How to sum: $sum_{n=0}^{infty} (n^2+2n+1)·(2·ln(x))^n$












0












$begingroup$


How to sum: $$sum_{n=0}^{infty} (n^2+2n+1)·(2·ln(x))^n$$



I've tried considering $$S_N=1+4(2ln(x))+9(2ln(x))^2+...+(N^2+2N+1)(2ln(x))^N$$
and $$(2ln(x))·S_N=1(2ln(x))+4(2ln(x))^2+...+((N-1)^2+2(N-1)+1)(2ln(x))^N+(N^2+2N+1)(2ln(x))^{N+1}$$



So $S_N-(2ln(x))S_N=1+3(2(ln(x))+5(2(ln(x))+...+$



But is this leading me to something? Is it the correct way to sum that?



Assume the series converges ($xin(e^{-1/2},e^{1/2})$)






real-analysis sequences-and-series







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    If you like, first let $y = 2 log x$ and compute the power series, then plug it back. Seems that your method could work, if no calculation errors.
    $endgroup$
    – xbh
    Jan 23 at 17:10












  • $begingroup$
    @xbh thanks. There would be a faster method? It's quite tedious doing it like this.
    $endgroup$
    – iggykimi
    Jan 23 at 17:13
















0












$begingroup$


How to sum: $$sum_{n=0}^{infty} (n^2+2n+1)·(2·ln(x))^n$$



I've tried considering $$S_N=1+4(2ln(x))+9(2ln(x))^2+...+(N^2+2N+1)(2ln(x))^N$$
and $$(2ln(x))·S_N=1(2ln(x))+4(2ln(x))^2+...+((N-1)^2+2(N-1)+1)(2ln(x))^N+(N^2+2N+1)(2ln(x))^{N+1}$$



So $S_N-(2ln(x))S_N=1+3(2(ln(x))+5(2(ln(x))+...+$



But is this leading me to something? Is it the correct way to sum that?



Assume the series converges ($xin(e^{-1/2},e^{1/2})$)






real-analysis sequences-and-series







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    If you like, first let $y = 2 log x$ and compute the power series, then plug it back. Seems that your method could work, if no calculation errors.
    $endgroup$
    – xbh
    Jan 23 at 17:10












  • $begingroup$
    @xbh thanks. There would be a faster method? It's quite tedious doing it like this.
    $endgroup$
    – iggykimi
    Jan 23 at 17:13














0












0








0





$begingroup$


How to sum: $$sum_{n=0}^{infty} (n^2+2n+1)·(2·ln(x))^n$$



I've tried considering $$S_N=1+4(2ln(x))+9(2ln(x))^2+...+(N^2+2N+1)(2ln(x))^N$$
and $$(2ln(x))·S_N=1(2ln(x))+4(2ln(x))^2+...+((N-1)^2+2(N-1)+1)(2ln(x))^N+(N^2+2N+1)(2ln(x))^{N+1}$$



So $S_N-(2ln(x))S_N=1+3(2(ln(x))+5(2(ln(x))+...+$



But is this leading me to something? Is it the correct way to sum that?



Assume the series converges ($xin(e^{-1/2},e^{1/2})$)






real-analysis sequences-and-series







share|cite|improve this question











$endgroup$




How to sum: $$sum_{n=0}^{infty} (n^2+2n+1)·(2·ln(x))^n$$



I've tried considering $$S_N=1+4(2ln(x))+9(2ln(x))^2+...+(N^2+2N+1)(2ln(x))^N$$
and $$(2ln(x))·S_N=1(2ln(x))+4(2ln(x))^2+...+((N-1)^2+2(N-1)+1)(2ln(x))^N+(N^2+2N+1)(2ln(x))^{N+1}$$



So $S_N-(2ln(x))S_N=1+3(2(ln(x))+5(2(ln(x))+...+$



But is this leading me to something? Is it the correct way to sum that?



