Proving that ( 1 0 ! + 1 1 ! + 1 2 ! + 1 3 ! + ⋯ + 1 n ! ) has a limit

Question

Proving that       (          1              0        !              +          1              1        !              +          1              2        !              +          1              3        !              +    ⋯    +          1              n        !              )   has a limit

Melina Glover · Accepted Answer

x    n    &amp;gt;  1  +  1  +      1    2    =  2.5Nowas David Mitra suggested,      1          3      !        =      1          1      ⋅      2      ⋅      3        &amp;lt;      1          2      ⋅      2        =      1          2      2        ,      1          4      !        =      1          1      ⋅      2      ⋅      3      ⋅      4        &amp;lt;      1          2      ⋅      2      ⋅      2        =      1          2      3            1          5      !        =      1          1      ⋅      2      ⋅      3      ⋅      4      ⋅      5        &amp;lt;      1          2      ⋅      2      ⋅      2      ⋅      2        =      1          2      4      and so on.Hence      x    n    &amp;lt;  1  +  1  +      1    2    +      1          2      2        +      1          2      3        +  ⋯  &amp;lt;  1  +      ∑          0      ≤      r      &amp;lt;      ∞            1          2      r        =  1  +      1          1      −              1        2              =  3

Akira Huang · Accepted Answer

Alternately you can use      x    n    =      1          0      !        +      1          1      !        +      1          2      !        +      1          3      !        +  .  .  .  +      1          n      !        ≤      x    n    =      1          0      !        +      1          1      !        +      1          1      ⋅      2        +      1          2      ⋅      3        +  .  .  .  +      1          (      n      −      1      )      ⋅      n      and      1          1      ⋅      2        +      1          2      ⋅      3        +  .  .  .  +      1          (      n      −      1      )      ⋅      n      is telescopic, since      1          k      (      k      +      1      )        =      1    k    −      1          k      +      1          .

Proving that ( 1 0 ! </mrow> </mfrac>

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