let-A-n-0-n-e-n-x-2-x-dx-with-n-integr-natural-1-calculate-A-n-2-find-lim-n-A-n-3-study-the-convergence-of-n-A-n-

Question Number 40040 by abdo mathsup 649 cc last updated on 15/Jul/18

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

1) c a l c u l a t e A_{n}

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

3) s t u d y t h e c o n v e r g e n c e o f \sum_{n} A_{n}

Commented by abdo mathsup 649 cc last updated on 17/Jul/18

1) w e h a v e A_{n} = \sum_{k = 0}^{n - 1} \int_{k}^{k + 1} e^{- n (x + 2 - k)} d x

= \sum_{k = 0}^{n - 1} e^{- n (2 - k)} \int_{k}^{k + 1} e^{- n x} d x

= \sum_{k = 0}^{n - 1} e^{n (k - 2)} {[- \frac{1}{n} e^{- n x}]}_{k}^{k + 1}

= \frac{1}{n} \sum_{k = 0}^{n - 1} e^{n (k - 2)} {e^{- n k} - e^{- (n + 1) k}}

= \frac{1}{n} \sum_{k = 0}^{n - 1} e^{- 2 n} - \frac{1}{n} \sum_{k = 0}^{n - 1} e^{- 2 n - k}

= e^{- 2 n} - \frac{e^{- 2 n}}{n} \sum_{k = 0}^{n - 1} {(e^{- 1})}^{k}

A_{n} = e^{- 2 n} - \frac{e^{- 2 n}}{n} \frac{1 - {(e^{- 1})}^{n}}{1 - e^{- 1}}

2) w e h a v e {l i m}_{n \to + \infty} e^{- 2 n} = 0

{l i m}_{n \to + \infty} \frac{e^{- 2 n}}{n} = 0 a n d {l i m}_{n \to + \infty} \frac{1 - e^{- n}}{2 - e^{- 1}} = \frac{1}{1 - e^{- 1}}

\Rightarrow {l i m}_{n \to + \infty} A_{n} = 0

3) w e h a v e \sum_{n = 1}^{\infty} A_{n} = \sum_{n = 1}^{\infty} e^{- 2 n} - \frac{1}{1 - e^{- 1}} \sum_{n = 1}^{\infty} \frac{e^{- 2 n}}{n}

+ \frac{1}{1 - e^{- 1}} \sum_{n = 1}^{\infty} \frac{e^{- 3 n}}{n} b u t w e h a v e

\sum_{n = 1}^{\infty} e^{- 2 n} = \sum_{n = 1}^{\infty} {(e^{- 2})}^{n} = \frac{1}{1 - e^{- 2}}

\sum_{n = 1}^{\infty} \frac{e^{- 2 n}}{n} = \sum_{n = 1}^{\infty} \frac{{(e^{- 2})}^{n}}{n} = - l n (1 - e^{- 2})

\sum_{n = 1}^{\infty} \frac{e^{- 3 n}}{n} = - l n (1 - e^{- 3}) \Rightarrow Σ A_{n} i s c o n v e r g e n t

a n d \sum_{n ⩾ 1} A_{n} = \frac{1}{1 - e^{- 2}} + \frac{1}{1 - e^{- 1}} l n (1 - e^{- 2})

- \frac{1}{1 - e^{- 1}} l n (1 - e^{- 3}) .