let-A-n-0-n-1-x-x-2-x-dx-1-calculate-A-n-and-lim-n-A-n-2-let-S-n-n-0-n-A-n-dtudy-the-convergence-of-S-n-4-let-W-n-n-1-n-1-A-n-study-th

Question Number 39703 by math khazana by abdo last updated on 10/Jul/18

l e t A_{n} = \int_{0}^{n} \frac{{(- 1)}^{[x]}}{x + 2 - [x]} d x

1) c a l c u l a t e A_{n} a n d {l i m}_{n \to + \infty} A_{n}

2) l e t S_{n} = \sum_{n = 0}^{n} A_{n} d t u d y t h e c o n v e r g e n c e o f S_{n}

4) l e t W_{n} = \sum_{n = 1}^{n} \frac{1}{A_{n}}

s t u d y t h e c o n v e r g e n c e o f W_{n}

Commented by math khazana by abdo last updated on 13/Jul/18

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

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

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

= \sum_{k = 0}^{n} {(- 1)}^{k} {l n (\frac{3}{2})} = l n (\frac{3}{2}) \frac{1 - {(- 1)}^{n + 1}}{2}

\Rightarrow A_{n} = \frac{1}{2} l n (\frac{3}{2}) (1 - {(- 1)}^{n + 1}) b u t t h e s e q u e n c e

{(- 1)}^{n} i s n o t c o n v e g e n t s o A_{n} i s n o t c o n v e r g e n t!

2) S_{n} = \sum_{n = 0}^{\infty} A_{n} = \sum_{n = 0}^{\infty} \frac{1}{2} l n (\frac{3}{2}) {1 + {(- 1)}^{n}}

S_{n} a l s o d i v e r g e s

4) w e h a v e A_{n} = 0 i f n e v e n a n d A_{n} = l n (\frac{3}{2}) i f n

i s o d d s o W_{n} i s n o t d e f i n e d!