Can someone shed some light on this inequality? The Next CEO of Stack OverflowShow that the sequence $left(frac2^nn!right)$ has a limit.Determine value the following: $L=sum_k=1^inftyfrac1k^k$Could someone help me clarify the steps for this solution?Understanding how to use $epsilon-delta$ definition of a limitHow can an imaginary equation give a real answer?Can someone claify on the work that was done in this question on Maclaurin SeriesConvergence of series $nq^n$.How does this limit converge to zeroUnderstanding part of a proof for Stolz-Cesaro TheoremAbout a statement of partial fraction in an answer

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Can someone shed some light on this inequality?



The Next CEO of Stack OverflowShow that the sequence $left(frac2^nn!right)$ has a limit.Determine value the following: $L=sum_k=1^inftyfrac1k^k$Could someone help me clarify the steps for this solution?Understanding how to use $epsilon-delta$ definition of a limitHow can an imaginary equation give a real answer?Can someone claify on the work that was done in this question on Maclaurin SeriesConvergence of series $nq^n$.How does this limit converge to zeroUnderstanding part of a proof for Stolz-Cesaro TheoremAbout a statement of partial fraction in an answer










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 $$fraca_n+1a_n>left (1-frac1n+1right ) left (fracn+1nright)$$



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



Ok, I have another relating question:



why $$fraca_n+1a_n> (1+frac1n)$$ ( 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
    2 days ago















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 $$fraca_n+1a_n>left (1-frac1n+1right ) left (fracn+1nright)$$



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



Ok, I have another relating question:



why $$fraca_n+1a_n> (1+frac1n)$$ ( 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
    2 days ago













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 $$fraca_n+1a_n>left (1-frac1n+1right ) left (fracn+1nright)$$



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



Ok, I have another relating question:



why $$fraca_n+1a_n> (1+frac1n)$$ ( 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 $$fraca_n+1a_n>left (1-frac1n+1right ) left (fracn+1nright)$$



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



Ok, I have another relating question:



why $$fraca_n+1a_n> (1+frac1n)$$ ( The expression of third line.



!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 2 days ago









Rodrigo de Azevedo

13.2k41960




13.2k41960










asked 2 days ago









Ieva BrakmaneIeva Brakmane

316




316











  • $begingroup$
    Please do not post necessary information only in a picture, not everyone can display and read it properly.
    $endgroup$
    – Carsten S
    2 days ago
















  • $begingroup$
    Please do not post necessary information only in a picture, not everyone can display and read it properly.
    $endgroup$
    – Carsten S
    2 days ago















$begingroup$
Please do not post necessary information only in a picture, not everyone can display and read it properly.
$endgroup$
– Carsten S
2 days ago




$begingroup$
Please do not post necessary information only in a picture, not everyone can display and read it properly.
$endgroup$
– Carsten S
2 days ago










3 Answers
3






active

oldest

votes


















4












$begingroup$

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



  • $fraca_n+1a_n = colorblueleft(1- frac1(n+1)^2 right)^n+1left( fracn+1nright)$

  • $colorblueleft(1- frac1(n+1)^2 right)^n+1 > colorgreen1 + (n+1)left( frac-1(n+1)^2right)$

$$Rightarrow fraca_n+1a_n > left(colorgreen1 + (n+1)left( frac-1(n+1)^2right)right)left( fracn+1nright) = left(underbrace1- frac1n+1_=fracnn+1right)left( fracn+1nright)$$






share|cite|improve this answer









$endgroup$




















    6












    $begingroup$

    From Bernoulli's inequality, we have



    $$left( 1- frac1(n+1)^2right) > 1+(n+1) left(frac-1(n+1)^2 right)=1-frac1n+1$$



    Hence,



    $$fraca_n+1a_n>left( 1- frac1(n+1)^2right)left( fracn+1nright)>left(1-frac1n+1 right)left( fracn+1nright)$$






    share|cite|improve this answer









    $endgroup$




















      2












      $begingroup$

      So, we have
      $$fraca_n+1a_n = left(1 - frac1(n+1)^2right)^n+1left(fracn+1nright).$$
      The author then applies Bernoulli's inequality to the first term on the RHS:
      $$left(1 - frac1(n+1)^2right)^n+1 > 1 + (n+1)left(frac-1(n+1)^2right) = 1 - frac1n+1.$$
      We can now return to the first equation and utilize this estimate; namely, we have
      $$fraca_n+1a_n = left(1 - frac1(n+1)^2right)^n+1left(fracn+1nright) > left(1-frac1n+1right)left(fracn+1nright).$$
      Finally, we multiply out the RHS of the inequality
      $$fraca_n+1a_n > left(1-frac1n+1right)left(fracn+1nright) = fracn+1n - frac1n = 1.$$
      So, we have
      $$fraca_n+1a_n > 1 implies a_n+1 > a_n,$$
      which means that $a_n$ is an increasing sequence.






