find A_n =∫_1 ^2 ( 1 +(1/x) +(1/x^2 ) +...+(1/x^n ))^2 dx


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

$f i n d A_{n} = \int_{1}^{2} {(1 + \frac{1}{x} + \frac{1}{x^{2}} + . . . + \frac{1}{x^{n}})}^{2} d x$
Commented by prof Abdo imad last updated on 18/Jun/18
$let a_i = (1/x^i ) with 0≤i≤n ⇒ (1+(1/x) +(1/x^2 ) +....+(1/x^n ))^2 =(Σ_(i=0) ^n a_i )^2 = Σ_(i=0) ^n a_i ^2 +2 Σ_(0≤i<j≤n) a_i .a_j =Σ_(i=0) ^n (1/x^(2i) ) +2 Σ_(0≤i<j≤n) (1/x^(i+j) ) ⇒ A_n = ∫_1 ^2 (Σ_(i=0) ^n (1/x^(2i) ))dx +2 ∫_1 ^2 Σ_(0≤i<j≤n) (1/x^(i+j) )dx =Σ_(i=0) ^n ∫_1 ^2 x^(−2i) dx +2 Σ_(0≤i<j≤n) ∫_1 ^2 x^(−(i+j)) dx Σ_(i=0) ^n (1/(−2i+1))[ x^(−2i+1) ]_1 ^2 +2 Σ_(0≤i<j≤n) (1/(1−(i+j)))[ x^(1−(i+j)) ]_1 ^2 =−Σ_(i=0) ^n (1/(2i−1)){ 2^(−2i+1) −1} −Σ_(0≤i<j≤n) (1/(i+j−1)){ 2^(1−(i+j)) −1}$
$l e t a_{i} = \frac{1}{x^{i}} w i t h 0 ⩽ i ⩽ n \Rightarrow$ ${(1 + \frac{1}{x} + \frac{1}{x^{2}} + . . . . + \frac{1}{x^{n}})}^{2} = {(\sum_{i = 0}^{n} a_{i})}^{2}$ $= \sum_{i = 0}^{n} a_{i}^{2} + 2 \sum_{0 ⩽ i < j ⩽ n} a_{i} . a_{j}$ $= \sum_{i = 0}^{n} \frac{1}{x^{2 i}} + 2 \sum_{0 ⩽ i < j ⩽ n} \frac{1}{x^{i + j}} \Rightarrow$ $A_{n} = \int_{1}^{2} (\sum_{i = 0}^{n} \frac{1}{x^{2 i}}) d x + 2 \int_{1}^{2} \sum_{0 ⩽ i < j ⩽ n} \frac{1}{x^{i + j}} d x$ $= \sum_{i = 0}^{n} \int_{1}^{2} x^{- 2 i} d x + 2 \sum_{0 ⩽ i < j ⩽ n} \int_{1}^{2} x^{- (i + j)} d x$ $\sum_{i = 0}^{n} \frac{1}{- 2 i + 1} {[x^{- 2 i + 1}]}_{1}^{2}$ $+ 2 \sum_{0 ⩽ i < j ⩽ n} \frac{1}{1 - (i + j)} {[x^{1 - (i + j)}]}_{1}^{2}$ $= - \sum_{i = 0}^{n} \frac{1}{2 i - 1} {2^{- 2 i + 1} - 1}$ $- \sum_{0 ⩽ i < j ⩽ n} \frac{1}{i + j - 1} {2^{1 - (i + j)} - 1}$
Commented by math khazana by abdo last updated on 18/Jun/18

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