Can someone clarify the logic behind the given equation?Recurrence relation: verify that the expression given...

How can I open an app using Terminal?

How do I go from 300 unfinished/half written blog posts, to published posts?

What deity do celestials/aasimars worship?

Class Action - which options I have?

Applicability of Single Responsibility Principle

Does The Brexit Deal Have To Be Agreed By Both Houses?

How is it posible to add a double into an ArrayList of Integer? (Java)

Why use "finir par" instead of "finir de" before an infinitive?

Can "Reverse Gravity" affect Meteor Swarm?

How can a function with a hole (removable discontinuity) equal a function with no hole?

ls Ordering[Ordering[list]] optimal?

How do scammers retract money, while you can’t?

How to safely derail a train during transit?

How do we know the LHC results are robust?

Magento 2 custom phtml file not calling to all products view pages

How to Reset Passwords on Multiple Websites Easily?

Why didn't Theresa May consult with Parliament before negotiating a deal with the EU?

How does the UK government determine the size of a mandate?

Go Pregnant or Go Home

How to write papers efficiently when English isn't my first language?

Customer Requests (Sometimes) Drive Me Bonkers!

Trouble understanding the speech of overseas colleagues

Rotate a column

Is a stroke of luck acceptable after a series of unfavorable events?



Can someone clarify the logic behind the given equation?


Recurrence relation: verify that the expression given for $a_n$Could someone help me clarify the steps for this solution?What is the logic behind this sequence limit proof?Understanding how to use $epsilon-delta$ definition of a limitProof question about infinite seriesHow can I tell if the sequence re-cycles?Show convergence for $a_n = frac{a_{n-1}}{2} + 1$ with $a_0 = 0$Proving a sequence converges $epsilon-N$Proof that convergent sequence in $Bbb R$ is bounded.Given the following non-increasing sequences of sets…













1












$begingroup$


I have a question relating to image that I've attached. It is a proof that the sequence is increasing. I don't understand the logic behind the third equation $$frac{a_{n+1}}{a_n}>left (1-frac{1}{n+1}right ) left (frac{n+1}{n}right)$$



where does the equation in the first and second parenthesis come from?



Ok, I have another relating question:



why $frac{a_{n+1}}{a_n}> (1+frac{1}{n})$ ? ( The expression of third line).



[![




  • enter image description here


]1]1



!The proof[1]






sequences-and-series limits eulers-constant







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    I have a question relating to image that I've attached. It is a proof that the sequence is increasing. I don't understand the logic behind the third equation $$frac{a_{n+1}}{a_n}>left (1-frac{1}{n+1}right ) left (frac{n+1}{n}right)$$



    where does the equation in the first and second parenthesis come from?



    Ok, I have another relating question:



    why $frac{a_{n+1}}{a_n}> (1+frac{1}{n})$ ? ( The expression of third line).



    [![




    • enter image description here


    ]1]1



    !The proof[1]






    sequences-and-series limits eulers-constant







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      I have a question relating to image that I've attached. It is a proof that the sequence is increasing. I don't understand the logic behind the third equation $$frac{a_{n+1}}{a_n}>left (1-frac{1}{n+1}right ) left (frac{n+1}{n}right)$$



      where does the equation in the first and second parenthesis come from?



      Ok, I have another relating question:



      why $frac{a_{n+1}}{a_n}> (1+frac{1}{n})$ ? ( The expression of third line).



      [![




      • enter image description here


      ]1]1



      !The proof[1]






      sequences-and-series limits eulers-constant







      share|cite|improve this question











      $endgroup$




      I have a question relating to image that I've attached. It is a proof that the sequence is increasing. I don't understand the logic behind the third equation $$frac{a_{n+1}}{a_n}>left (1-frac{1}{n+1}right ) left (frac{n+1}{n}right)$$



      where does the equation in the first and second parenthesis come from?



      Ok, I have another relating question:



      why $frac{a_{n+1}}{a_n}> (1+frac{1}{n})$ ? ( The expression of third line).