Assume the series converges ($xin(e^{-1/2},e^{1/2})$)






real-analysis sequences-and-series




real-analysis sequences-and-series






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 23 at 17:11









Martin Sleziak

44.9k10119273




44.9k10119273










asked Jan 23 at 17:08









iggykimiiggykimi

30710




30710








  • 1




    $begingroup$
    If you like, first let $y = 2 log x$ and compute the power series, then plug it back. Seems that your method could work, if no calculation errors.
    $endgroup$
    – xbh
    Jan 23 at 17:10












  • $begingroup$
    @xbh thanks. There would be a faster method? It's quite tedious doing it like this.
    $endgroup$
    – iggykimi
    Jan 23 at 17:13














  • 1




    $begingroup$
    If you like, first let $y = 2 log x$ and compute the power series, then plug it back. Seems that your method could work, if no calculation errors.
    $endgroup$
    – xbh
    Jan 23 at 17:10












  • $begingroup$
    @xbh thanks. There would be a faster method? It's quite tedious doing it like this.
    $endgroup$
    – iggykimi
    Jan 23 at 17:13








1




1




$begingroup$
If you like, first let $y = 2 log x$ and compute the power series, then plug it back. Seems that your method could work, if no calculation errors.
$endgroup$
– xbh
Jan 23 at 17:10






$begingroup$
If you like, first let $y = 2 log x$ and compute the power series, then plug it back. Seems that your method could work, if no calculation errors.
$endgroup$
– xbh
Jan 23 at 17:10














$begingroup$
@xbh thanks. There would be a faster method? It's quite tedious doing it like this.
$endgroup$
– iggykimi
Jan 23 at 17:13




$begingroup$
@xbh thanks. There would be a faster method? It's quite tedious doing it like this.
$endgroup$
– iggykimi
Jan 23 at 17:13










2 Answers
2






active

oldest

votes


















5












$begingroup$

$$sum_{nge0}(n^2+2n+1)y^n = sum_{nge0}(n+2)(n+1)y^n-sum_{nge0}(n+1)y^n = frac{2}{(1-y)^3}-frac{1}{(1-y)}^2$$
Plug $y=2ln x$ in the solution.






share|cite|improve this answer









$endgroup$





















    3












    $begingroup$

    Starting from $frac{1}{1-u}=sum_{n=0}^{infty}u^n$ you get




    • $frac{d}{du}left(frac{u}{1-u}right) = sum_{n=0}^{infty}(n+1)u^n$


    • $Rightarrow frac{d}{du}left(ufrac{d}{du}left(frac{u}{1-u}right)right) = boxed{sum_{n=0}^{infty}(n+1)^2u^n}$ (This is your sum for $u =2 ln x$.)

    • Differentiating gives:
      $$frac{d}{du}left(ufrac{d}{du}left(frac{u}{1-u}right)right) = frac{1+u}{(1-u)^3} stackrel{|ln x| < frac{1}{2}}{Rightarrow} boxed{sum_{n=0}^{infty}(n+1)^2(2ln x)^n = frac{1+2ln x}{(1-2ln x)^3}}$$






    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%2f3084769%2fhow-to-sum-sum-n-0-infty-n22n1-2-lnxn%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









      5












      $begingroup$

      $$sum_{nge0}(n^2+2n+1)y^n = sum_{nge0}(n+2)(n+1)y^n-sum_{nge0}(n+1)y^n = frac{2}{(1-y)^3}-frac{1}{(1-y)}^2$$
      Plug $y=2ln x$ in the solution.






      share|cite|improve this answer









      $endgroup$


















        5












        $begingroup$

        $$sum_{nge0}(n^2+2n+1)y^n = sum_{nge0}(n+2)(n+1)y^n-sum_{nge0}(n+1)y^n = frac{2}{(1-y)^3}-frac{1}{(1-y)}^2$$
        Plug $y=2ln x$ in the solution.






        share|cite|improve this answer









        $endgroup$
















          5












          5








          5





          $begingroup$

          $$sum_{nge0}(n^2+2n+1)y^n = sum_{nge0}(n+2)(n+1)y^n-sum_{nge0}(n+1)y^n = frac{2}{(1-y)^3}-frac{1}{(1-y)}^2$$
          Plug $y=2ln x$ in the solution.






          share|cite|improve this answer









          $endgroup$



          $$sum_{nge0}(n^2+2n+1)y^n = sum_{nge0}(n+2)(n+1)y^n-sum_{nge0}(n+1)y^n = frac{2}{(1-y)^3}-frac{1}{(1-y)}^2$$
          Plug $y=2ln x$ in the solution.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 23 at 17:13









