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Question Number 41408 by maxmathsup by imad last updated on 06/Aug/18

calculate Σ_(n=1) ^∞ (((−1)^n )/(n(n+1)(n+2)(n+3)))

$${calculate}\:\:\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)\left({n}+\mathrm{3}\right)} \\ $$

Commented by maxmathsup by imad last updated on 07/Aug/18

let decompose F(x)= (1/(x(x+1)(x+2)(x+3))) F(x)=(a/x) +(b/(x+1)) +(c/(x+2)) +(d/(x+3)) a =lim_(x→0) xF(x)=(1/6) b =lim_(x→−1) (x+1)F(x) = −(1/2) c =lim_(x→−2) (x+2)F(x) = (1/2) d =lim_(x→−3) (x+3)F(x)=−(1/6) ⇒ F(x)=(1/(6x)) −(1/(2(x+1))) +(1/(2(x+2))) −(1/(6(x+3))) ⇒ S =(1/6) Σ_(n=1) ^∞ (((−1)^n )/n) −(1/2)Σ_(n=1) ^∞ (((−1)^n )/(n+1)) +(1/2) Σ_(n=1) ^∞ (((−1)^n )/(n+2)) −(1/6)Σ_(n=1) ^∞ (((−1)^n )/(n+3)) but Σ_(n=1) ^∞ (((−1)^n )/n) =−ln(2) Σ_(n=1) ^∞ (((−1)^n )/(n+1)) =Σ_(n=2) ^∞ (((−1)^(n−1) )/n) =ln(2)−1 Σ_(n=1) ^∞ (((−1)^n )/(n+2)) =Σ_(n=3) ^∞ (((−1)^(n−2) )/n) =Σ_(n=3) ^∞ (((−1)^n )/n) =−ln(2)−{−1+(1/2)} =−ln(2)+(1/2) Σ_(n=1) ^∞ (((−1)^n )/(n+3)) =Σ_(n=4) ^∞ (((−1)^(n−3) )/n) =Σ_(n=4) ^∞ (((−1)^(n−1) )/n) =ln(2)−{1−(1/2) +(1/3)} =ln(2)−((1/2) +(1/3))=ln(2)−(5/6) ⇒ S =−(1/6)ln(2) −(1/2)ln(2) +(1/2) −(1/2)ln(2) +(1/4) −(1/6)ln(2) +(5/(36)) S =(−(1/3) −1)ln(2) +(3/4) +(5/(36)) =−(4/3)ln(2) + ((27+5)/(36)) =−(4/3)ln(2) + ((16)/(18)) S =(8/9) −(4/3)ln(2)

$${let}\:{decompose}\:{F}\left({x}\right)=\:\frac{\mathrm{1}}{{x}\left({x}+\mathrm{1}\right)\left({x}+\mathrm{2}\right)\left({x}+\mathrm{3}\right)} \\ $$$${F}\left({x}\right)=\frac{{a}}{{x}}\:+\frac{{b}}{{x}+\mathrm{1}}\:+\frac{{c}}{{x}+\mathrm{2}}\:+\frac{{d}}{{x}+\mathrm{3}} \\ $$$${a}\:={lim}_{{x}\rightarrow\mathrm{0}} {xF}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{6}} \\ $$$${b}\:={lim}_{{x}\rightarrow−\mathrm{1}} \left({x}+\mathrm{1}\right){F}\left({x}\right)\:=\:−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${c}\:={lim}_{{x}\rightarrow−\mathrm{2}} \left({x}+\mathrm{2}\right){F}\left({x}\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${d}\:={lim}_{{x}\rightarrow−\mathrm{3}} \left({x}+\mathrm{3}\right){F}\left({x}\right)=−\frac{\mathrm{1}}{\mathrm{6}}\:\Rightarrow \\ $$$${F}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{6}{x}}\:−\frac{\mathrm{1}}{\mathrm{2}\left({x}+\mathrm{1}\right)}\:+\frac{\mathrm{1}}{\mathrm{2}\left({x}+\mathrm{2}\right)}\:−\frac{\mathrm{1}}{\mathrm{6}\left({x}+\mathrm{3}\right)}\:\Rightarrow \\ $$$${S}\:=\frac{\mathrm{1}}{\mathrm{6}}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}}\:−\frac{\mathrm{1}}{\mathrm{2}}\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}+\mathrm{1}}\:+\frac{\mathrm{1}}{\mathrm{2}}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}+\mathrm{2}}\:−\frac{\mathrm{1}}{\mathrm{6}}\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}+\mathrm{3}}\:{but} \\ $$$$\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}}\:=−{ln}\left(\mathrm{2}\right) \\ $$$$\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}+\mathrm{1}}\:=\sum_{{n}=\mathrm{2}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{{n}}\:={ln}\left(\mathrm{2}\right)−\mathrm{1} \\ $$$$\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}+\mathrm{2}}\:=\sum_{{n}=\mathrm{3}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{2}} }{{n}}\:=\sum_{{n}=\mathrm{3}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}}\:=−{ln}\left(\mathrm{2}\right)−\left\{−\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\right\} \\ $$$$=−{ln}\left(\mathrm{2}\right)+\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}+\mathrm{3}}\:=\sum_{{n}=\mathrm{4}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{3}} }{{n}}\:=\sum_{{n}=\mathrm{4}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{{n}} \\ $$$$={ln}\left(\mathrm{2}\right)−\left\{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{3}}\right\}\:={ln}\left(\mathrm{2}\right)−\left(\frac{\mathrm{1}}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{3}}\right)={ln}\left(\mathrm{2}\right)−\frac{\mathrm{5}}{\mathrm{6}}\:\Rightarrow \\ $$$${S}\:=−\frac{\mathrm{1}}{\mathrm{6}}{ln}\left(\mathrm{2}\right)\:−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{2}\right)\:+\frac{\mathrm{1}}{\mathrm{2}}\:−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{2}\right)\:+\frac{\mathrm{1}}{\mathrm{4}}\:−\frac{\mathrm{1}}{\mathrm{6}}{ln}\left(\mathrm{2}\right)\:+\frac{\mathrm{5}}{\mathrm{36}} \\ $$$${S}\:=\left(−\frac{\mathrm{1}}{\mathrm{3}}\:−\mathrm{1}\right){ln}\left(\mathrm{2}\right)\:\:+\frac{\mathrm{3}}{\mathrm{4}}\:+\frac{\mathrm{5}}{\mathrm{36}}\:=−\frac{\mathrm{4}}{\mathrm{3}}{ln}\left(\mathrm{2}\right)\:+\:\frac{\mathrm{27}+\mathrm{5}}{\mathrm{36}}\:=−\frac{\mathrm{4}}{\mathrm{3}}{ln}\left(\mathrm{2}\right)\:+\:\frac{\mathrm{16}}{\mathrm{18}} \\ $$$${S}\:=\frac{\mathrm{8}}{\mathrm{9}}\:−\frac{\mathrm{4}}{\mathrm{3}}{ln}\left(\mathrm{2}\right) \\ $$

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