1-calculate-U-n-0-e-nx-x-dx-2-find-lim-n-n-U-n-3-determine-nsture-of-the-serie-U-n-

Question Number 74342 by mathmax by abdo last updated on 22/Nov/19

1) c a l c u l a t e U_{n} = \int_{0}^{\infty} e^{- n x} [x] d x

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

3) d e t e r m i n e n s t u r e o f t h e s e r i e Σ U_{n}

Commented by mathmax by abdo last updated on 23/Nov/19

1) w e h a v e U_{n} = \int_{0}^{\infty} e^{- n x} [x] d x = \sum_{k = 0}^{\infty} \int_{k}^{k + 1} e^{- n x} k d x

= \sum_{k = 0}^{\infty} k {[- \frac{1}{n} e^{- n x}]}_{k}^{k + 1} = \frac{1}{n} \sum_{k = 0}^{\infty} k {e^{- n k} - e^{- n (k + 1)}}

\Rightarrow {n U}_{n} = \sum_{k = 0}^{\infty} k e^{- n k} - \sum_{k = 0}^{\infty} k e^{- n (k + 1)}

= \sum_{k = 0}^{\infty} k e^{- n k} - \sum_{k = 1}^{\infty} (k - 1) e^{- n k}

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

= \sum_{k = 0}^{\infty} {(e^{- n})}^{k} - 1 = \frac{1}{1 - e^{- n}} - 1 = \frac{1 - 1 + e^{- n}}{1 - e^{- n}} \Rightarrow {n U}_{n} = \frac{e^{- n}}{1 - e^{- n}}

= \frac{1}{e^{n} - 1} \Rightarrow U_{n} = \frac{1}{n (e^{n} - 1)} (n > 0)

2) w e h a v e U_{n} = \frac{1}{n (e^{n} - 1)} \Rightarrow U_{n} \sim \frac{1}{{n e}^{n}} \to 0 \Rightarrow {l i m}_{n \to + \infty} U_{n} = 0

3) U_{n} \to 0 a n d d e c r e a s e s o Σ U_{n} a n d \int_{1}^{+ \infty} \frac{d t}{{t e}^{t}} h a v e t h e s a m e

n a t u r e a n d \int_{1}^{+ \infty} \frac{e^{- t}}{t} d t c o n v e r g e s \Rightarrow Σ U_{n} c o n v e r g e s .

Commented by mathmax by abdo last updated on 23/Nov/19

{n U}_{n} \sim e^{- n} \to 0 \Rightarrow {l i m}_{n \to + \infty} n U_{n} = 0

Answered by mind is power last updated on 22/Nov/19

U_{n} = \underset{k = 0}{\sum^{+ \infty}} \int_{k}^{k + 1} e^{- n x} [x] d x

= \underset{k = 0}{\sum^{+ \infty}} \int_{k}^{k + 1} e^{- n x} . k = \sum_{k ⩾ 0} {k e}^{- n x} . - \frac{1}{n}

= Σ - \frac{k}{n} (e^{- n (k + 1)} - e^{- n k})

= \sum_{k ⩾ 0} \frac{- {k e}^{- n k} (e^{- n} - 1)}{n} \Rightarrow {n U}_{n} = - (e^{- n} - 1) (\sum_{k ⩾ 1} {k e}^{- n k})

= - (e^{- n} - 1) . (e^{- n} + \sum_{k ⩾ 2} {k e}^{- n k})

Σ {k e}^{- n k} ⩽ e^{- n} Σ {k e}^{- k}

\forall n, k w i t h e n ⩾ 2, k ⩾ 2

n k ⩾ n + k

\Leftrightarrow n (k - 1) - k ⩾ 0 \Leftrightarrow (n - 1) (k - 1) ⩾ 1 c l e a r

\Rightarrow e^{- n k} ⩽ e^{- n} . e^{- k}

\Rightarrow Σ {k e}^{- n k} ⩽ e^{- n} . (Σ {k e}^{- k}) \to 0

\Rightarrow l i m {n U}_{n} \to 0