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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∑n=1∞(−1)nn(n+1)(n+2)(n+3)
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)
letdecomposeF(x)=1x(x+1)(x+2)(x+3)F(x)=ax+bx+1+cx+2+dx+3a=limx→0xF(x)=16b=limx→−1(x+1)F(x)=−12c=limx→−2(x+2)F(x)=12d=limx→−3(x+3)F(x)=−16⇒F(x)=16x−12(x+1)+12(x+2)−16(x+3)⇒S=16∑n=1∞(−1)nn−12∑n=1∞(−1)nn+1+12∑n=1∞(−1)nn+2−16∑n=1∞(−1)nn+3but∑n=1∞(−1)nn=−ln(2)∑n=1∞(−1)nn+1=∑n=2∞(−1)n−1n=ln(2)−1∑n=1∞(−1)nn+2=∑n=3∞(−1)n−2n=∑n=3∞(−1)nn=−ln(2)−{−1+12}=−ln(2)+12∑n=1∞(−1)nn+3=∑n=4∞(−1)n−3n=∑n=4∞(−1)n−1n=ln(2)−{1−12+13}=ln(2)−(12+13)=ln(2)−56⇒S=−16ln(2)−12ln(2)+12−12ln(2)+14−16ln(2)+536S=(−13−1)ln(2)+34+536=−43ln(2)+27+536=−43ln(2)+1618S=89−43ln(2)

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