Can someone shed some light on this inequality?












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.



!The proof[1]






sequences-and-series limits eulers-constant







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    Please do not post necessary information only in a picture, not everyone can display and read it properly.
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    – Carsten S
    Mar 28 at 16:04
















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.



!The proof[1]






sequences-and-series limits eulers-constant







share|cite|improve this question











$endgroup$












  • $begingroup$
    Please do not post necessary information only in a picture, not everyone can display and read it properly.
    $endgroup$
    – Carsten S
    Mar 28 at 16:04














1












1








1


2



$begingroup$


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.



!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.



!The proof[1]






sequences-and-series limits eulers-constant




sequences-and-series limits eulers-constant






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edited Mar 28 at 16:34









Rodrigo de Azevedo

13.2k41962




13.2k41962










asked Mar 28 at 11:44









Ieva BrakmaneIeva Brakmane

537




537












  • $begingroup$
    Please do not post necessary information only in a picture, not everyone can display and read it properly.
    $endgroup$
    – Carsten S
    Mar 28 at 16:04


















  • $begingroup$
    Please do not post necessary information only in a picture, not everyone can display and read it properly.
    $endgroup$
    – Carsten S
    Mar 28 at 16:04
















$begingroup$
Please do not post necessary information only in a picture, not everyone can display and read it properly.
$endgroup$
– Carsten S
Mar 28 at 16:04




$begingroup$
Please do not post necessary information only in a picture, not everyone can display and read it properly.
$endgroup$
– Carsten S
Mar 28 at 16:04










3 Answers
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$begingroup$

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




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

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


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






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    6












    $begingroup$

    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,



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






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      2












      $begingroup$

      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









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        3 Answers
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        3 Answers
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        4












        $begingroup$

        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









        $endgroup$


















          4












          $begingroup$

          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









          $endgroup$
















            4












            4








            4





            $begingroup$

            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












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            share|cite|improve this answer










            answered Mar 28 at 11:57









            trancelocationtrancelocation

            13.9k1829




            13.9k1829























                6












                $begingroup$

                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









                $endgroup$


















                  6












                  $begingroup$

                  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









                  $endgroup$
















                    6












                    6








                    6





                    $begingroup$

                    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,



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







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                    answered Mar 28 at 11:52









                    Siong Thye GohSiong Thye Goh

                    104k1468120




                    104k1468120























                        2












                        $begingroup$

                        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









                        $endgroup$


















                          2












                          $begingroup$

                          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









                          $endgroup$
















                            2












                            2








                            2





                            $begingroup$

                            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 Mar 28 at 12:10









                            Gary MoonGary Moon

                            921127




                            921127






























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