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for-a-gt-0-and-b-gt-a-2-verify-the-follwing-claim-n-1-n-a-a-1-a-2-a-n-1-b-b-1-b-2-b-n-1-a-b-1-b-a-1-b-a-2-

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Question Number 78456 by arkanmath7@gmail.com last updated on 17/Jan/20
for a>0 and b>a+2 , verify the follwing claim: Σ_(n=1) ^( ∞) n ((a(a+1)(a+2)...(a+n−1))/(b(b+1)(b+2)...(b+n−1))) =((a(b−1))/((b−a−1)(b−a−2)))
fora>0andb>a+2,verifythefollwingclaim:∑n=1∞na(a+1)(a+2)…(a+n−1)b(b+1)(b+2)…(b+n−1)=a(b−1)(b−a−1)(b−a−2)
Answered by mind is power last updated on 17/Jan/20
=Σ_(n≥1) (((n+1−1)a_n )/b_n )=Σ_(n≥1) (((n+1)!.a_n )/(b_n .n!))−Σ_(n≥1) ((n!.a_n )/(n!.b_n ))=S a_n =a(a+1)...(a+n−1) _2 F_1 (a,b,c,z)=1+Σ_(n≥1) ((a_n .b_n )/c_n ).(z^n /(n!)) = lim_(z→1) { _2 F_1 (a,2,b,z)−_2 F_1 (a,1,b,z)} we have β(b,c−b).2F_1 (a,b,c,z)=∫_0 ^1 x^(b−1) (1−x)^(c−b−1) (1−zx)^(−a) dx β(b,c−b) _2 F_1 (a,b,c,1)=∫_0 ^1 x^(b−1) (1−x)^(c−b−a−1) dx ⇒β(b,c−b) _2 F_1 (a,b,c,1)=β(b,c−b−a) ⇒_2 F_1 =((β(b,c−b−a))/(β(b,c−b)))=((Γ(b)Γ(c−b−a)Γ(c))/(Γ(c−a)Γ(b)Γ(c−b)))⇔ 2F_1 (a,b,c,1)=((Γ(c).Γ(c−a−b))/(Γ(c−a)Γ(c−b))) 2F_1 (a,2,b,1)=((Γ(b)Γ(b−a−2))/(Γ(b−a)Γ(b−2)))=(((b−1)(b−2)Γ(b−2)Γ(b−a−2))/((b−a−1)(b−a−2)Γ(b−a−2)Γ(b−2))) =(((b−1)(b−2))/((b−a−1)(b−a−2))) 2F_1 (a,1,b,1)=((Γ(b)Γ(b−a−1))/(Γ(b−a)Γ(b−1)))=(((b−1)Γ(b−1)Γ(b−a−1))/((b−a−1)Γ(b−a−1)Γ(b−1))) =((b−1)/(b−a−1)) S=2F_1 (a,2,b,1)−2F_1 (a,1,b,1)= S=(((b−1)(b−2))/((b−a−1)(b−a−2)))−((b−1)/((b−a−1)))=((b−1)/(b−a−1)).(((b−2)/(b−a−2))−1) =(((b−1))/((b−a−1)))((a/(b−a−2)))=((a(b−1))/((b−a−1)(b−a−2))) what i used 2F_1 (a,b,c,z)=Σ((a_n .b_n )/c_n ).(z^n /(n!)) (n+1)!=1.2......(n+2−1) n=1.2.....(n+1−1) β(x,y)=∫_0 ^1 t^(x−1) .(1−t)^(y−1) dt=((Γ(x).Γ(y))/(Γ(x+y)))
=∑n⩾1(n+1−1)anbn=∑n⩾1(n+1)!.anbn.n!−∑n⩾1n!.ann!.bn=San=a(a+1)…(a+n−1)2F1(a,b,c,z)=1+∑n⩾1an.bncn.znn!=limz→1{2F1(a,2,b,z)−2F1(a,1,b,z)}wehaveβ(b,c−b).2F1(a,b,c,z)=∫01xb−1(1−x)c−b−1(1−zx)−adxβ(b,c−b)2F1(a,b,c,1)=∫01xb−1(1−x)c−b−a−1dx⇒β(b,c−b)2F1(a,b,c,1)=β(b,c−b−a)⇒2F1=β(b,c−b−a)β(b,c−b)=Γ(b)Γ(c−b−a)Γ(c)Γ(c−a)Γ(b)Γ(c−b)⇔2F1(a,b,c,1)=Γ(c).Γ(c−a−b)Γ(c−a)Γ(c−b)2F1(a,2,b,1)=Γ(b)Γ(b−a−2)Γ(b−a)Γ(b−2)=(b−1)(b−2)Γ(b−2)Γ(b−a−2)(b−a−1)(b−a−2)Γ(b−a−2)Γ(b−2)=(b−1)(b−2)(b−a−1)(b−a−2)2F1(a,1,b,1)=Γ(b)Γ(b−a−1)Γ(b−a)Γ(b−1)=(b−1)Γ(b−1)Γ(b−a−1)(b−a−1)Γ(b−a−1)Γ(b−1)=b−1b−a−1S=2F1(a,2,b,1)−2F1(a,1,b,1)=S=(b−1)(b−2)(b−a−1)(b−a−2)−b−1(b−a−1)=b−1b−a−1.(b−2b−a−2−1)=(b−1)(b−a−1)(ab−a−2)=a(b−1)(b−a−1)(b−a−2)whatiused2F1(a,b,c,z)=Σan.bncn.znn!(n+1)!=1.2……(n+2−1)n=1.2…..(n+1−1)β(x,y)=∫01tx−1.(1−t)y−1dt=Γ(x).Γ(y)Γ(x+y)

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