Correctness of the induction step












1












$begingroup$


Prove by induction that for all $ n geq 2$ that the following:



$sum_{i = 1}^n frac{i}{i+1} < frac{n^2}{n+1}$



Looking at the $n+1$ step, is it safe to assume the following ?



$sum_{i = 1}^{n+1} frac{i}{i+1} < frac{(n+1)^2}{n+2} iff frac{n+1}{n+2} leq frac{(n+1)^2}{n+2} - frac{n^2}{n+1}$



Assuming the inequality is true for $n$ we can derive that the inequality for $n+1$ stays true if the growth rate of the left side is lesser or equal to the growth rate of the right side.



We then solve the inequality on the right:



$(n+1)^2 leq (n+1)^3 -n^2(n+2)$



$n^2+2n+1 leq n^3+3n^2+3n+1 -n^3-2n^2$



$0 leq n$



Since the questions assumes $ n geq 2$ q.e.d.






proof-verification inequality induction







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    Prove by induction that for all $ n geq 2$ that the following:



    $sum_{i = 1}^n frac{i}{i+1} < frac{n^2}{n+1}$



    Looking at the $n+1$ step, is it safe to assume the following ?



    $sum_{i = 1}^{n+1} frac{i}{i+1} < frac{(n+1)^2}{n+2} iff frac{n+1}{n+2} leq frac{(n+1)^2}{n+2} - frac{n^2}{n+1}$



    Assuming the inequality is true for $n$ we can derive that the inequality for $n+1$ stays true if the growth rate of the left side is lesser or equal to the growth rate of the right side.



    We then solve the inequality on the right:



    $(n+1)^2 leq (n+1)^3 -n^2(n+2)$



    $n^2+2n+1 leq n^3+3n^2+3n+1 -n^3-2n^2$



    $0 leq n$



    Since the questions assumes $ n geq 2$ q.e.d.






    proof-verification inequality induction







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      Prove by induction that for all $ n geq 2$ that the following:



      $sum_{i = 1}^n frac{i}{i+1} < frac{n^2}{n+1}$



      Looking at the $n+1$ step, is it safe to assume the following ?



      $sum_{i = 1}^{n+1} frac{i}{i+1} < frac{(n+1)^2}{n+2} iff frac{n+1}{n+2} leq frac{(n+1)^2}{n+2} - frac{n^2}{n+1}$



      Assuming the inequality is true for $n$ we can derive that the inequality for $n+1$ stays true if the growth rate of the left side is lesser or equal to the growth rate of the right side.



      We then solve the inequality on the right:



      $(n+1)^2 leq (n+1)^3 -n^2(n+2)$



      $n^2+2n+1 leq n^3+3n^2+3n+1 -n^3-2n^2$



      $0 leq n$



      Since the questions assumes $ n geq 2$ q.e.d.






      proof-verification inequality induction







      share|cite|improve this question











      $endgroup$




      Prove by induction that for all $ n geq 2$ that the following:



      $sum_{i = 1}^n frac{i}{i+1} < frac{n^2}{n+1}$



      Looking at the $n+1$ step, is it safe to assume the following ?



      $sum_{i = 1}^{n+1} frac{i}{i+1} < frac{(n+1)^2}{n+2} iff frac{n+1}{n+2} leq frac{(n+1)^2}{n+2} - frac{n^2}{n+1}$



      Assuming the inequality is true for $n$ we can derive that the inequality for $n+1$ stays true if the growth rate of the left side is lesser or equal to the growth rate of the right side.



      We then solve the inequality on the right:



      $(n+1)^2 leq (n+1)^3 -n^2(n+2)$



      $n^2+2n+1 leq n^3+3n^2+3n+1 -n^3-2n^2$



      $0 leq n$



      Since the questions assumes $ n geq 2$ q.e.d.






