Question Number 103633 by Lordose last updated on 16/Jul/20

  The question is  Σ_(n=1) ^∞ (((2n−1)/(n(n+1)(n+2)))=...

Commented bybobhans last updated on 16/Jul/20

well well

Answered by Dwaipayan Shikari last updated on 16/Jul/20

S=Σ^∞ (((n+n+2−3)/(n(n+1)(n+2))))=Σ^∞ (1/((n+1)(n+2)))+Σ^∞ (1/(n(n+1)))−3S_1   S=Σ^∞ (1/(n+1))−(1/(n+2))+Σ^∞ (1/n)−(1/(n+1))−3S_1 =(3/2)−3S_1   S_1 =(1/2)Σ^∞ ((2+n−n)/(n(n+1)(n+2)))=(1/2)Σ^∞ (1/n)−(1/(n+1))−(1/2)Σ^∞ (1/(n+1))−(1/(n+2))  S_1 =(1/2)−(1/4)=(1/4)  {   Σ^∞ (1/n)−(1/(n+1))=lim_(n→∞) 1−(1/(n+1))=1  S=(3/2)−(3/4)=(3/4)     {Σ^∞ (1/(n+1))−(1/(n+2))=lim_(n→∞) (1/2)−(1/(n+2))=(1/2)

Commented bymathmax by abdo last updated on 16/Jul/20

not correct

Commented byDwaipayan Shikari last updated on 16/Jul/20

kindly can you show my mistake?

Answered by Worm_Tail last updated on 16/Jul/20

Σ_(n=1) ^∞ (((2n−1)/(n(n+1)(n+2)))=A      pfd     A=Σ((3/(n+1))−(5/(2(n+2)))−(1/(2n)))      A=   3Σ((1/(n+1)))−(5/2)Σ((1/(n+2)))−(1/2)Σ((1/n))      A=   3((1/2)+Σ_(n=2) ((1/(n+1))))−(5/2)Σ_(n=1) ((1/(n+2)))−(1/2)(1+(1/2)+Σ_(n=3) ((1/n)))      Σ_(n=2) ((1/(n+1)))=Σ_(n=1) ((1/(n+2)))=Σ_(n=3) ((1/n))=s      A=   3((1/2)+s)−(5/2)s−(1/2)(1+(1/2)+s)      A=   (3/2)+3s−(5/2)s−(1/2)−(1/4)−(1/2)s      A=   3s−(5/2)s−(1/2)s+(3/4)      A=   (3/4)

Commented bymathmax by abdo last updated on 16/Jul/20

correct answer thanks