Proving the uniform convergence of $sum_{i=1}^{infty}nx^n$ where $x in [0,1)$ using Weirstrass M test.












1












$begingroup$


I want to prove the uniform convergence of $sum_{i=1}^{infty}nx^n$ where $x in [0,1)$. I know this sum has been asked a lot of times before but Im supposed to prove this with tools I have seen and this means not using any Theorem involving derivatives. Im trying to prove this using M Weierstrass test, so I want to bound each term of the sum. This means finding a sequence $lbrace M_{n} rbrace geq 0$ such $|nx^n| leq M_{n}$ for every $n in mathbb{x}$ and then proving $sum_{n=1}^{infty} M_{n}< infty$. Im stuck finding a sequence which each term bound the original sum terms. I know that for every $x in [0,1)$ we have$0 leq x^{n} leq 1$ then $0 leq n x^{n} leq n$ but the series $sum_{i=1}^{infty} n$ is not convergent :(






real-analysis calculus sequences-and-series complex-analysis functional-analysis







share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    I want to prove the uniform convergence of $sum_{i=1}^{infty}nx^n$ where $x in [0,1)$. I know this sum has been asked a lot of times before but Im supposed to prove this with tools I have seen and this means not using any Theorem involving derivatives. Im trying to prove this using M Weierstrass test, so I want to bound each term of the sum. This means finding a sequence $lbrace M_{n} rbrace geq 0$ such $|nx^n| leq M_{n}$ for every $n in mathbb{x}$ and then proving $sum_{n=1}^{infty} M_{n}< infty$. Im stuck finding a sequence which each term bound the original sum terms. I know that for every $x in [0,1)$ we have$0 leq x^{n} leq 1$ then $0 leq n x^{n} leq n$ but the series $sum_{i=1}^{infty} n$ is not convergent :(






    real-analysis calculus sequences-and-series complex-analysis functional-analysis







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      I want to prove the uniform convergence of $sum_{i=1}^{infty}nx^n$ where $x in [0,1)$. I know this sum has been asked a lot of times before but Im supposed to prove this with tools I have seen and this means not using any Theorem involving derivatives. Im trying to prove this using M Weierstrass test, so I want to bound each term of the sum. This means finding a sequence $lbrace M_{n} rbrace geq 0$ such $|nx^n| leq M_{n}$ for every $n in mathbb{x}$ and then proving $sum_{n=1}^{infty} M_{n}< infty$. Im stuck finding a sequence which each term bound the original sum terms. I know that for every $x in [0,1)$ we have$0 leq x^{n} leq 1$ then $0 leq n x^{n} leq n$ but the series $sum_{i=1}^{infty} n$ is not convergent :(






      real-analysis calculus sequences-and-series complex-analysis functional-analysis







      share|cite|improve this question









      $endgroup$




      I want to prove the uniform convergence of $sum_{i=1}^{infty}nx^n$ where $x in [0,1)$. I know this sum has been asked a lot of times before but Im supposed to prove this with tools I have seen and this means not using any Theorem involving derivatives. Im trying to prove this using M Weierstrass test, so I want to bound each term of the sum. This means finding a sequence $lbrace M_{n} rbrace geq 0$ such $|nx^n| leq M_{n}$ for every $n in mathbb{x}$ and then proving $sum_{n=1}^{infty} M_{n}< infty$. Im stuck finding a sequence which each term bound the original sum terms. I know that for every $x in [0,1)$ we have$0 leq x^{n} leq 1$ then $0 leq n x^{n} leq n$ but the series $sum_{i=1}^{infty} n$ is not convergent :(






      real-analysis calculus sequences-and-series complex-analysis functional-analysis




      real-analysis calculus sequences-and-series complex-analysis functional-analysis






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 23 at 3:08









      CosCos

      23027




      23027






















          2 Answers
          2






          active

          oldest

          votes


















          2












          $begingroup$

          Hints:



          Consider separately $[0,a]$ with $a < 1$ where convergence is uniform and $[0,1)$ where it is not.



          For $x in [0,a]$ with $a < 1$ take $x = frac{1}{1+y}$ where $y geqslant frac{1}{a}-1 $



          We have $nx^n = frac{n}{(1+y)^n} $ and by the binomial theorem
          $$(1+y)^n > frac{n(n-1)(n-2)}{6}y^3,$$



          so for $n > 2$, $nx^n < ldots$



          For $x in [0,1)$ note that



          $$sum_{k=n}^{2n} k x^k > (n+1)(n)x^{2n}$$






          share|cite|improve this answer











          $endgroup$





















            0












            $begingroup$

            The series is not uniformly convergent on $[0,1)$. If it is, then $nx^{n}$ must tend to $0$ uniformly as $ n to infty$. In particular there exists $n_0$ such that $|nx^{n}| <1$ for all $n geq n_0$ for all $x in [0,1)$. Take $x=frac 1 {n^{1/n}}$ with $n=n_0$ to get a contradiction .






