find-A-n-1-2-1-1-x-1-x-2-1-x-n-2-dx-

Question Number 37271 by abdo.msup.com last updated on 11/Jun/18

f i n d A_{n} = \int_{1}^{2} {(1 + \frac{1}{x} + \frac{1}{x^{2}} + \dots + \frac{1}{x^{n}})}^{2} d x

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

l e t a_{i} = \frac{1}{x^{i}} w i t h 0 ⩽ i ⩽ n \Rightarrow

{(1 + \frac{1}{x} + \frac{1}{x^{2}} + \dots . + \frac{1}{x^{n}})}^{2} = {(\sum_{i = 0}^{n} a_{i})}^{2}

= \sum_{i = 0}^{n} a_{i}^{2} + 2 \sum_{0 ⩽ i < j ⩽ n} a_{i} . a_{j}

= \sum_{i = 0}^{n} \frac{1}{x^{2 i}} + 2 \sum_{0 ⩽ i < j ⩽ n} \frac{1}{x^{i + j}} \Rightarrow

A_{n} = \int_{1}^{2} (\sum_{i = 0}^{n} \frac{1}{x^{2 i}}) d x + 2 \int_{1}^{2} \sum_{0 ⩽ i < j ⩽ n} \frac{1}{x^{i + j}} d x

= \sum_{i = 0}^{n} \int_{1}^{2} x^{- 2 i} d x + 2 \sum_{0 ⩽ i < j ⩽ n} \int_{1}^{2} x^{- (i + j)} d x

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

+ 2 \sum_{0 ⩽ i < j ⩽ n} \frac{1}{1 - (i + j)} {[x^{1 - (i + j)}]}_{1}^{2}

= - \sum_{i = 0}^{n} \frac{1}{2 i - 1} {2^{- 2 i + 1} - 1}

- \sum_{0 ⩽ i < j ⩽ n} \frac{1}{i + j - 1} {2^{1 - (i + j)} - 1}

Commented by math khazana by abdo last updated on 18/Jun/18

A_{n} = \sum_{i = 0}^{n} \frac{1 - 2^{- 2 i + 1}}{2 i - 1} + \sum_{0 ⩽ i < j ⩽ n} \frac{1 - 2^{1 - i - j}}{i + j - 1} .