Question Number 37271 by abdo.msup.com last updated on 11/Jun/18
$${find}\:\:{A}_{{n}} =\int_{\mathrm{1}} ^{\mathrm{2}} \left(\:\mathrm{1}\:+\frac{\mathrm{1}}{{x}}\:+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\:+…+\frac{\mathrm{1}}{{x}^{{n}} }\right)^{\mathrm{2}} {dx} \\ $$
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}](https://www.tinkutara.com/question/Q37831.png)
$${let}\:\:{a}_{{i}} =\:\frac{\mathrm{1}}{{x}^{{i}} }\:\:{with}\:\mathrm{0}\leqslant{i}\leqslant{n}\:\Rightarrow \\ $$$$\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\:+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\:+….+\frac{\mathrm{1}}{{x}^{{n}} }\right)^{\mathrm{2}} \:=\left(\sum_{{i}=\mathrm{0}} ^{{n}} \:{a}_{{i}} \right)^{\mathrm{2}} \\ $$$$=\:\sum_{{i}=\mathrm{0}} ^{{n}} \:{a}_{{i}} ^{\mathrm{2}} \:\:\:+\mathrm{2}\:\sum_{\mathrm{0}\leqslant{i}<{j}\leqslant{n}} {a}_{{i}} .{a}_{{j}} \\ $$$$=\sum_{{i}=\mathrm{0}} ^{{n}} \:\:\frac{\mathrm{1}}{{x}^{\mathrm{2}{i}} }\:+\mathrm{2}\:\sum_{\mathrm{0}\leqslant{i}<{j}\leqslant{n}} \:\:\frac{\mathrm{1}}{{x}^{{i}+{j}} }\:\Rightarrow \\ $$$${A}_{{n}} \:=\:\int_{\mathrm{1}} ^{\mathrm{2}} \:\left(\sum_{{i}=\mathrm{0}} ^{{n}} \:\frac{\mathrm{1}}{{x}^{\mathrm{2}{i}} }\right){dx}\:+\mathrm{2}\:\int_{\mathrm{1}} ^{\mathrm{2}} \sum_{\mathrm{0}\leqslant{i}<{j}\leqslant{n}} \:\frac{\mathrm{1}}{{x}^{{i}+{j}} }{dx} \\ $$$$=\sum_{{i}=\mathrm{0}} ^{{n}} \:\int_{\mathrm{1}} ^{\mathrm{2}} \:\:\:{x}^{−\mathrm{2}{i}} {dx}\:+\mathrm{2}\:\sum_{\mathrm{0}\leqslant{i}<{j}\leqslant{n}} \int_{\mathrm{1}} ^{\mathrm{2}} \:{x}^{−\left({i}+{j}\right)} {dx} \\ $$$$\sum_{{i}=\mathrm{0}} ^{{n}} \:\frac{\mathrm{1}}{−\mathrm{2}{i}+\mathrm{1}}\left[\:{x}^{−\mathrm{2}{i}+\mathrm{1}} \right]_{\mathrm{1}} ^{\mathrm{2}} \\ $$$$+\mathrm{2}\:\sum_{\mathrm{0}\leqslant{i}<{j}\leqslant{n}} \:\frac{\mathrm{1}}{\mathrm{1}−\left({i}+{j}\right)}\left[\:{x}^{\mathrm{1}−\left({i}+{j}\right)} \right]_{\mathrm{1}} ^{\mathrm{2}} \\ $$$$=−\sum_{{i}=\mathrm{0}} ^{{n}} \:\:\frac{\mathrm{1}}{\mathrm{2}{i}−\mathrm{1}}\left\{\:\mathrm{2}^{−\mathrm{2}{i}+\mathrm{1}} \:−\mathrm{1}\right\} \\ $$$$−\sum_{\mathrm{0}\leqslant{i}<{j}\leqslant{n}} \:\frac{\mathrm{1}}{{i}+{j}−\mathrm{1}}\left\{\:\mathrm{2}^{\mathrm{1}−\left({i}+{j}\right)} \:−\mathrm{1}\right\} \\ $$
Commented by math khazana by abdo last updated on 18/Jun/18
$${A}_{{n}} =\:\sum_{{i}=\mathrm{0}} ^{{n}} \:\frac{\mathrm{1}−\mathrm{2}^{−\mathrm{2}{i}+\mathrm{1}} }{\mathrm{2}{i}−\mathrm{1}}\:\:+\sum_{\mathrm{0}\leqslant{i}<{j}\leqslant{n}} \frac{\mathrm{1}−\mathrm{2}^{\mathrm{1}−{i}−{j}} }{{i}+{j}−\mathrm{1}}\:\:. \\ $$