      share|cite|improve this answer









      $endgroup$













        Your Answer





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        3 Answers
        3






        active

        oldest

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        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        4












        $begingroup$

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



        • $fraca_n+1a_n = colorblueleft(1- frac1(n+1)^2 right)^n+1left( fracn+1nright)$

        • $colorblueleft(1- frac1(n+1)^2 right)^n+1 > colorgreen1 + (n+1)left( frac-1(n+1)^2right)$

        $$Rightarrow fraca_n+1a_n > left(colorgreen1 + (n+1)left( frac-1(n+1)^2right)right)left( fracn+1nright) = left(underbrace1- frac1n+1_=fracnn+1right)left( fracn+1nright)$$






        share|cite|improve this answer









        $endgroup$

















          4












          $begingroup$

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



          • $fraca_n+1a_n = colorblueleft(1- frac1(n+1)^2 right)^n+1left( fracn+1nright)$

          • $colorblueleft(1- frac1(n+1)^2 right)^n+1 > colorgreen1 + (n+1)left( frac-1(n+1)^2right)$

          $$Rightarrow fraca_n+1a_n > left(colorgreen1 + (n+1)left( frac-1(n+1)^2right)right)left( fracn+1nright) = left(underbrace1- frac1n+1_=fracnn+1right)left( fracn+1nright)$$






          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:



            • $fraca_n+1a_n = colorblueleft(1- frac1(n+1)^2 right)^n+1left( fracn+1nright)$

            • $colorblueleft(1- frac1(n+1)^2 right)^n+1 > colorgreen1 + (n+1)left( frac-1(n+1)^2right)$

            $$Rightarrow fraca_n+1a_n > left(colorgreen1 + (n+1)left( frac-1(n+1)^2right)right)left( fracn+1nright) = left(underbrace1- frac1n+1_=fracnn+1right)left( fracn+1nright)$$






            share|cite|improve this answer









            $endgroup$



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



            • $fraca_n+1a_n = colorblueleft(1- frac1(n+1)^2 right)^n+1left( fracn+1nright)$

            • $colorblueleft(1- frac1(n+1)^2 right)^n+1 > colorgreen1 + (n+1)left( frac-1(n+1)^2right)$

            $$Rightarrow fraca_n+1a_n > left(colorgreen1 + (n+1)left( frac-1(n+1)^2right)right)left( fracn+1nright) = left(underbrace1- frac1n+1_=fracnn+1right)left( fracn+1nright)$$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 2 days ago









            trancelocationtrancelocation

            13.4k1827




            13.4k1827





















                6












                $begingroup$

                From Bernoulli's inequality, we have



                $$left( 1- frac1(n+1)^2right) > 1+(n+1) left(frac-1(n+1)^2 right)=1-frac1n+1$$



                Hence,



                $$fraca_n+1a_n>left( 1- frac1(n+1)^2right)left( fracn+1nright)>left(1-frac1n+1 right)left( fracn+1nright)$$






                share|cite|improve this answer









                $endgroup$

















                  6












                  $begingroup$

                  From Bernoulli's inequality, we have



                  $$left( 1- frac1(n+1)^2right) > 1+(n+1) left(frac-1(n+1)^2 right)=1-frac1n+1$$



                  Hence,



                  $$fraca_n+1a_n>left( 1- frac1(n+1)^2right)left( fracn+1nright)>left(1-frac1n+1 right)left( fracn+1nright)$$






                  share|cite|improve this answer









                  $endgroup$















                    6












                    6








                    6





                    $begingroup$

                    From Bernoulli's inequality, we have



                    $$left( 1- frac1(n+1)^2right) > 1+(n+1) left(frac-1(n+1)^2 right)=1-frac1n+1$$



                    Hence,



                    $$fraca_n+1a_n>left( 1- frac1(n+1)^2right)left( fracn+1nright)>left(1-frac1n+1 right)left( fracn+1nright)$$






                    share|cite|improve this answer









                    $endgroup$



                    From Bernoulli's inequality, we have



                    $$left( 1- frac1(n+1)^2right) > 1+(n+1) left(frac-1(n+1)^2 right)=1-frac1n+1$$



                    Hence,



                    $$fraca_n+1a_n>left( 1- frac1(n+1)^2right)left( fracn+1nright)>left(1-frac1n+1 right)left( fracn+1nright)$$







