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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,bN1n(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=1A=1abB=2a+bab2C=2b2n=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)+ban=1(Bn+a)+n=b+1(An)+n=ba+1(Bn+a)+n=1(Cn+b)=bn=1(An)+ban=1(Bn+a)+n=b+1(An)+n=b+1(Bn)+n=b+1(Cn)=bn=1(An)+ban=1(Bn+a)+n=b+1(A+B+Cn)=bn=1(An)+ban=1(Bn+a)=bn=1(An)+bn=a+1(Bn)=1ab[bn=11n(2ab+1)bn=a+11n]=1ab(an=11n2abbn=a+11n)=1aban=11n2b2bn=a+11n

Commented by tri26112004 last updated on 03/Jul/23

Thank you so much

Thankyousomuch

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