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Blast into Math!                                 Analatic nummer theora: ants, ghosts and giants



                        Let’s look at the same ratio for the sequence

                                                              x n
                                                     x n =         ,
                                                           (n +1)!



                        then

                                                        x n+1  n!      x
                                              x n+1
                                                   =              =       .
                                               x n    (n +1)! x n    n +1


                        We know that  x ∈ R , which means that  x  is either a rational number or the least upper
                        bound of a bounded set of rational numbers. In either case, there is some


                                                m ∈ N such that m> x.


                        Then, if n ≥ 2m ,

                                                         x       m       1
                                              x n+1
                                                    =        <        < .
                                               x n     n +1     n +1     2


                        We can re-arrange this to

                                                         x n
                                                 x n+1 <   ,   ∀n ≥ 2m.
                                                         2


                     Let’s think about this a bit more, and try to relate it further to a geometric series. We know that


                                                  x 2m            x 2m+1    x 2m
                                        x 2m+1 <      ,  x 2m+2 <        <      .
                                                   2                 2       2 2

                        Now you can prove by induction that


                                                          x 2m
                                                 x 2m+n <       ∀n ∈ N.
                                                           2 n

                        So, now you can prove that the partial sums are all bounded, by splitting up the sum,


                                   N    n    m   n     N   n     m   n    N−m
                                     x        x        x        x         x m
                            S N =         =        +         ≤         +          ,  N ≥ m.
                                      n!        n!        n!        n!         2 n
                                  n=0       n=0       m+1       n=0       n=1











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