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Question Number 194241 by tri26112004 last updated on 01/Jul/23

Answered by mr W last updated on 02/Jul/23

assumed a,b∈N (1/(n(n+a)(n+b)))=(A/n)+(B/(n+a))+(C/(n+b)) (A+B+C)n^2 +[(2a+b)A+bB]n+abA=1 A+B+C=0 (2a+b)A+bB=0 abA=1 ⇒A=(1/(ab)) ⇒B=−((2a+b)/(ab^2 )) ⇒C=(2/b^2 ) Σ_(n=1) ^∞ (1/(n(n+a)(n+b))) =Σ_(n=1) ^∞ ((A/n)+(B/(n+a))+(C/(n+b))) =Σ_(n=1) ^∞ ((A/n))+Σ_(n=1) ^∞ ((B/(n+a)))+Σ_(n=1) ^∞ ((C/(n+b))) =Σ_(n=1) ^b ((A/n))+Σ_(n=1) ^(b−a) ((B/(n+a)))+Σ_(n=b+1) ^∞ ((A/n))+Σ_(n=b−a+1) ^∞ ((B/(n+a)))+Σ_(n=1) ^∞ ((C/(n+b))) =Σ_(n=1) ^b ((A/n))+Σ_(n=1) ^(b−a) ((B/(n+a)))+Σ_(n=b+1) ^∞ ((A/n))+Σ_(n=b+1) ^∞ ((B/n))+Σ_(n=b+1) ^∞ ((C/n)) =Σ_(n=1) ^b ((A/n))+Σ_(n=1) ^(b−a) ((B/(n+a)))+Σ_(n=b+1) ^∞ (((A+B+C)/n)) =Σ_(n=1) ^b ((A/n))+Σ_(n=1) ^(b−a) ((B/(n+a))) =Σ_(n=1) ^b ((A/n))+Σ_(n=a+1) ^b ((B/n)) =(1/(ab))[Σ_(n=1) ^b (1/n)−(((2a)/b)+1)Σ_(n=a+1) ^b (1/n)] =(1/(ab))(Σ_(n=1) ^a (1/n)−((2a)/b)Σ_(n=a+1) ^b (1/n)) =(1/(ab))Σ_(n=1) ^a (1/n)−(2/b^2 )Σ_(n=a+1) ^b (1/n) ✓

assumeda,b∈N1n(n+a)(n+b)=An+Bn+a+Cn+b(A+B+C)n2+[(2a+b)A+bB]n+abA=1A+B+C=0(2a+b)A+bB=0abA=1⇒A=1ab⇒B=−2a+bab2⇒C=2b2∑∞n=11n(n+a)(n+b)=∑∞n=1(An+Bn+a+Cn+b)=∑∞n=1(An)+∑∞n=1(Bn+a)+∑∞n=1(Cn+b)=∑bn=1(An)+∑b−an=1(Bn+a)+∑∞n=b+1(An)+∑∞n=b−a+1(Bn+a)+∑∞n=1(Cn+b)=∑bn=1(An)+∑b−an=1(Bn+a)+∑∞n=b+1(An)+∑∞n=b+1(Bn)+∑∞n=b+1(Cn)=∑bn=1(An)+∑b−an=1(Bn+a)+∑∞n=b+1(A+B+Cn)=∑bn=1(An)+∑b−an=1(Bn+a)=∑bn=1(An)+∑bn=a+1(Bn)=1ab[∑bn=11n−(2ab+1)∑bn=a+11n]=1ab(∑an=11n−2ab∑bn=a+11n)=1ab∑an=11n−2b2∑bn=a+11n✓

Commented by tri26112004 last updated on 03/Jul/23

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