Can someone clarify the logic behind the given equation?












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 27 mins ago







      Ieva Brakmane

















      asked 59 mins ago









      Ieva BrakmaneIeva Brakmane

      265




      265






















          3 Answers
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          $begingroup$

          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









              $endgroup$













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                3 Answers
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                active

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                3 Answers
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                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 52 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 47 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 33 mins ago









                                    Gary MoonGary Moon

                                    84116




                                    84116






























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Hall Of Fame””Slayer Wins 'Best Metal' Grammy Award””Slayer Guitarist Jeff Hanneman Dies””Bullet-For My Valentine booed at Metal Hammer Golden Gods Awards””Unholy Aliance””The End Of Slayer?””Slayer: We Could Thrash Out Two More Albums If We're Fast Enough...””'The Unholy Alliance: Chapter III' UK Dates Added”originalet”Megadeth And Slayer To Co-Headline 'Canadian Carnage' Trek”originalet”World Painted Blood””Release “World Painted Blood” by Slayer””Metallica Heading To Cinemas””Slayer, Megadeth To Join Forces For 'European Carnage' Tour - Dec. 18, 2010”originalet”Slayer's Hanneman Contracts Acute Infection; Band To Bring In Guest Guitarist””Cannibal Corpse's Pat O'Brien Will Step In As Slayer's Guest Guitarist”originalet”Slayer’s Jeff Hanneman Dead at 49””Dave Lombardo Says He Made Only $67,000 In 2011 While Touring With Slayer””Slayer: We Do Not Agree With Dave Lombardo's Substance Or Timeline Of Events””Slayer Welcomes Drummer Paul Bostaph Back To The Fold””Slayer Hope to Unveil Never-Before-Heard Jeff Hanneman Material on Next Album””Slayer Debut New Song 'Implode' During Surprise Golden Gods Appearance””Release group Repentless by Slayer””Repentless - Slayer - Credits””Slayer””Metal Storm Awards 2015””Slayer - to release comic book "Repentless #1"””Slayer To Release 'Repentless' 6.66" Vinyl Box Set””BREAKING NEWS: Slayer Announce Farewell Tour””Slayer Recruit Lamb of God, Anthrax, Behemoth + Testament for Final Tour””Slayer lägger ner efter 37 år””Slayer Announces Second North American Leg Of 'Final' Tour””Final World Tour””Slayer Announces Final European Tour With Lamb of God, Anthrax And Obituary””Slayer To Tour Europe With Lamb of God, Anthrax And Obituary””Slayer To Play 'Last French Show Ever' At Next Year's Hellfst””Slayer's Final World Tour Will Extend Into 2019””Death Angel's Rob Cavestany On Slayer's 'Farewell' Tour: 'Some Of Us Could See This Coming'””Testament Has No Plans To Retire Anytime Soon, Says Chuck Billy””Anthrax's Scott Ian On Slayer's 'Farewell' Tour Plans: 'I Was Surprised And I Wasn't Surprised'””Slayer””Slayer's Morbid Schlock””Review/Rock; For Slayer, the Mania Is the Message””Slayer - Biography””Slayer - Reign In Blood”originalet”Dave Lombardo””An exclusive oral history of Slayer”originalet”Exclusive! Interview With Slayer Guitarist Jeff Hanneman”originalet”Thinking Out Loud: Slayer's Kerry King on hair metal, Satan and being polite””Slayer Lyrics””Slayer - Biography””Most influential artists for extreme metal music””Slayer - Reign in Blood””Slayer guitarist Jeff Hanneman dies aged 49””Slatanic Slaughter: A Tribute to Slayer””Gateway to Hell: A Tribute to Slayer””Covered In Blood””Slayer: The Origins of Thrash in San Francisco, CA.””Why They Rule - #6 Slayer”originalet”Guitar World's 100 Greatest Heavy Metal Guitarists Of All Time”originalet”The fans have spoken: Slayer comes out on top in readers' polls”originalet”Tribute to Jeff Hanneman (1964-2013)””Lamb Of God Frontman: We Sound Like A Slayer Rip-Off””BEHEMOTH Frontman Pays Tribute To SLAYER's JEFF HANNEMAN””Slayer, Hatebreed Doing Double Duty On This Year's Ozzfest””System of a Down””Lacuna Coil’s Andrea Ferro Talks Influences, Skateboarding, Band Origins + More””Slayer - Reign in Blood””Into The Lungs of Hell””Slayer rules - en utställning om fans””Slayer and Their Fans Slashed Through a No-Holds-Barred Night at Gas Monkey””Home””Slayer””Gold & Platinum - The Big 4 Live from Sofia, Bulgaria””Exclusive! Interview With Slayer Guitarist Kerry King””2008-02-23: Wiltern, Los Angeles, CA, USA””Slayer's Kerry King To Perform With Megadeth Tonight! - Oct. 21, 2010”originalet”Dave Lombardo - Biography”Slayer Case DismissedArkiveradUltimate Classic Rock: Slayer guitarist Jeff Hanneman dead at 49.”Slayer: "We could never do any thing like Some Kind Of Monster..."””Cannibal Corpse'S Pat O'Brien Will Step In As Slayer'S Guest Guitarist | The Official Slayer Site”originalet”Slayer Wins 'Best Metal' Grammy Award””Slayer Guitarist Jeff Hanneman Dies””Kerrang! Awards 2006 Blog: Kerrang! Hall Of Fame””Kerrang! Awards 2013: Kerrang! Legend”originalet”Metallica, Slayer, Iron Maien Among Winners At Metal Hammer Awards””Metal Hammer Golden Gods Awards””Bullet For My Valentine Booed At Metal Hammer Golden Gods Awards””Metal Storm Awards 2006””Metal Storm Awards 2015””Slayer's Concert History””Slayer - Relationships””Slayer - Releases”Slayers officiella webbplatsSlayer på MusicBrainzOfficiell webbplatsSlayerSlayerr1373445760000 0001 1540 47353068615-5086262726cb13906545x(data)6033143kn20030215029

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                                        SlayerMusikgrupper bildade 1981Amerikanska thrash metal-grupper amerikansktthrash metalHuntington Park1981gitarristernaJeff HannemanKerry KingbasistenTom ArayatrummisenDave LombardoGary HoltPaul Bostaphcertifierats med guldMetallicaAnthraxMegadeththrash metal1986extrema metalgenrerdeath metal22 januari2018USAKanadaEuropaHuntington ParkKalifornien1981gitarristernaJeff HannemanKerry KingbassångDave Lombardoheavy metalbrittiskaIron MaidenJudas PriestthrashMetallicaRattSteeler1982demoundergroundMetal Blade1983high schoolGene HoglanDeathStrapping Young Ladkassettvinyl1984LondonMarquee Club1984Eldorado StudiosTrack Record StudiosLos Angelesmastrade1985CDCombat Recordsextreme metalVenomExodusStudio 54 club1985VHSCombat RecordsBelgienUFORick RubinDef Jam Recordship hoprockakterThe Black CrowesRun–D.M.C.LL Cool JBeastie BoysthrashbpmAngel of DeathRaining BloodAndy WallaceHowie Weinberg1986bildskivaBillboard 200RIAAMetal HammerKerrang!Angela Gossowdeath metalArch Enemythrash metalOverk...
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