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Question Number 37342 by math khazana by abdo last updated on 12/Jun/18
calculate Σ_(n=1) ^∞ (((−1)^n )/(n^2 (n+1))) x^n with ∣x∣<1 2) find the value of Σ_(n=1) ^∞ (1/(n^2 (n+1)2^n )) .
$${calculate}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{\mathrm{2}} \left({n}+\mathrm{1}\right)}\:{x}^{{n}} \:\:\:{with}\:\mid{x}\mid<\mathrm{1} \\ $$$$\left.\mathrm{2}\right)\:{find}\:{the}\:{value}\:{of}\:\:\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\:\frac{\mathrm{1}}{{n}^{\mathrm{2}} \left({n}+\mathrm{1}\right)\mathrm{2}^{{n}} }\:. \\ $$
Commented by prof Abdo imad last updated on 15/Jun/18
let decompose F(x)=(1/(x^2 (x+1))) F(x)=(a/x) +(b/x^2 ) +(c/(x+1)) b=lim_(x→0) x^2 F(x)=1 c =lim_(x→−1) (x+1)F(x)=1 ⇒ F(x)= (a/x) +(1/x^2 ) +(1/(x+1)) F(1)=(1/2) =a +1 +(1/2) ⇒a=−1 ⇒ F(x)=−(1/x) +(1/x^2 ) +(1/(x+1)) ⇒ S(x)=Σ_(n=1) ^∞ (−1)^n {−(1/n) +(1/n^2 ) +(1/(n+1))}x^n =−Σ_(n=1) ^∞ (((−1)^n )/n) x^n +Σ_(n=1) ^∞ (((−1)^n )/n^2 ) x^n + Σ_(n=1) ^∞ (((−1)^n )/(n+1))x^n let w(x)=Σ_(n=1) ^∞ (((−1)^n )/n)x^n w^′ (x)= Σ_(n=1) ^∞ (−1)^n x^(n−1) =(1/x){ Σ_(n=0) ^∞ (−x)^n −1} = (1/(x(x+1))) −(1/x) =(1/x){(1/(x+1)) −1}=(1/x)((−x)/(1+x)) = ((−1)/(1+x)) ⇒w(x)=−ln(1+x) +c butc=w(1)=0 ⇒w(x)= −ln(1+x) Σ_(n=1) ^∞ (((−1)^n )/(n+1)) x^n = Σ_(n=2) ^∞ (((−1)^(n−1) )/n) x^(n−1) =−(1/x)Σ_(n=2) ^∞ (((−1)^n )/n) x^n =−(1/x){ Σ_(n=1) ^∞ (((−1)^n )/n)x^n +1} =−(1/x){−ln(1+x)} −(1/x) = (1/x)ln(1+x) −(1/x) let v(x)= Σ_(n=1) ^∞ (((−1)^n )/n^2 )x^n v^′ (x) =Σ_(n=1) ^∞ (((−1)^n )/n)x^n =−ln(1+x) ⇒ v(x)= −∫ln(1+x)dx +λ =−{ xln(1+x)−∫ (x/(1+x))dx} +λ =xln(1+x) +∫ ((1+x−1)/(1+x))dx +λ =xln(1+x) +x −ln(1+x) +λ =(x−1)ln(1+x) +x+λ but λ=v(0)=0 ⇒ v(x)=(x−1)ln(1+x) +x ⇒ S(x)=ln(1+x)+(1/x){ln(1+x)−1} +(x−1)ln(1+x) +x ⇒ S(x)=x ln(1+x) +(1/x)ln(1+x) +x−(1/x) = ((x^2 +1)/x)ln(1+x) +((x^2 −1)/x) with −1<x<1 and x≠0
$${let}\:{decompose}\:{F}\left({x}\right)=\frac{\mathrm{1}}{{x}^{\mathrm{2}} \left({x}+\mathrm{1}\right)} \\ $$$${F}\left({x}\right)=\frac{{a}}{{x}}\:+\frac{{b}}{{x}^{\mathrm{2}} }\:+\frac{{c}}{{x}+\mathrm{1}} \\ $$$${b}={lim}_{{x}\rightarrow\mathrm{0}} {x}^{\mathrm{2}} {F}\left({x}\right)=\mathrm{1} \\ $$$${c}\:={lim}_{{x}\rightarrow−\mathrm{1}} \left({x}+\mathrm{1}\right){F}\left({x}\right)=\mathrm{1}\:\Rightarrow \\ $$$${F}\left({x}\right)=\:\frac{{a}}{{x}}\:+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\:+\frac{\mathrm{1}}{{x}+\mathrm{1}} \\ $$$${F}\left(\mathrm{1}\right)=\frac{\mathrm{1}}{\mathrm{2}}\:={a}\:+\mathrm{1}\:+\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow{a}=−\mathrm{1}\:\Rightarrow \\ $$$${F}\left({x}\right)=−\frac{\mathrm{1}}{{x}}\:+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\:+\frac{\mathrm{1}}{{x}+\mathrm{1}}\:\Rightarrow \\ $$$${S}\left({x}\right)=\sum_{{n}=\mathrm{1}} ^{\infty} \left(−\mathrm{1}\right)^{{n}} \left\{−\frac{\mathrm{1}}{{n}}\:+\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\:+\frac{\mathrm{1}}{{n}+\mathrm{1}}\right\}{x}^{{n}} \\ $$$$=−\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}}\:{x}^{{n}} \:+\sum_{{n}=\mathrm{1}} ^{\infty} \frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{\mathrm{2}} }\:{x}^{{n}} \\ $$$$\:+\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}+\mathrm{1}}{x}^{{n}} \:\:\:{let}\:{w}\left({x}\right)=\sum_{{n}=\mathrm{1}} ^{\infty} \frac{\left(−\mathrm{1}\right)^{{n}} }{{n}}{x}^{{n}} \\ $$$${w}^{'} \left({x}\right)=\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\left(−\mathrm{1}\right)^{{n}} {x}^{{n}−\mathrm{1}} =\frac{\mathrm{1}}{{x}}\left\{\:\sum_{{n}=\mathrm{0}} ^{\infty} \left(−{x}\right)^{{n}} \:−\mathrm{1}\right\} \\ $$$$=\:\frac{\mathrm{1}}{{x}\left({x}+\mathrm{1}\right)}\:−\frac{\mathrm{1}}{{x}}\:=\frac{\mathrm{1}}{{x}}\left\{\frac{\mathrm{1}}{{x}+\mathrm{1}}\:−\mathrm{1}\right\}=\frac{\mathrm{1}}{{x}}\frac{−{x}}{\mathrm{1}+{x}} \\ $$$$=\:\frac{−\mathrm{1}}{\mathrm{1}+{x}}\:\Rightarrow{w}\left({x}\right)=−{ln}\left(\mathrm{1}+{x}\right)\:+{c}\:{butc}={w}\left(\mathrm{1}\right)=\mathrm{0} \\ $$$$\Rightarrow{w}\left({x}\right)=\:−{ln}\left(\mathrm{1}+{x}\right) \\ $$$$\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}+\mathrm{1}}\:{x}^{{n}} \:=\:\sum_{{n}=\mathrm{2}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{{n}}\:{x}^{{n}−\mathrm{1}} \\ $$$$=−\frac{\mathrm{1}}{{x}}\sum_{{n}=\mathrm{2}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}}\:{x}^{{n}} \\ $$$$=−\frac{\mathrm{1}}{{x}}\left\{\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}}{x}^{{n}} \:+\mathrm{1}\right\} \\ $$$$=−\frac{\mathrm{1}}{{x}}\left\{−{ln}\left(\mathrm{1}+{x}\right)\right\}\:−\frac{\mathrm{1}}{{x}}\:=\:\frac{\mathrm{1}}{{x}}{ln}\left(\mathrm{1}+{x}\right)\:−\frac{\mathrm{1}}{{x}} \\ $$$${let}\:{v}\left({x}\right)=\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{\mathrm{2}} }{x}^{{n}} \\ $$$${v}^{'} \left({x}\right)\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}}{x}^{{n}} =−{ln}\left(\mathrm{1}+{x}\right)\:\Rightarrow \\ $$$${v}\left({x}\right)=\:−\int{ln}\left(\mathrm{1}+{x}\right){dx}\:+\lambda \\ $$$$=−\left\{\:{xln}\left(\mathrm{1}+{x}\right)−\int\:\frac{{x}}{\mathrm{1}+{x}}{dx}\right\}\:+\lambda \\ $$$$={xln}\left(\mathrm{1}+{x}\right)\:+\int\:\:\frac{\mathrm{1}+{x}−\mathrm{1}}{\mathrm{1}+{x}}{dx}\:+\lambda \\ $$$$={xln}\left(\mathrm{1}+{x}\right)\:\:+{x}\:−{ln}\left(\mathrm{1}+{x}\right)\:+\lambda \\ $$$$=\left({x}−\mathrm{1}\right){ln}\left(\mathrm{1}+{x}\right)\:+{x}+\lambda\:{but}\:\lambda={v}\left(\mathrm{0}\right)=\mathrm{0}\:\Rightarrow \\ $$$${v}\left({x}\right)=\left({x}−\mathrm{1}\right){ln}\left(\mathrm{1}+{x}\right)\:+{x}\:\Rightarrow \\ $$$${S}\left({x}\right)={ln}\left(\mathrm{1}+{x}\right)+\frac{\mathrm{1}}{{x}}\left\{{ln}\left(\mathrm{1}+{x}\right)−\mathrm{1}\right\} \\ $$$$+\left({x}−\mathrm{1}\right){ln}\left(\mathrm{1}+{x}\right)\:+{x}\:\:\:\Rightarrow \\ $$$${S}\left({x}\right)={x}\:{ln}\left(\mathrm{1}+{x}\right)\:+\frac{\mathrm{1}}{{x}}{ln}\left(\mathrm{1}+{x}\right)\:\:+{x}−\frac{\mathrm{1}}{{x}} \\ $$$$=\:\frac{{x}^{\mathrm{2}} \:+\mathrm{1}}{{x}}{ln}\left(\mathrm{1}+{x}\right)\:+\frac{{x}^{\mathrm{2}} \:−\mathrm{1}}{{x}}\:\:{with}\:−\mathrm{1}<{x}<\mathrm{1}\:{and} \\ $$$${x}\neq\mathrm{0} \\ $$$$ \\ $$$$ \\ $$
Commented by prof Abdo imad last updated on 15/Jun/18
Σ_(n=1) ^∞ (1/(n^2 (n+1)2^n )) =S((1/2)) =(((1/4)+1)/(1/2))ln((3/2)) +(((1/4)−1)/(1/2)) =2.(5/4)ln((3/2)) +2.(((−3)/4)) =(5/2)ln((3/2)) −(3/2)
$$\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\mathrm{1}}{{n}^{\mathrm{2}} \left({n}+\mathrm{1}\right)\mathrm{2}^{{n}} }\:={S}\left(\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$=\frac{\frac{\mathrm{1}}{\mathrm{4}}+\mathrm{1}}{\frac{\mathrm{1}}{\mathrm{2}}}{ln}\left(\frac{\mathrm{3}}{\mathrm{2}}\right)\:\:+\frac{\frac{\mathrm{1}}{\mathrm{4}}−\mathrm{1}}{\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$=\mathrm{2}.\frac{\mathrm{5}}{\mathrm{4}}{ln}\left(\frac{\mathrm{3}}{\mathrm{2}}\right)\:+\mathrm{2}.\left(\frac{−\mathrm{3}}{\mathrm{4}}\right) \\ $$$$=\frac{\mathrm{5}}{\mathrm{2}}{ln}\left(\frac{\mathrm{3}}{\mathrm{2}}\right)\:−\frac{\mathrm{3}}{\mathrm{2}} \\ $$

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