      proof-verification inequality induction




      proof-verification inequality induction






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 30 at 10:20









      José Carlos Santos

      171k23132240




      171k23132240










      asked Jan 29 at 19:37









      YuonYuon

      83




      83






















          2 Answers
          2






          active

          oldest

          votes


















          1












          $begingroup$

          No, it is not safe to assume that. Since this is supposed to be an induction proof, what you should assume is that $sum_{i=1}^nfrac i{i+1}<frac{n^2}{n+1}$. Then$$sum_{i=1}^{n+1}frac i{i+1}=left(sum_{i=1}^nfrac i{i+1}right)+frac{n+1}{n+2}<frac{n^2}{n+1}+frac{n+1}{n+2}.$$So, all that remains to be proved is that$$frac{n^2}{n+1}+frac{n+1}{n+2}leqslantfrac{(n+1)^2}{n+2}.$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you ! We do arrive to the exact same inequality, does it mean the way I showed it was not rigorous enough ?
            $endgroup$
            – Yuon
            Jan 29 at 20:17










          • $begingroup$
            Yes, that is my opinion.
            $endgroup$
            – José Carlos Santos
            Jan 30 at 10:19



















          0












          $begingroup$

          Hint: You must show that $$frac{1}{2}+frac{2}{3}+frac{3}{4}+...+frac{n}{n+1}+frac{n+1}{n+2}<frac{(n+1)^2}{n+2}$$ and this is (using your inequality)
          $$frac{n^2}{n+1}+frac{n+1}{n+2}<frac{(n+1)^2}{n+2}$$
          And we get
          $$frac{(n+1)^2}{n+2}-frac{n+1}{n+2}-frac{n^2}{n+1}=frac{n}{(n+1) (n+2)}$$






          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%2f3092624%2fcorrectness-of-the-induction-step%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









            1












            $begingroup$

            No, it is not safe to assume that. Since this is supposed to be an induction proof, what you should assume is that $sum_{i=1}^nfrac i{i+1}<frac{n^2}{n+1}$. Then$$sum_{i=1}^{n+1}frac i{i+1}=left(sum_{i=1}^nfrac i{i+1}right)+frac{n+1}{n+2}<frac{n^2}{n+1}+frac{n+1}{n+2}.$$So, all that remains to be proved is that$$frac{n^2}{n+1}+frac{n+1}{n+2}leqslantfrac{(n+1)^2}{n+2}.$$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Thank you ! We do arrive to the exact same inequality, does it mean the way I showed it was not rigorous enough ?
              $endgroup$
              – Yuon
              Jan 29 at 20:17










            • $begingroup$
              Yes, that is my opinion.
              $endgroup$
              – José Carlos Santos
              Jan 30 at 10:19
















            1












            $begingroup$

            No, it is not safe to assume that. Since this is supposed to be an induction proof, what you should assume is that $sum_{i=1}^nfrac i{i+1}<frac{n^2}{n+1}$. Then$$sum_{i=1}^{n+1}frac i{i+1}=left(sum_{i=1}^nfrac i{i+1}right)+frac{n+1}{n+2}<frac{n^2}{n+1}+frac{n+1}{n+2}.$$So, all that remains to be proved is that$$frac{n^2}{n+1}+frac{n+1}{n+2}leqslantfrac{(n+1)^2}{n+2}.$$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Thank you ! We do arrive to the exact same inequality, does it mean the way I showed it was not rigorous enough ?
              $endgroup$
              – Yuon
              Jan 29 at 20:17










            • $begingroup$
              Yes, that is my opinion.
              $endgroup$
              – José Carlos Santos
              Jan 30 at 10:19














            1












            1








            1





            $begingroup$

            No, it is not safe to assume that. Since this is supposed to be an induction proof, what you should assume is that $sum_{i=1}^nfrac i{i+1}<frac{n^2}{n+1}$. Then$$sum_{i=1}^{n+1}frac i{i+1}=left(sum_{i=1}^nfrac i{i+1}right)+frac{n+1}{n+2}<frac{n^2}{n+1}+frac{n+1}{n+2}.$$So, all that remains to be proved is that$$frac{n^2}{n+1}+frac{n+1}{n+2}leqslantfrac{(n+1)^2}{n+2}.$$






            share|cite|improve this answer









            $endgroup$



            No, it is not safe to assume that. Since this is supposed to be an induction proof, what you should assume is that $sum_{i=1}^nfrac i{i+1}<frac{n^2}{n+1}$. Then$$sum_{i=1}^{n+1}frac i{i+1}=left(sum_{i=1}^nfrac i{i+1}right)+frac{n+1}{n+2}<frac{n^2}{n+1}+frac{n+1}{n+2}.$$So, all that remains to be proved is that$$frac{n^2}{n+1}+frac{n+1}{n+2}leqslantfrac{(n+1)^2}{n+2}.$$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 29 at 19:44