            share|cite|improve this answer









            $endgroup$













              Your Answer





              StackExchange.ifUsing("editor", function () {
              return StackExchange.using("mathjaxEditing", function () {
              StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
              StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
              });
              });
              }, "mathjax-editing");

              StackExchange.ready(function() {
              var channelOptions = {
              tags: "".split(" "),
              id: "69"
              };
              initTagRenderer("".split(" "), "".split(" "), channelOptions);

              StackExchange.using("externalEditor", function() {
              // Have to fire editor after snippets, if snippets enabled
              if (StackExchange.settings.snippets.snippetsEnabled) {
              StackExchange.using("snippets", function() {
              createEditor();
              });
              }
              else {
              createEditor();
              }
              });

              function createEditor() {
              StackExchange.prepareEditor({
              heartbeatType: 'answer',
              autoActivateHeartbeat: false,
              convertImagesToLinks: true,
              noModals: true,
              showLowRepImageUploadWarning: true,
              reputationToPostImages: 10,
              bindNavPrevention: true,
              postfix: "",
              imageUploader: {
              brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
              contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
              allowUrls: true
              },
              noCode: true, onDemand: true,
              discardSelector: ".discard-answer"
              ,immediatelyShowMarkdownHelp:true
              });


              }
              });














              draft saved

              draft discarded


















              StackExchange.ready(
              function () {
              StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3084029%2fproving-the-uniform-convergence-of-sum-i-1-inftynxn-where-x-in-0-1%23new-answer', 'question_page');
              }
              );

              Post as a guest















              Required, but never shown

























              2 Answers
              2






              active

              oldest

              votes








              2 Answers
              2






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              2












              $begingroup$

              Hints:



              Consider separately $[0,a]$ with $a < 1$ where convergence is uniform and $[0,1)$ where it is not.



              For $x in [0,a]$ with $a < 1$ take $x = frac{1}{1+y}$ where $y geqslant frac{1}{a}-1 $



              We have $nx^n = frac{n}{(1+y)^n} $ and by the binomial theorem
              $$(1+y)^n > frac{n(n-1)(n-2)}{6}y^3,$$



              so for $n > 2$, $nx^n < ldots$



              For $x in [0,1)$ note that



              $$sum_{k=n}^{2n} k x^k > (n+1)(n)x^{2n}$$






              share|cite|improve this answer











              $endgroup$


















                2












                $begingroup$

                Hints:



                Consider separately $[0,a]$ with $a < 1$ where convergence is uniform and $[0,1)$ where it is not.



                For $x in [0,a]$ with $a < 1$ take $x = frac{1}{1+y}$ where $y geqslant frac{1}{a}-1 $



                We have $nx^n = frac{n}{(1+y)^n} $ and by the binomial theorem
                $$(1+y)^n > frac{n(n-1)(n-2)}{6}y^3,$$



                so for $n > 2$, $nx^n < ldots$



                For $x in [0,1)$ note that



                $$sum_{k=n}^{2n} k x^k > (n+1)(n)x^{2n}$$






                share|cite|improve this answer











                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  Hints:



                  Consider separately $[0,a]$ with $a < 1$ where convergence is uniform and $[0,1)$ where it is not.



                  For $x in [0,a]$ with $a < 1$ take $x = frac{1}{1+y}$ where $y geqslant frac{1}{a}-1 $



                  We have $nx^n = frac{n}{(1+y)^n} $ and by the binomial theorem
                  $$(1+y)^n > frac{n(n-1)(n-2)}{6}y^3,$$



                  so for $n > 2$, $nx^n < ldots$



                  For $x in [0,1)$ note that



                  $$sum_{k=n}^{2n} k x^k > (n+1)(n)x^{2n}$$






                  share|cite|improve this answer











                  $endgroup$



                  Hints:



                  Consider separately $[0,a]$ with $a < 1$ where convergence is uniform and $[0,1)$ where it is not.



                  For $x in [0,a]$ with $a < 1$ take $x = frac{1}{1+y}$ where $y geqslant frac{1}{a}-1 $



                  We have $nx^n = frac{n}{(1+y)^n} $ and by the binomial theorem
                  $$(1+y)^n > frac{n(n-1)(n-2)}{6}y^3,$$



                  so for $n > 2$, $nx^n < ldots$



                  For $x in [0,1)$ note that



                  $$sum_{k=n}^{2n} k x^k > (n+1)(n)x^{2n}$$







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Jan 23 at 7:14

























                  answered Jan 23 at 3:25









                  RRLRRL

                  52.5k42573




                  52.5k42573























                      0












                      $begingroup$

                      The series is not uniformly convergent on $[0,1)$. If it is, then $nx^{n}$ must tend to $0$ uniformly as $ n to infty$. In particular there exists $n_0$ such that $|nx^{n}| <1$ for all $n geq n_0$ for all $x in [0,1)$. Take $x=frac 1 {n^{1/n}}$ with $n=n_0$ to get a contradiction .






                      share|cite|improve this answer









                      $endgroup$


















                        0












                        $begingroup$

                        The series is not uniformly convergent on $[0,1)$. If it is, then $nx^{n}$ must tend to $0$ uniformly as $ n to infty$. In particular there exists $n_0$ such that $|nx^{n}| <1$ for all $n geq n_0$ for all $x in [0,1)$. Take $x=frac 1 {n^{1/n}}$ with $n=n_0$ to get a contradiction .






