
                      $$|n+z|^2ge|n-|z||^2.$$
                      Then
                      $$frac1^2lefrac1$$
                      provided that $|z|ne n$.






                      share|cite|improve this answer









                      $endgroup$



                      The triangle inequality gives
                      $$|n+z|ge n-|-z|=n-|z|$$
                      and
                      $$|n+z|ge |z|-|-n|=-(n-|z|).$$
                      In both cases
                      $$|n+z|ge|n-|z||$$ and
                      so
                      $$|n+z|^2ge|n-|z||^2.$$
                      Then
                      $$frac1^2lefrac1$$
                      provided that $|z|ne n$.







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered Mar 23 at 16:41









                      Lord Shark the UnknownLord Shark the Unknown

                      108k1162136




                      108k1162136





















                          4












                          $begingroup$

                          HINT:



                          Rather than using the proof based on the triangle inequality, $|x+y|ge ||x|-|y||$, simply write



                          $$|n+z|^2=n^2+|z|^2+2ntextRe(z)ge n^2+1+2n=(n+1)^2$$






                          share|cite|improve this answer









                          $endgroup$

















                            4












                            $begingroup$

                            HINT:



                            Rather than using the proof based on the triangle inequality, $|x+y|ge ||x|-|y||$, simply write



                            $$|n+z|^2=n^2+|z|^2+2ntextRe(z)ge n^2+1+2n=(n+1)^2$$






                            share|cite|improve this answer









                            $endgroup$















                              4












                              4








                              4





                              $begingroup$

                              HINT:



                              Rather than using the proof based on the triangle inequality, $|x+y|ge ||x|-|y||$, simply write



                              $$|n+z|^2=n^2+|z|^2+2ntextRe(z)ge n^2+1+2n=(n+1)^2$$






                              share|cite|improve this answer









                              $endgroup$



                              HINT:



                              Rather than using the proof based on the triangle inequality, $|x+y|ge ||x|-|y||$, simply write



                              $$|n+z|^2=n^2+|z|^2+2ntextRe(z)ge n^2+1+2n=(n+1)^2$$







                              share|cite|improve this answer












                              share|cite|improve this answer



                              share|cite|improve this answer










                              answered Mar 23 at 16:44









                              Mark ViolaMark Viola

                              134k1278177




                              134k1278177





















                                  3












                                  $begingroup$

                                  What was used is the reverse triangle inequality $|z-w|geq ||z|-|w||$.






                                  share|cite|improve this answer









                                  $endgroup$

















                                    3












                                    $begingroup$

                                    What was used is the reverse triangle inequality $|z-w|geq ||z|-|w||$.






                                    share|cite|improve this answer









                                    $endgroup$















                                      3












                                      3








                                      3





                                      $begingroup$

                                      What was used is the reverse triangle inequality $|z-w|geq ||z|-|w||$.






                                      share|cite|improve this answer









                                      $endgroup$



                                      What was used is the reverse triangle inequality $|z-w|geq ||z|-|w||$.







                                      share|cite|improve this answer












                                      share|cite|improve this answer



                                      share|cite|improve this answer










                                      answered Mar 23 at 16:41









                                      MarkMark

                                      10.5k1622




                                      10.5k1622





















                                          2












                                          $begingroup$

                                          If $mathrmRe,zge 1$, then
                                          $$
                                          |z+n|ge |mathrmRe,z+n|=mathrmRe,z+nge n+1
                                          $$
                                          and hence
                                          $$
                                          left|frac1(z+n)^2right|le frac1(n+1)^2.
                                          $$
                                          Comparison Test implies that $sum frac1(z+n)^2$ converges ABSOLUTELY, and since that right hand side does not depend on $z$, then it converges uniformly on $z$, for $mathrmRe,zge 1$.



