lim-n-n-e-x-n-where-x-n-1-1-1-1-2-1-n-

Question Number 204879 by universe last updated on 09/Aug/24

\lim_{n \to \infty} n! (e - x_{n}) = ?

where x_{n} = 1 + \frac{1}{1!} + \frac{1}{2!} + \dots + \frac{1}{n!}

Commented by mr W last updated on 01/Mar/24

y o u m e a n n \to \infty ?

Commented by universe last updated on 01/Mar/24

y e s s i r

Answered by Frix last updated on 01/Mar/24

? ? ?

x_{n} = \underset{k = 0}{\sum^{n}} \frac{1}{k!} \Rightarrow x_{0} = 1

0! (e - x_{0}) = 1 \times (e - 1) = e - 1

Answered by witcher3 last updated on 02/Mar/24

e - x_{n} = \sum_{k ⩾ n + 1} \frac{1}{k!} = \frac{1}{(n + 1)!} \sum_{k ⩾ n + 1} \frac{(n + 1)!}{k!} = \frac{1}{(n + 1)!} (\sum_{k ⩾ n + 1} \frac{(n + 1)!}{k!})

\sum_{k ⩾ n + 1} \frac{(n + 1)!}{k!} = (1 + \frac{1}{(n + 2)} + \frac{1}{(n + 2) (n + 3}) =

\sum_{k ⩾ n + 1} \frac{(n + 1)!}{k!} ⩽ 1 + \sum_{k ⩾ n + 2} \frac{1}{{(n + 2)}^{k - n}} ⩽ 1 + \sum_{k ⩾ 2} \frac{1}{2^{k}} = \frac{3}{2}; k + 1 ⩾ 2; n ⩾ 1

e - x_{n} = \frac{1}{(n + 1)!} Σ \frac{(n + 1)!}{k!} ⩽ \frac{3}{2 (n + 1)!}

\Rightarrow n! ∣ e - x_{n} ∣⩽ \frac{3}{2 (n + 1)}

0 ⩽ \lim_{n \to \infty} n! ∣ e - x_{n} ∣⩽ \lim_{n \to \infty} \frac{3}{2 (n + 1)} = 0

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