      [![




      • enter image description here


      ]1]1



      !The proof[1]






      sequences-and-series limits eulers-constant




      sequences-and-series limits eulers-constant






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited 14 mins ago







      Ieva Brakmane

















      asked 46 mins ago









      Ieva BrakmaneIeva Brakmane

      265




      265






















          3 Answers
          3






          active

          oldest

          votes


















          4












          $begingroup$

          From Bernoulli's inequality, we have



          $$left( 1- frac{1}{(n+1)^2}right) > 1+(n+1) left(frac{-1}{(n+1)^2} right)=1-frac1{n+1}$$



          Hence,



          $$frac{a_{n+1}}{a_n}>left( 1- frac{1}{(n+1)^2}right)left( frac{n+1}{n}right)>left(1-frac1{n+1} right)left( frac{n+1}{n}right)$$






          share|cite|improve this answer









          $endgroup$





















            2












            $begingroup$

            It is putting together the result from the first red box with the second one:




            • $frac{a_{n+1}}{a_n} = color{blue}{left(1- frac{1}{(n+1)^2} right)^{n+1}}left( frac{n+1}{n}right)$

            • $color{blue}{left(1- frac{1}{(n+1)^2} right)^{n+1}} > color{green}{1 + (n+1)left( frac{-1}{(n+1)^2}right)}$


            $$Rightarrow frac{a_{n+1}}{a_n} > left(color{green}{1 + (n+1)left( frac{-1}{(n+1)^2}right)}right)left( frac{n+1}{n}right) = left(underbrace{1- frac{1}{n+1}}_{=frac{n}{n+1}}right)left( frac{n+1}{n}right)$$






            share|cite|improve this answer









            $endgroup$





















              1












              $begingroup$

              So, we have
              $$frac{a_{n+1}}{a_n} = left(1 - frac{1}{(n+1)^2}right)^{n+1}left(frac{n+1}{n}right).$$
              The author then applies Bernoulli's inequality to the first term on the RHS:
              $$left(1 - frac{1}{(n+1)^2}right)^{n+1} > 1 + (n+1)left(frac{-1}{(n+1)^2}right) = 1 - frac{1}{n+1}.$$
              We can now return to the first equation and utilize this estimate; namely, we have
              $$frac{a_{n+1}}{a_n} = left(1 - frac{1}{(n+1)^2}right)^{n+1}left(frac{n+1}{n}right) > left(1-frac{1}{n+1}right)left(frac{n+1}{n}right).$$
              Finally, we multiply out the RHS of the inequality
              $$frac{a_{n+1}}{a_n} > left(1-frac{1}{n+1}right)left(frac{n+1}{n}right) = frac{n+1}{n} - frac{1}{n} = 1.$$
              So, we have
              $$frac{a_{n+1}}{a_n} > 1 implies a_{n+1} > a_n,$$
              which means that ${a_n}$ is an increasing sequence.






              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%2f3165778%2fcan-someone-clarify-the-logic-behind-the-given-equation%23new-answer', 'question_page');
                }
                );

                Post as a guest















                Required, but never shown

























                3 Answers
                3






                active

                oldest

                votes








                3 Answers
                3






                active

                oldest

                votes









                active

                oldest

                votes






                active

                oldest

                votes









                4












                $begingroup$

                From Bernoulli's inequality, we have



                $$left( 1- frac{1}{(n+1)^2}right) > 1+(n+1) left(frac{-1}{(n+1)^2} right)=1-frac1{n+1}$$



                Hence,



                $$frac{a_{n+1}}{a_n}>left( 1- frac{1}{(n+1)^2}right)left( frac{n+1}{n}right)>left(1-frac1{n+1} right)left( frac{n+1}{n}right)$$






                share|cite|improve this answer









                $endgroup$


















                  4












                  $begingroup$

                  From Bernoulli's inequality, we have



                  $$left( 1- frac{1}{(n+1)^2}right) > 1+(n+1) left(frac{-1}{(n+1)^2} right)=1-frac1{n+1}$$



                  Hence,



                  $$frac{a_{n+1}}{a_n}>left( 1- frac{1}{(n+1)^2}right)left( frac{n+1}{n}right)>left(1-frac1{n+1} right)left( frac{n+1}{n}right)$$






                  share|cite|improve this answer









                  $endgroup$
















                    4












                    4








                    4





                    $begingroup$

                    From Bernoulli's inequality, we have



                    $$left( 1- frac{1}{(n+1)^2}right) > 1+(n+1) left(frac{-1}{(n+1)^2} right)=1-frac1{n+1}$$



