I am attempting to justify the expansion 1 + x = 1 + x 2 + ∑ n = 2 ∞ ( − 1 ) n 1 2 n ( 1 − 1 2 ) ⋯ ( ( n − 1 ) − 1 2 ) ( n − 1 ) ! x n

Question

I am attempting to justify the expansion      1    +    x    =  1  +      x    2    +      ∑          n      =      2              ∞            (    −    1          )      n              1              2        n                            (        1        −                  1          2                )        ⋯        (        (        n        −        1        )        −                  1          2                )                    (        n        −        1        )        !                    x      n

Nathen Austin · Accepted Answer

Consider the absolute value      a    n    =      1          2      n                  (      1      −              1        2            )      ⋯      (      (      n      −      1      )      −              1        2            )              (      n      −      1      )      !      of the coefficient of       x    n   in the series expansion of       1    +    x  . Then            a              n        +        1                    a      n        =            n      −              1        2                    n      +      1        =  1      −          3      2            1    n    +  o      (          1      n        )    ,and the heuristics in such cases is that       a    n   behaves roughly like       n    s   with   s  =      −          3      2       since, for any s,            (      n      +      1              )        s                    n      s        =  1  +      s    n    +  o      (          1      n        )    .To continue the proof more rigorously, choose any   t  &amp;lt;      3    2   and consider       b    n    =      n    t        a    n  . Then            b              n        +        1                    b      n        =            (      n      +      1              )        t                    n      t                  a              n        +        1                    a      n        =      (    1    +          t      n        +    o          (              1        n            )        )    ⋅      (    1    −          3      2              1      n        +    o          (              1        n            )        )    ,hence            b              n        +        1                    b      n        =  1  +      (    t    −          3      2        )        1    n    +  o      (          1      n        )    .Since   t  −      3    2    &amp;lt;  0, this implies that   (      b    n    ) is ultimately decreasing, in particular       b    n    ⩽      c    t   for every n, for some finite       c    t  .Finally,       a    n    ⩽      c    t        n          −      t       uniformly over n for every   t  &amp;lt;      3    2  , and one can choose   t  &amp;gt;  1. Then       ∑    n        n          −      t       converges (absolutely) hence       ∑    n        a    n   converges (absolutely) and the series expansion of       1    +    x   converges (absolutely) for every       |    x      |    ⩽  1

I am attempting to justify the expansion <msqrt> 1 + x </msqrt> = 1 +

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