let-I-n-0-n-1-x-2x-1-2-dx-1-calculate-I-n-interms-of-n-2-find-lim-n-I-n-

Question Number 37896 by abdo mathsup 649 cc last updated on 19/Jun/18

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

1) c a l c u l a t e I_{n} i n t e r m s o f n

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

Commented by prof Abdo imad last updated on 19/Jun/18

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

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

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

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

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

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

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

I_{n} = - \frac{1}{6} + \frac{1}{2} \sum_{k = 0}^{n - 2} \frac{{(- 1)}^{k}}{2 k + 1} + \frac{1}{2} \sum_{k = 0}^{n - 2} \frac{{(- 1)}^{k}}{2 k + 1} + \frac{{(- 1)}^{n - 1}}{2 n - 1}

I_{n} = - \frac{1}{6} + \frac{{(- 1)}^{n - 1}}{2 n - 1} + \sum_{k = 0}^{n - 2} \frac{{(- 1)}^{k}}{2 k + 1}

2) {l i m}_{n \to + \infty} I_{n} = - \frac{1}{6} + \sum_{k = 0}^{\infty} \frac{{(- 1)}^{k}}{2 k + 1}

= \frac{π}{4} - \frac{1}{6} .

Answered by tanmay.chaudhury50@gmail.com last updated on 19/Jun/18

[x] = 0 1 > x ⩾ 0

[x] = 1 2 > x ⩾ 1

[x] = 2 3 > x ⩾ 2

t h u s t h e v a l u e o f {(- 1)}^{[x]} i s e i t h e r + 1 o r - 1

\int_{0}^{1} \frac{1}{{(2 x + 1)}^{2}} d x + \int_{1}^{2} \frac{- 1}{{(2 x + 1)}^{2}} + \int_{2}^{3} \frac{1}{{(2 x + 1)}^{2}} + \dots

+ \int_{n - 1}^{n} \frac{{(- 1)}^{n - 1}}{{(2 x + 1)}^{2}} d x

c o n t d

Answered by tanmay.chaudhury50@gmail.com last updated on 19/Jun/18

Commented by tanmay.chaudhury50@gmail.com last updated on 19/Jun/18

Commented by tanmay.chaudhury50@gmail.com last updated on 19/Jun/18