Question Number 148203 by liberty last updated on 26/Jul/21

Σ_(n=1) ^∞ ((8/(n^2 +n)))=?

Answered by mathmax by abdo last updated on 26/Jul/21

S=8Σ_(n=1) ^∞ (1/(n(n+1))) =8lim_(n→+∞) Σ_(k=1) ^n (1/(k(k+1))) but Σ_(k=1) ^n (1/(k(k+1)))=Σ_(k=1) ^n ((1/k)−(1/(k+1)))=1−(1/2)+(1/2)−(1/3)+...+(1/n)−(1/(n+1)) =1−(1/(n+1)) ⇒S=8lim_(n→+∞) (1−(1/(n+1)))=8×1=8

Answered by puissant last updated on 26/Jul/21

posons f(x)=8Σ_(n=1) ^∞ (x^(n−1) /(n(n+1))) ⇒f(x)=8Σ_(n=1) ^∞ (x^(n−1) /n) − 8Σ_(n=1) ^∞ (x^(n−1) /(n+1)) or 8Σ_(n=1) ^∞ (x^(n−1) /n)=(8/x)Σ_(n=1) ^∞ (x^n /n)=−(8/x)ln(1−x) d′autre part, 8Σ_(n=1) ^∞ (x^(n−1) /(n+1))= 8Σ_(n=2) ^∞ (x^(n−2) /n)= (8/x^2 )Σ_(n=2) ^∞ (x^n /n) =−(8/x^2 )(ln(1−x)+x) ⇒f(x)=8(−(1/x)ln(1−x)+(1/x^2 )(x+ln(1−x))) ⇒f(x)=8((1/x)+((1/x^2 )−(1/x))ln(1−x)) alors, S=f(1) ⇒ S=8×1=8....