          Nicolas FRANCOISNicolas FRANCOIS

          3,8021516




          3,8021516























              3












              $begingroup$

              Starting from $frac{1}{1-u}=sum_{n=0}^{infty}u^n$ you get




              • $frac{d}{du}left(frac{u}{1-u}right) = sum_{n=0}^{infty}(n+1)u^n$


              • $Rightarrow frac{d}{du}left(ufrac{d}{du}left(frac{u}{1-u}right)right) = boxed{sum_{n=0}^{infty}(n+1)^2u^n}$ (This is your sum for $u =2 ln x$.)

              • Differentiating gives:
                $$frac{d}{du}left(ufrac{d}{du}left(frac{u}{1-u}right)right) = frac{1+u}{(1-u)^3} stackrel{|ln x| < frac{1}{2}}{Rightarrow} boxed{sum_{n=0}^{infty}(n+1)^2(2ln x)^n = frac{1+2ln x}{(1-2ln x)^3}}$$






              share|cite|improve this answer









              $endgroup$


















                3












                $begingroup$

                Starting from $frac{1}{1-u}=sum_{n=0}^{infty}u^n$ you get




                • $frac{d}{du}left(frac{u}{1-u}right) = sum_{n=0}^{infty}(n+1)u^n$


                • $Rightarrow frac{d}{du}left(ufrac{d}{du}left(frac{u}{1-u}right)right) = boxed{sum_{n=0}^{infty}(n+1)^2u^n}$ (This is your sum for $u =2 ln x$.)

                • Differentiating gives:
                  $$frac{d}{du}left(ufrac{d}{du}left(frac{u}{1-u}right)right) = frac{1+u}{(1-u)^3} stackrel{|ln x| < frac{1}{2}}{Rightarrow} boxed{sum_{n=0}^{infty}(n+1)^2(2ln x)^n = frac{1+2ln x}{(1-2ln x)^3}}$$






                share|cite|improve this answer









                $endgroup$
















                  3












                  3








                  3





                  $begingroup$

                  Starting from $frac{1}{1-u}=sum_{n=0}^{infty}u^n$ you get




                  • $frac{d}{du}left(frac{u}{1-u}right) = sum_{n=0}^{infty}(n+1)u^n$


                  • $Rightarrow frac{d}{du}left(ufrac{d}{du}left(frac{u}{1-u}right)right) = boxed{sum_{n=0}^{infty}(n+1)^2u^n}$ (This is your sum for $u =2 ln x$.)

                  • Differentiating gives:
                    $$frac{d}{du}left(ufrac{d}{du}left(frac{u}{1-u}right)right) = frac{1+u}{(1-u)^3} stackrel{|ln x| < frac{1}{2}}{Rightarrow} boxed{sum_{n=0}^{infty}(n+1)^2(2ln x)^n = frac{1+2ln x}{(1-2ln x)^3}}$$






                  share|cite|improve this answer









                  $endgroup$



                  Starting from $frac{1}{1-u}=sum_{n=0}^{infty}u^n$ you get




                  • $frac{d}{du}left(frac{u}{1-u}right) = sum_{n=0}^{infty}(n+1)u^n$


                  • $Rightarrow frac{d}{du}left(ufrac{d}{du}left(frac{u}{1-u}right)right) = boxed{sum_{n=0}^{infty}(n+1)^2u^n}$ (This is your sum for $u =2 ln x$.)

                  • Differentiating gives:
                    $$frac{d}{du}left(ufrac{d}{du}left(frac{u}{1-u}right)right) = frac{1+u}{(1-u)^3} stackrel{|ln x| < frac{1}{2}}{Rightarrow} boxed{sum_{n=0}^{infty}(n+1)^2(2ln x)^n = frac{1+2ln x}{(1-2ln x)^3}}$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 23 at 17:51









                  trancelocationtrancelocation

                  12.6k1826




                  12.6k1826






























                      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%2f3084769%2fhow-to-sum-sum-n-0-infty-n22n1-2-lnxn%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