                    Hence,



                    $$frac{a_{n+1}}{a_n}>left( 1- frac{1}{(n+1)^2}right)left( frac{n+1}{n}right)>left(1-frac1{n+1} right)left( frac{n+1}{n}right)$$






                    share|cite|improve this answer









                    $endgroup$



                    From Bernoulli's inequality, we have



                    $$left( 1- frac{1}{(n+1)^2}right) > 1+(n+1) left(frac{-1}{(n+1)^2} right)=1-frac1{n+1}$$



                    Hence,



                    $$frac{a_{n+1}}{a_n}>left( 1- frac{1}{(n+1)^2}right)left( frac{n+1}{n}right)>left(1-frac1{n+1} right)left( frac{n+1}{n}right)$$







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered 39 mins ago









                    Siong Thye GohSiong Thye Goh

                    103k1468119




                    103k1468119























                        2












                        $begingroup$

                        It is putting together the result from the first red box with the second one:




                        • $frac{a_{n+1}}{a_n} = color{blue}{left(1- frac{1}{(n+1)^2} right)^{n+1}}left( frac{n+1}{n}right)$

                        • $color{blue}{left(1- frac{1}{(n+1)^2} right)^{n+1}} > color{green}{1 + (n+1)left( frac{-1}{(n+1)^2}right)}$


                        $$Rightarrow frac{a_{n+1}}{a_n} > left(color{green}{1 + (n+1)left( frac{-1}{(n+1)^2}right)}right)left( frac{n+1}{n}right) = left(underbrace{1- frac{1}{n+1}}_{=frac{n}{n+1}}right)left( frac{n+1}{n}right)$$






                        share|cite|improve this answer









                        $endgroup$


















                          2












                          $begingroup$

                          It is putting together the result from the first red box with the second one:




                          • $frac{a_{n+1}}{a_n} = color{blue}{left(1- frac{1}{(n+1)^2} right)^{n+1}}left( frac{n+1}{n}right)$

                          • $color{blue}{left(1- frac{1}{(n+1)^2} right)^{n+1}} > color{green}{1 + (n+1)left( frac{-1}{(n+1)^2}right)}$


                          $$Rightarrow frac{a_{n+1}}{a_n} > left(color{green}{1 + (n+1)left( frac{-1}{(n+1)^2}right)}right)left( frac{n+1}{n}right) = left(underbrace{1- frac{1}{n+1}}_{=frac{n}{n+1}}right)left( frac{n+1}{n}right)$$






                          share|cite|improve this answer









                          $endgroup$
















                            2












                            2








                            2





                            $begingroup$

                            It is putting together the result from the first red box with the second one:




                            • $frac{a_{n+1}}{a_n} = color{blue}{left(1- frac{1}{(n+1)^2} right)^{n+1}}left( frac{n+1}{n}right)$

                            • $color{blue}{left(1- frac{1}{(n+1)^2} right)^{n+1}} > color{green}{1 + (n+1)left( frac{-1}{(n+1)^2}right)}$


                            $$Rightarrow frac{a_{n+1}}{a_n} > left(color{green}{1 + (n+1)left( frac{-1}{(n+1)^2}right)}right)left( frac{n+1}{n}right) = left(underbrace{1- frac{1}{n+1}}_{=frac{n}{n+1}}right)left( frac{n+1}{n}right)$$






                            share|cite|improve this answer









                            $endgroup$



                            It is putting together the result from the first red box with the second one:




                            • $frac{a_{n+1}}{a_n} = color{blue}{left(1- frac{1}{(n+1)^2} right)^{n+1}}left( frac{n+1}{n}right)$

                            • $color{blue}{left(1- frac{1}{(n+1)^2} right)^{n+1}} > color{green}{1 + (n+1)left( frac{-1}{(n+1)^2}right)}$


                            $$Rightarrow frac{a_{n+1}}{a_n} > left(color{green}{1 + (n+1)left( frac{-1}{(n+1)^2}right)}right)left( frac{n+1}{n}right) = left(underbrace{1- frac{1}{n+1}}_{=frac{n}{n+1}}right)left( frac{n+1}{n}right)$$







