Prove-that-lim-n-1-n-n-0-t-n-e-t-dt-1-2-

Question Number 193951 by Erico last updated on 23/Jun/23

Prove that :

\lim_{n \to + \infty} \frac{1}{n!} {\int_{0}}^{n} t^{n} e^{- t} d t = \frac{1}{2}

Answered by senestro last updated on 24/Jun/23

\lim_{n \to + \infty} \frac{1}{n!} \int_{0}^{n} t^{n} e^{- t} d t = \lim_{n \to + \infty} - e^{- t} (1 + \frac{t}{1!} + \frac{t^{2}}{2!} + \dots + \frac{t^{n}}{n!}) ∣_{0}^{n}

= \lim_{n \to + \infty} - e^{- n} (1 + \frac{n}{1!} + \frac{n^{2}}{2!} + \dots + \frac{n^{n}}{n!}) + \lim_{n \to + \infty} e^{- 0} (1 + \frac{0}{1!} + \frac{0^{2}}{2!} + \dots + \frac{0^{n}}{n!})

= \lim_{n \to + \infty} - e^{- n} (1 + \frac{n}{1!} + \frac{n^{2}}{2!} + \dots + \frac{n^{n}}{n!}) + \underset{n \to + \infty}{\lim 1} (1)

= \lim_{n \to + \infty} - e^{- n} e^{n} + \underset{n \to + \infty}{\lim 1}

= \lim_{n \to + \infty} - 1 + \underset{n \to + \infty}{\lim 1}

= - 1 + 1 = 0

h e n c e t h e q u e s t i o n i s w r o n g

\lim_{n \to + \infty} \frac{1}{n!} \int_{0}^{n} t^{n} e^{- t} d t = 0

b u t

\lim_{n \to + \infty} \frac{1}{n!} \int_{0}^{\infty} t^{n} e^{- t} d t = 1

a n d

\lim_{n \to + \infty} \frac{1}{2 n!} \int_{0}^{\infty} t^{n} e^{- t} d t = \frac{1}{2}

Answered by aba last updated on 24/Jun/23

\lim_{n \to \infty} \frac{1}{2 n!} \int_{0}^{\infty} t^{n} e^{- t} dt = \lim_{n \to \infty} \frac{Γ (n + 1)}{2 n!} = \lim_{n \to \infty} \frac{n!}{2 n!} = \frac{1}{2} ✓

Commented by Erico last updated on 26/Jun/23

Regarde :

Posons f (t) = \frac{t^{n}}{n!} e^{- t}

\forall n \in N {\int_{n}}^{2 n} f (t) dt = {\int_{0}}^{n} f (2 n - t) dt

or \int_{2 n}^{x} f (t) dt = {\int_{n}}^{x - n} \frac{(u + n)}{n!} e^{- u - n} du

or u ⩾ n \Rightarrow \frac{{(u + n)}^{n}}{n!} e^{- u - n} ⩽ \frac{{(2 u)}^{n}}{n!} e^{- u - n}

\Rightarrow Si x ⩾ 2 n, {\int_{n}}^{x - n} \frac{{(u + n)}^{n}}{n!} e^{- u - n} du ⩽ {(\frac{2}{e})}^{n} {\int_{n}}^{x - n} \frac{u^{n}}{n!} e^{- u} du ⩽ {(\frac{2}{e})}^{n} {\int_{0}}^{+ \infty} \frac{u^{n}}{n!} e^{- u} du

⩽ {(\frac{2}{e})}^{n}

donc \lim_{n \to + \infty} \frac{1}{n!} {\int_{2 n}}^{+ \infty} u^{n} e^{- u} du = 0

\Rightarrow \lim_{n \to + \infty} \frac{1}{n!} {\int_{0}}^{2 n} t^{n} e^{- t} dt = 1

\Rightarrow \lim_{n \to + \infty} [\frac{1}{n!} {\int_{0}}^{n} t^{n} e^{- t} dt + \frac{1}{n!} {\int_{n}}^{2 n} t^{n} e^{- t} dt] = 1

donc \lim_{n \to + \infty} \frac{1}{n!} {\int_{0}}^{n} t^{n} e^{- t} dt \neq 0

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