Question Number 103623 by Lordose last updated on 16/Jul/20

find the sum of the series whose nth  term is ((2n−1)/(n(n+1)(n+2)).  i have a problem with this and i need  help please

Answered by OlafThorendsen last updated on 16/Jul/20

find the sum of the series whose nth  term is ((2n−1)/(n(n+1)(n+2)).  i have a problem with this and i need  help please  u_n  = ((2n−1)/(n(n+1)(n+2)))  u_n  = −(1/2).(1/n)+(3/(n+1))−(5/2).(1/(n+2))  Σ_(n=1) ^p u_n  = −(1/2)Σ_(n=1) ^p (1/n)+3Σ_(n=1) ^p (1/(n+1))−(5/2)Σ_(n=1) ^p (1/(n+2))  Σ_(n=1) ^p u_n  = −(1/2)Σ_(n=1) ^p (1/n)+3Σ_(n=2) ^(p+1) (1/n)−(5/2)Σ_(n=3) ^(p+2) (1/n)  Σ_(n=1) ^p u_n  = −(1/2)(1+(1/2))+3((1/2))+3((1/(p+1)))−(5/2)((1/(p+1))+(1/(p+2)))  Σ_(n=1) ^p u_n  = (3/4)+(1/2)((1/(p+1)))−(5/2)((1/(p+2)))  Σ_(n=1) ^p u_n  = (3/4)−((2p+(3/2))/((p+1)(p+2)))  and Σ_(n=1) ^∞ u_n  = (3/4)

Commented byLordose last updated on 16/Jul/20

sir i don′t understand the 4th line  can you explain pls

Answered by bobhans last updated on 16/Jul/20

Σ_(n=1) ^∞ ((2n−1)/(n(n+1)(n+2)))  =Σ_(n=1) ^∞  (3/(n+1))−Σ_(n=1) ^∞  (1/(2n))−Σ_(n=1) ^∞  (5/(2(n+2)))  = Σ_(n=2) ^∞ (3/n)−(1/2)Σ_(n=1) ^∞ (1/n)−(5/2)Σ_(n=3) ^∞ (1/n)  =Σ_(n=1) ^∞ (3/n)−3−(1/2)Σ_(n=1) ^∞ (1/n)−(5/2){Σ_(n=1) ^∞ (1/n)−1−(1/2)}  =−3+((15)/4)+(3−(1/2)−(5/2))Σ_(n=1) ^∞ (1/n)  = (3/4) +0 lim_(k→∞)  Σ_(n=1) ^k  (1/n) = (3/4) + 0 = (3/4)

Commented byLordose last updated on 16/Jul/20

can anyone make me understand pls

Commented bybobhans last updated on 16/Jul/20

what part you don′t understand?