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered 34 mins ago









                            trancelocationtrancelocation

                            13.2k1827




                            13.2k1827























                                1












                                $begingroup$

                                So, we have
                                $$frac{a_{n+1}}{a_n} = left(1 - frac{1}{(n+1)^2}right)^{n+1}left(frac{n+1}{n}right).$$
                                The author then applies Bernoulli's inequality to the first term on the RHS:
                                $$left(1 - frac{1}{(n+1)^2}right)^{n+1} > 1 + (n+1)left(frac{-1}{(n+1)^2}right) = 1 - frac{1}{n+1}.$$
                                We can now return to the first equation and utilize this estimate; namely, we have
                                $$frac{a_{n+1}}{a_n} = left(1 - frac{1}{(n+1)^2}right)^{n+1}left(frac{n+1}{n}right) > left(1-frac{1}{n+1}right)left(frac{n+1}{n}right).$$
                                Finally, we multiply out the RHS of the inequality
                                $$frac{a_{n+1}}{a_n} > left(1-frac{1}{n+1}right)left(frac{n+1}{n}right) = frac{n+1}{n} - frac{1}{n} = 1.$$
                                So, we have
                                $$frac{a_{n+1}}{a_n} > 1 implies a_{n+1} > a_n,$$
                                which means that ${a_n}$ is an increasing sequence.






                                share|cite|improve this answer









                                $endgroup$


















                                  1












                                  $begingroup$

                                  So, we have
                                  $$frac{a_{n+1}}{a_n} = left(1 - frac{1}{(n+1)^2}right)^{n+1}left(frac{n+1}{n}right).$$
                                  The author then applies Bernoulli's inequality to the first term on the RHS:
                                  $$left(1 - frac{1}{(n+1)^2}right)^{n+1} > 1 + (n+1)left(frac{-1}{(n+1)^2}right) = 1 - frac{1}{n+1}.$$
                                  We can now return to the first equation and utilize this estimate; namely, we have
                                  $$frac{a_{n+1}}{a_n} = left(1 - frac{1}{(n+1)^2}right)^{n+1}left(frac{n+1}{n}right) > left(1-frac{1}{n+1}right)left(frac{n+1}{n}right).$$
                                  Finally, we multiply out the RHS of the inequality
                                  $$frac{a_{n+1}}{a_n} > left(1-frac{1}{n+1}right)left(frac{n+1}{n}right) = frac{n+1}{n} - frac{1}{n} = 1.$$
                                  So, we have
                                  $$frac{a_{n+1}}{a_n} > 1 implies a_{n+1} > a_n,$$
                                  which means that ${a_n}$ is an increasing sequence.






                                  share|cite|improve this answer









                                  $endgroup$
















                                    1












                                    1








                                    1





                                    $begingroup$

                                    So, we have
                                    $$frac{a_{n+1}}{a_n} = left(1 - frac{1}{(n+1)^2}right)^{n+1}left(frac{n+1}{n}right).$$
                                    The author then applies Bernoulli's inequality to the first term on the RHS:
                                    $$left(1 - frac{1}{(n+1)^2}right)^{n+1} > 1 + (n+1)left(frac{-1}{(n+1)^2}right) = 1 - frac{1}{n+1}.$$
                                    We can now return to the first equation and utilize this estimate; namely, we have
                                    $$frac{a_{n+1}}{a_n} = left(1 - frac{1}{(n+1)^2}right)^{n+1}left(frac{n+1}{n}right) > left(1-frac{1}{n+1}right)left(frac{n+1}{n}right).$$
                                    Finally, we multiply out the RHS of the inequality
                                    $$frac{a_{n+1}}{a_n} > left(1-frac{1}{n+1}right)left(frac{n+1}{n}right) = frac{n+1}{n} - frac{1}{n} = 1.$$
                                    So, we have
                                    $$frac{a_{n+1}}{a_n} > 1 implies a_{n+1} > a_n,$$
                                    which means that ${a_n}$ is an increasing sequence.






