L = n → ∞ lim [ ( n + 1 ) ! ( e x n + 1 − e x n ) ]
If x n = k = 0 ∑ n k ! 1 , then find the value of ln ( L ) . Give your answer to 3 decimal places.
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Is this correct? The first limit in your answer does not tend to 1 and the equation in the second line is not true.
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L = n → ∞ lim [ ( n + 1 ) ! ( e x n + 1 − e x n ) ]
= n → ∞ lim [ ( n + 1 ) ! ⋅ e x n ( e x n + 1 − x n − 1 ) ]
= n → ∞ lim [ ( n + 1 ) ! ⋅ e x n ( e ( n + 1 ) ! 1 − 1 ) ]
= n → ∞ lim e x n × n → ∞ lim ( ( n + 1 ) ! 1 e ( n + 1 ) ! 1 − 1 )
= e e × 1 = e e
Therefore, ln ( L ) = ln ( e e ) = e ≈ 2 . 7 1 8 .