            José Carlos SantosJosé Carlos Santos

            171k23132240




            171k23132240












            • $begingroup$
              Thank you ! We do arrive to the exact same inequality, does it mean the way I showed it was not rigorous enough ?
              $endgroup$
              – Yuon
              Jan 29 at 20:17










            • $begingroup$
              Yes, that is my opinion.
              $endgroup$
              – José Carlos Santos
              Jan 30 at 10:19


















            • $begingroup$
              Thank you ! We do arrive to the exact same inequality, does it mean the way I showed it was not rigorous enough ?
              $endgroup$
              – Yuon
              Jan 29 at 20:17










            • $begingroup$
              Yes, that is my opinion.
              $endgroup$
              – José Carlos Santos
              Jan 30 at 10:19
















            $begingroup$
            Thank you ! We do arrive to the exact same inequality, does it mean the way I showed it was not rigorous enough ?
            $endgroup$
            – Yuon
            Jan 29 at 20:17




            $begingroup$
            Thank you ! We do arrive to the exact same inequality, does it mean the way I showed it was not rigorous enough ?
            $endgroup$
            – Yuon
            Jan 29 at 20:17












            $begingroup$
            Yes, that is my opinion.
            $endgroup$
            – José Carlos Santos
            Jan 30 at 10:19




            $begingroup$
            Yes, that is my opinion.
            $endgroup$
            – José Carlos Santos
            Jan 30 at 10:19











            0












            $begingroup$

            Hint: You must show that $$frac{1}{2}+frac{2}{3}+frac{3}{4}+...+frac{n}{n+1}+frac{n+1}{n+2}<frac{(n+1)^2}{n+2}$$ and this is (using your inequality)
            $$frac{n^2}{n+1}+frac{n+1}{n+2}<frac{(n+1)^2}{n+2}$$
            And we get
            $$frac{(n+1)^2}{n+2}-frac{n+1}{n+2}-frac{n^2}{n+1}=frac{n}{(n+1) (n+2)}$$






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              Hint: You must show that $$frac{1}{2}+frac{2}{3}+frac{3}{4}+...+frac{n}{n+1}+frac{n+1}{n+2}<frac{(n+1)^2}{n+2}$$ and this is (using your inequality)
              $$frac{n^2}{n+1}+frac{n+1}{n+2}<frac{(n+1)^2}{n+2}$$
              And we get
              $$frac{(n+1)^2}{n+2}-frac{n+1}{n+2}-frac{n^2}{n+1}=frac{n}{(n+1) (n+2)}$$






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                Hint: You must show that $$frac{1}{2}+frac{2}{3}+frac{3}{4}+...+frac{n}{n+1}+frac{n+1}{n+2}<frac{(n+1)^2}{n+2}$$ and this is (using your inequality)
                $$frac{n^2}{n+1}+frac{n+1}{n+2}<frac{(n+1)^2}{n+2}$$
                And we get
                $$frac{(n+1)^2}{n+2}-frac{n+1}{n+2}-frac{n^2}{n+1}=frac{n}{(n+1) (n+2)}$$






                share|cite|improve this answer









                $endgroup$



                Hint: You must show that $$frac{1}{2}+frac{2}{3}+frac{3}{4}+...+frac{n}{n+1}+frac{n+1}{n+2}<frac{(n+1)^2}{n+2}$$ and this is (using your inequality)
                $$frac{n^2}{n+1}+frac{n+1}{n+2}<frac{(n+1)^2}{n+2}$$
                And we get
                $$frac{(n+1)^2}{n+2}-frac{n+1}{n+2}-frac{n^2}{n+1}=frac{n}{(n+1) (n+2)}$$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 29 at 19:45









                Dr. Sonnhard GraubnerDr. Sonnhard Graubner

                78.4k42867




                78.4k42867






























                    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%2f3092624%2fcorrectness-of-the-induction-step%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