                                    share|cite|improve this answer









                                    $endgroup$



                                    So, we have
                                    $$frac{a_{n+1}}{a_n} = left(1 - frac{1}{(n+1)^2}right)^{n+1}left(frac{n+1}{n}right).$$
                                    The author then applies Bernoulli's inequality to the first term on the RHS:
                                    $$left(1 - frac{1}{(n+1)^2}right)^{n+1} > 1 + (n+1)left(frac{-1}{(n+1)^2}right) = 1 - frac{1}{n+1}.$$
                                    We can now return to the first equation and utilize this estimate; namely, we have
                                    $$frac{a_{n+1}}{a_n} = left(1 - frac{1}{(n+1)^2}right)^{n+1}left(frac{n+1}{n}right) > left(1-frac{1}{n+1}right)left(frac{n+1}{n}right).$$
                                    Finally, we multiply out the RHS of the inequality
                                    $$frac{a_{n+1}}{a_n} > left(1-frac{1}{n+1}right)left(frac{n+1}{n}right) = frac{n+1}{n} - frac{1}{n} = 1.$$
                                    So, we have
                                    $$frac{a_{n+1}}{a_n} > 1 implies a_{n+1} > a_n,$$
                                    which means that ${a_n}$ is an increasing sequence.







                                    share|cite|improve this answer












                                    share|cite|improve this answer



                                    share|cite|improve this answer










                                    answered 20 mins ago









                                    Gary MoonGary Moon

                                    84116




                                    84116






























                                        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%2f3165778%2fcan-someone-clarify-the-logic-behind-the-given-equation%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

                                        is 'sed' thread safeWhat should someone know about using Python scripts in the shell?Nexenta bash script uses...

                                        Image
                                        1970s scifi/horror novel where protagonist is used by a crablike creature to feed its larvae, goes mad, and is defeated by retraumatising him Practical reasons to have both a large police force and bounty hunting network? Is there a frame of reference in which I was born before I was conceived? A peculiar integral identity Is there a way to find out the age of climbing ropes? PTIJ: Aharon, King of Egypt Specific Chinese carabiner QA? Can we carry rice to Japan? Would the melodic leap of the opening phrase of Mozart's K545 be considered dissonant? Can an earth elemental drown/bury its opponent underground using earth glide?
                                        Read more

                                        How do i solve the “ No module named 'mlxtend' ” issue on Jupyter?

                                        Image
                                        Dilemma of explaining to interviewer that he is the reason for declining second interview Meaning of すきっとした How much time does it take for a broken magnet to recover its poles? Can a hotel cancel a confirmed reservation? Removing debris from PCB Using AWS Fargate as web server When does coming up with an idea constitute sufficient contribution for authorship? If a druid in Wild Shape swallows a creature whole, then turns back to her normal form, what happens? How Should I Define/Declare String Constants How to add multiple differently colored borders around a node? Can I retract my name from a
                                        Read more

                                        Pilgersdorf Inhaltsverzeichnis Geografie | Geschichte | Bevölkerungsentwicklung | Politik | Kultur...

                                        Image
                                        Gemeinde im BurgenlandOrt im Bezirk OberpullendorfPilgersdorfErsterwähnung 1225 ungarischkroatischBurgenlandBezirk OberpullendorfÖsterreichZöbernbachKirchschlag in der Buckligen WeltLockenhausKatastralgemeindenNoricumkeltischen Höhensiedlung BurgBurgbergPannoniaLudwigs des DeutschenDominicusRatpotsRihherisMagyarenGüssinger FehdeAlbrechtFranziskanernKanizsayKirchschlagKrumbachTürkenProtestantenSteinbachHaidukenAugustiner-EremitenKuruzenPestVolkszählungGemeinderatBürgermeisterdirektwahlErstenZweiten WeltkriegsDeutschkreutzDraßmarktFrankenau-UnterpullendorfGroßwarasdorfHoritschonKaisersdorfKobersdorfLackenbachLackendorfLockenhausLutzmannsburgMannersdorf an der RabnitzMarkt Sankt MartinNeckenmarktNeutalNikitschOberloisdorfOberpullendorfPilgersdorfPiringsdorfRaidingRitzingSteinberg-DörflStoobUnterfrauenhaidUnterrabnitz-SchwendgrabenWeingrabenWeppersdorf Dieser Artikel erläutert die burgenländische Gemeinde. Siehe auch Pilgersdorf (Gemeinde Lasberg), Oberösterreich bzw. Pilgersdorf,
                                        Read more