                                          To understand this, observe that if
                                          $$
                                          s_n(z)=sum_k=1^nfrac1(z+k)^2
                                          $$
                                          then
                                          $$
                                          |s_m(z)-s_n(z)|le sum_k=n+1^mfrac1z+kle sum_k=n+1^mfrac1(k+1)^2
                                          $$
                                          and the right hand side becomes less that $varepsilon$, for $m,nge N$, when $N$ is sufficiently large, and this $N$ clearly depends only on $varepsilon$ and not on $z$.






                                          share|cite|improve this answer









                                          $endgroup$

















                                            2












                                            $begingroup$

                                            If $mathrmRe,zge 1$, then
                                            $$
                                            |z+n|ge |mathrmRe,z+n|=mathrmRe,z+nge n+1
                                            $$
                                            and hence
                                            $$
                                            left|frac1(z+n)^2right|le frac1(n+1)^2.
                                            $$
                                            Comparison Test implies that $sum frac1(z+n)^2$ converges ABSOLUTELY, and since that right hand side does not depend on $z$, then it converges uniformly on $z$, for $mathrmRe,zge 1$.



                                            To understand this, observe that if
                                            $$
                                            s_n(z)=sum_k=1^nfrac1(z+k)^2
                                            $$
                                            then
                                            $$
                                            |s_m(z)-s_n(z)|le sum_k=n+1^mfrac1z+kle sum_k=n+1^mfrac1(k+1)^2
                                            $$
                                            and the right hand side becomes less that $varepsilon$, for $m,nge N$, when $N$ is sufficiently large, and this $N$ clearly depends only on $varepsilon$ and not on $z$.






                                            share|cite|improve this answer









                                            $endgroup$















                                              2












                                              2








                                              2





                                              $begingroup$

                                              If $mathrmRe,zge 1$, then
                                              $$
                                              |z+n|ge |mathrmRe,z+n|=mathrmRe,z+nge n+1
                                              $$
                                              and hence
                                              $$
                                              left|frac1(z+n)^2right|le frac1(n+1)^2.
                                              $$
                                              Comparison Test implies that $sum frac1(z+n)^2$ converges ABSOLUTELY, and since that right hand side does not depend on $z$, then it converges uniformly on $z$, for $mathrmRe,zge 1$.



                                              To understand this, observe that if
                                              $$
                                              s_n(z)=sum_k=1^nfrac1(z+k)^2
                                              $$
                                              then
                                              $$
                                              |s_m(z)-s_n(z)|le sum_k=n+1^mfrac1z+kle sum_k=n+1^mfrac1(k+1)^2
                                              $$
                                              and the right hand side becomes less that $varepsilon$, for $m,nge N$, when $N$ is sufficiently large, and this $N$ clearly depends only on $varepsilon$ and not on $z$.






                                              share|cite|improve this answer









                                              $endgroup$



                                              If $mathrmRe,zge 1$, then
                                              $$
                                              |z+n|ge |mathrmRe,z+n|=mathrmRe,z+nge n+1
                                              $$
                                              and hence
                                              $$
                                              left|frac1(z+n)^2right|le frac1(n+1)^2.
                                              $$
                                              Comparison Test implies that $sum frac1(z+n)^2$ converges ABSOLUTELY, and since that right hand side does not depend on $z$, then it converges uniformly on $z$, for $mathrmRe,zge 1$.



                                              To understand this, observe that if
                                              $$
                                              s_n(z)=sum_k=1^nfrac1(z+k)^2
                                              $$
                                              then
                                              $$
                                              |s_m(z)-s_n(z)|le sum_k=n+1^mfrac1z+kle sum_k=n+1^mfrac1(k+1)^2
                                              $$
                                              and the right hand side becomes less that $varepsilon$, for $m,nge N$, when $N$ is sufficiently large, and this $N$ clearly depends only on $varepsilon$ and not on $z$.







                                              share|cite|improve this answer












                                              share|cite|improve this answer



                                              share|cite|improve this answer










                                              answered Mar 23 at 16:47









                                              Yiorgos S. SmyrlisYiorgos S. Smyrlis

                                              63.7k1385165




                                              63.7k1385165



























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