1-calculte-A-n-0-e-nx-e-x-dx-with-n-integr-and-n-2-2-find-lim-n-n-n-A-n-

Question Number 74498 by mathmax by abdo last updated on 25/Nov/19

1) c a l c u l t e A_{n} = \int_{0}^{\infty} e^{- n x} [e^{x}] d x w i t h n i n t e g r a n d n ⩾ 2

2) f i n d {l i m}_{n \to + \infty} n^{n} A_{n}

Commented by mathmax by abdo last updated on 25/Nov/19

1) A_{n} = \int_{0}^{\infty} e^{- n x} [e^{x}] d x c h a n g e m e n t e^{x} = t g i v e

A_{n} = \int_{1}^{+ \infty} e^{- n l n t} [t] \frac{d t}{t} = \int_{1}^{+ \infty} \frac{[t]}{t^{n + 1}} d t = \sum_{k = 1}^{\infty} \int_{k}^{k + 1} \frac{k}{t^{n + 1}} d t

= \sum_{k = 1}^{\infty} k \int_{k}^{k +} t^{- n - 1} d t = \sum_{k = 1}^{\infty} k {[- \frac{1}{n} t^{- n}]}_{k}^{k + 1}

= \sum_{k = 1}^{\infty} \frac{k}{n} {\frac{1}{k^{n}} - \frac{1}{k^{n + 1}}} = \frac{1}{n} \sum_{k = 1}^{\infty} \frac{1}{k^{n - 1}} - \frac{1}{n} \sum_{k = 1}^{\infty} \frac{1}{k^{n}}

w e h a v e ξ (x) = \sum_{k = 1}^{\infty} \frac{1}{k^{x}} (x > 1) \Rightarrow A_{n} = \frac{1}{n} ξ (n - 1) - \frac{1}{n} ξ (n)

2) w e h a v e n^{n} A_{n} = n^{n - 1} {ξ (n - 1) - ξ (n)} \Rightarrow n^{n} A_{n} \sim c n^{n - 1} \Rightarrow

{l i m}_{n \to + \infty} n^{n} A_{n} = \infty

Answered by mind is power last updated on 25/Nov/19

A_{n} = \sum_{k ⩾ 1} \int_{l n (k)}^{l n (k + 1)} e^{- n x} [e^{x}] d x

= \sum_{k ⩾ 1} \int_{l n (k)}^{l n (k + 1)} . {k e}^{- n x} d x

= \sum_{k ⩾ 1} \frac{- k}{n} {[e^{- n x}]}_{l n (k)}^{l n (k + 1)} = \sum_{k ⩾ 1} \frac{- k}{n {(k + 1)}^{n}} + \frac{k}{{n k}^{n}}

= \sum_{k ⩾ 1} {\frac{- 1}{n {(k + 1)}^{n - 1}} + \frac{1}{{n k}^{n - 1}} + \frac{1}{n {(k + 1)}^{n}}}

= \frac{1}{n} + \sum_{k ⩾ 1} \frac{1}{n {(k + 1)}^{n}} = A_{n}

n^{n} A n ⩾ n^{n - 1} \to + \infty