Can someone clarify the logic behind the given equation?












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$begingroup$


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
                3






                active

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                active

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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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