                        share|cite|improve this answer









                        $endgroup$
















                          0












                          0








                          0





                          $begingroup$

                          The series is not uniformly convergent on $[0,1)$. If it is, then $nx^{n}$ must tend to $0$ uniformly as $ n to infty$. In particular there exists $n_0$ such that $|nx^{n}| <1$ for all $n geq n_0$ for all $x in [0,1)$. Take $x=frac 1 {n^{1/n}}$ with $n=n_0$ to get a contradiction .






                          share|cite|improve this answer









                          $endgroup$



                          The series is not uniformly convergent on $[0,1)$. If it is, then $nx^{n}$ must tend to $0$ uniformly as $ n to infty$. In particular there exists $n_0$ such that $|nx^{n}| <1$ for all $n geq n_0$ for all $x in [0,1)$. Take $x=frac 1 {n^{1/n}}$ with $n=n_0$ to get a contradiction .







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Jan 23 at 6:17









                          Kavi Rama MurthyKavi Rama Murthy

                          66.7k52867




                          66.7k52867






























                              draft saved

                              draft discarded




















































                              Thanks for contributing an answer to Mathematics Stack Exchange!


                              • Please be sure to answer the question. Provide details and share your research!

                              But avoid …



                              • Asking for help, clarification, or responding to other answers.

                              • Making statements based on opinion; back them up with references or personal experience.


                              Use MathJax to format equations. MathJax reference.


                              To learn more, see our tips on writing great answers.




                              draft saved


                              draft discarded














                              StackExchange.ready(
                              function () {
                              StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3084029%2fproving-the-uniform-convergence-of-sum-i-1-inftynxn-where-x-in-0-1%23new-answer', 'question_page');
                              }
                              );

                              Post as a guest















                              Required, but never shown





















































                              Required, but never shown














                              Required, but never shown












                              Required, but never shown







                              Required, but never shown

































                              Required, but never shown














                              Required, but never shown












                              Required, but never shown







                              Required, but never shown







                              Popular posts from this blog

                              MongoDB - Not Authorized To Execute Command

                              Image
                              .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 0 I have successfully enabled authorization on MongoDB and I have created an account on the admin database and then I created an account for my database called test. The following connection string to connect to my test database works successfully: mongo --host 192.168.17.52 --port 27017 -u user1 -p password --authenticationDatabase test Only problem I have now is, I cannot execute commands such as: show dbs. I get the following error when I try to do so: "errmsg" : "not authorized on admin to execute command { listDatabases: 1.0, lsid: { id: UUID("a1d5bc0d-bc58-485e-b232-270758a89455") }, $db: "admin...
                              Read more

                              How to fix TextFormField cause rebuild widget in Flutter

                              Image
                              .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 0 I've got the same issue at #9280. I tried all solutions from that topic with/without GlobalKey for Scaffold and Form but the problem wasn't fixed. I tried to build a authentication form with login and signup. But the keyboard doesn't show when I tap the TextFormField at the first time. Then I try to input some letter from physical keyboard, the widget is rebuilt. Like this: login screen Nothing to be log when I run flutter run --verbose, so I logged manually. Every tapping on field, the widget called dispose -> init -> build. [√] Flutter (Channel beta, v1.0.0, on Microsoft Windows [Version 10.0.17763.195], ...
                              Read more

                              in spring boot 2.1 many test slices are not allowed anymore due to multiple @BootstrapWith

                              Image
                              1 I tried to upgrade a yummy sandwich made of two test slices (@JsonTest and @JdbcTest in my case, crunchy test code in between) adding spring boot 2.1 flavour to it. But it seems it was not much of a success. I cannot annotate my tests with many @...Test since they are now each bringing their own XxxTestContextBootstrapper. It used to work when they all used same SpringBootTestContextBootstrapper. @RunWith(SpringRunner.class) @JdbcTest @JsonTest public class Test { @Test public void test() { System.out.printn("Hello, World !"); } } The error I get from BootstrapUtils is illegalStateException : Configuration error: found multiple declarations of @BootstrapWith for test class I understand I might be doing something wrong here but is there an easy way I could load both Json and Jdbc contex...
                              Read more