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered 2 days ago









                    Siong Thye GohSiong Thye Goh

                    103k1468119




                    103k1468119





















                        2












                        $begingroup$

                        So, we have
                        $$fraca_n+1a_n = left(1 - frac1(n+1)^2right)^n+1left(fracn+1nright).$$
                        The author then applies Bernoulli's inequality to the first term on the RHS:
                        $$left(1 - frac1(n+1)^2right)^n+1 > 1 + (n+1)left(frac-1(n+1)^2right) = 1 - frac1n+1.$$
                        We can now return to the first equation and utilize this estimate; namely, we have
                        $$fraca_n+1a_n = left(1 - frac1(n+1)^2right)^n+1left(fracn+1nright) > left(1-frac1n+1right)left(fracn+1nright).$$
                        Finally, we multiply out the RHS of the inequality
                        $$fraca_n+1a_n > left(1-frac1n+1right)left(fracn+1nright) = fracn+1n - frac1n = 1.$$
                        So, we have
                        $$fraca_n+1a_n > 1 implies a_n+1 > a_n,$$
                        which means that $a_n$ is an increasing sequence.






                        share|cite|improve this answer









                        $endgroup$

















                          2












                          $begingroup$

                          So, we have
                          $$fraca_n+1a_n = left(1 - frac1(n+1)^2right)^n+1left(fracn+1nright).$$
                          The author then applies Bernoulli's inequality to the first term on the RHS:
                          $$left(1 - frac1(n+1)^2right)^n+1 > 1 + (n+1)left(frac-1(n+1)^2right) = 1 - frac1n+1.$$
                          We can now return to the first equation and utilize this estimate; namely, we have
                          $$fraca_n+1a_n = left(1 - frac1(n+1)^2right)^n+1left(fracn+1nright) > left(1-frac1n+1right)left(fracn+1nright).$$
                          Finally, we multiply out the RHS of the inequality
                          $$fraca_n+1a_n > left(1-frac1n+1right)left(fracn+1nright) = fracn+1n - frac1n = 1.$$
                          So, we have
                          $$fraca_n+1a_n > 1 implies a_n+1 > a_n,$$
                          which means that $a_n$ is an increasing sequence.






                          share|cite|improve this answer









                          $endgroup$















                            2












                            2








                            2





                            $begingroup$

                            So, we have
                            $$fraca_n+1a_n = left(1 - frac1(n+1)^2right)^n+1left(fracn+1nright).$$
                            The author then applies Bernoulli's inequality to the first term on the RHS:
                            $$left(1 - frac1(n+1)^2right)^n+1 > 1 + (n+1)left(frac-1(n+1)^2right) = 1 - frac1n+1.$$
                            We can now return to the first equation and utilize this estimate; namely, we have
                            $$fraca_n+1a_n = left(1 - frac1(n+1)^2right)^n+1left(fracn+1nright) > left(1-frac1n+1right)left(fracn+1nright).$$
                            Finally, we multiply out the RHS of the inequality
                            $$fraca_n+1a_n > left(1-frac1n+1right)left(fracn+1nright) = fracn+1n - frac1n = 1.$$
                            So, we have
                            $$fraca_n+1a_n > 1 implies a_n+1 > a_n,$$
                            which means that $a_n$ is an increasing sequence.






                            share|cite|improve this answer









                            $endgroup$



                            So, we have
                            $$fraca_n+1a_n = left(1 - frac1(n+1)^2right)^n+1left(fracn+1nright).$$
                            The author then applies Bernoulli's inequality to the first term on the RHS:
                            $$left(1 - frac1(n+1)^2right)^n+1 > 1 + (n+1)left(frac-1(n+1)^2right) = 1 - frac1n+1.$$
                            We can now return to the first equation and utilize this estimate; namely, we have
                            $$fraca_n+1a_n = left(1 - frac1(n+1)^2right)^n+1left(fracn+1nright) > left(1-frac1n+1right)left(fracn+1nright).$$
                            Finally, we multiply out the RHS of the inequality
                            $$fraca_n+1a_n > left(1-frac1n+1right)left(fracn+1nright) = fracn+1n - frac1n = 1.$$
                            So, we have
                            $$fraca_n+1a_n > 1 implies a_n+1 > a_n,$$
                            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 2 days ago









                            Gary MoonGary Moon

                            92127




                            92127



























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