This is a question from a GRE math subject test practice material. ∑ n = 1 ∞ n ! x 2 n n n ( 1 + x 2 n )

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This is a question from a GRE math subject test practice material.      ∑          n      =      1              ∞                  n      !              x                  2          n                                    n        n            (      1      +              x                  2          n                    )

Braedon Rivas · Accepted Answer

Applying the root rest:      lim sup          n      →      ∞                  (                        n          !                      x                          2              n                                                            n            n                    (          1          +                      x                          2              n                                )                    )              1              /            n        =      lim sup          n      →      ∞                  x      2              n      ⋅      max      (      1      ,              x        2            )                  (      n      !      )              1              /            n        =            e              −      1        ⋅            x      2              max      (      1      ,              x        2            )      The right-hand-side is bounded from about by             e              −      1        &amp;lt;  1for all x, that is it converges for all   x  ∈      R  And if one really wants to use the ratio test, one can, though it certainly qualifies as doing things the hard way:                              |                      lim                          n              →              ∞                                                                                            (                  n                  +                  1                  )                  !                                      x                                          2                      n                      +                      2                                                                                        (                  n                  +                  1                                      )                                          n                      +                      1                                                        (                  1                  +                                      x                                          2                      n                      +                      2                                                        )                                                                                                      n                  !                                      x                                          2                      n                                                                                                            n                    n                                    (                  1                  +                                      x                                          2                      n                                                        )                                                              |                                    =                  lim                      n            →            ∞                                                              n              n                        (            n            +            1            )            !                          x                              2                n                +                2                                      (            1            +                          x                              2                n                                      )                                (            n            +            1                          )                              n                +                1                                      n            !                          x                              2                n                                      (            1            +                          x                              2                n                +                2                                      )                                                            =                  lim                      n            →            ∞                                                              n              n                        (            n            +            1            )                          x              2                        (            1            +                          x                              2                n                                      )                                (            n            +            1                          )                              n                +                1                                      (            1            +                          x                              2                n                +                2                                      )                                                            =                  lim                      n            →            ∞                                                (                          n                              n                +                1                                      )                    n                                      1            +                          x                              2                n                                                          1            +                          x                              2                n                +                2                                                                                      =                  lim                      n            →            ∞                                                (            1            −                          1                              n                +                1                                      )                    n                                      1            +                          x                              2                n                                                          1            +                          x                              2                n                +                2                                                                                      =                  lim                      n            →            ∞                                                (                                          (                1                −                                  1                                      n                    +                    1                                                  )                                            n                +                1                                      )                                              n                              n                +                1                                                                          1            +                          x                              2                n                                                          1            +                          x                              2                n                +                2                                                                                      =                  1          e                          lim                      n            →            ∞                                                1            +                          x                              2                n                                                          1            +                          x                              2                n                +                2                                                                                      =                  {                                                                      1                  e                                ,                                            if                                   |                                x                                  |                                ≤                1                                                                                                          0                ,                                            if                                   |                                x                                  |                                &amp;gt;                1                                                                                                &amp;lt;        1                .

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