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)))


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Question Number 78456 by arkanmath7@gmail.com last updated on 17/Jan/20

$f o r a > 0 a n d b > a + 2, v e r i f y t h e f o l l w i n g$ $c l a i m :$ $\sum_{n = 1}^{\infty} n \frac{a (a + 1) (a + 2) . . . (a + n - 1)}{b (b + 1) (b + 2) . . . (b + n - 1)} = \frac{a (b - 1)}{(b - a - 1) (b - a - 2)}$
	Answered by mind is power last updated on 17/Jan/20
	$=Σ_(n≥1) (((n+1−1)a_n )/b_n )=Σ_(n≥1) (((n+1)!.a_n )/(b_n .n!))−Σ_(n≥1) ((n!.a_n )/(n!.b_n ))=S a_n =a(a+1)...(a+n−1) _2 F_1 (a,b,c,z)=1+Σ_(n≥1) ((a_n .b_n )/c_n ).(z^n /(n!)) = lim_(z→1) { _2 F_1 (a,2,b,z)−_2 F_1 (a,1,b,z)} we have β(b,c−b).2F_1 (a,b,c,z)=∫_0 ^1 x^(b−1) (1−x)^(c−b−1) (1−zx)^(−a) dx β(b,c−b) _2 F_1 (a,b,c,1)=∫_0 ^1 x^(b−1) (1−x)^(c−b−a−1) dx ⇒β(b,c−b) _2 F_1 (a,b,c,1)=β(b,c−b−a) ⇒_2 F_1 =((β(b,c−b−a))/(β(b,c−b)))=((Γ(b)Γ(c−b−a)Γ(c))/(Γ(c−a)Γ(b)Γ(c−b)))⇔ 2F_1 (a,b,c,1)=((Γ(c).Γ(c−a−b))/(Γ(c−a)Γ(c−b))) 2F_1 (a,2,b,1)=((Γ(b)Γ(b−a−2))/(Γ(b−a)Γ(b−2)))=(((b−1)(b−2)Γ(b−2)Γ(b−a−2))/((b−a−1)(b−a−2)Γ(b−a−2)Γ(b−2))) =(((b−1)(b−2))/((b−a−1)(b−a−2))) 2F_1 (a,1,b,1)=((Γ(b)Γ(b−a−1))/(Γ(b−a)Γ(b−1)))=(((b−1)Γ(b−1)Γ(b−a−1))/((b−a−1)Γ(b−a−1)Γ(b−1))) =((b−1)/(b−a−1)) S=2F_1 (a,2,b,1)−2F_1 (a,1,b,1)= S=(((b−1)(b−2))/((b−a−1)(b−a−2)))−((b−1)/((b−a−1)))=((b−1)/(b−a−1)).(((b−2)/(b−a−2))−1) =(((b−1))/((b−a−1)))((a/(b−a−2)))=((a(b−1))/((b−a−1)(b−a−2))) what i used 2F_1 (a,b,c,z)=Σ((a_n .b_n )/c_n ).(z^n /(n!)) (n+1)!=1.2......(n+2−1) n=1.2.....(n+1−1) β(x,y)=∫_0 ^1 t^(x−1) .(1−t)^(y−1) dt=((Γ(x).Γ(y))/(Γ(x+y)))$
	$= \sum_{n ⩾ 1} \frac{(n + 1 - 1) a_{n}}{b_{n}} = \sum_{n ⩾ 1} \frac{(n + 1)! . a_{n}}{b_{n} . n!} - \sum_{n ⩾ 1} \frac{n! . a_{n}}{n! . b_{n}} = S$ $a_{n} = a (a + 1) . . . (a + n - 1)$ $_{2} F_{1} (a, b, c, z) = 1 + \sum_{n ⩾ 1} \frac{a_{n} . b_{n}}{c_{n}} . \frac{z^{n}}{n!}$ $= \lim_{z \to 1} {_{2} F_{1} (a, 2, b, z) -_{2} F_{1} (a, 1, b, z)}$ $we have$ $β (b, c - b) . {2 F}_{1} (a, b, c, z) = \int_{0}^{1} x^{b - 1} {(1 - x)}^{c - b - 1} {(1 - zx)}^{- a} dx$ $β (b, c - b)_{2} F_{1} (a, b, c, 1) = \int_{0}^{1} x^{b - 1} {(1 - x)}^{c - b - a - 1} dx$ $\Rightarrow β (b, c - b)_{2} F_{1} (a, b, c, 1) = β (b, c - b - a)$ $\Rightarrow_{2} F_{1} = \frac{β (b, c - b - a)}{β (b, c - b)} = \frac{Γ (b) Γ (c - b - a) Γ (c)}{Γ (c - a) Γ (b) Γ (c - b)} \Leftrightarrow$ ${2 F}_{1} (a, b, c, 1) = \frac{Γ (c) . Γ (c - a - b)}{Γ (c - a) Γ (c - b)}$ ${2 F}_{1} (a, 2, b, 1) = \frac{Γ (b) Γ (b - a - 2)}{Γ (b - a) Γ (b - 2)} = \frac{(b - 1) (b - 2) Γ (b - 2) Γ (b - a - 2)}{(b - a - 1) (b - a - 2) Γ (b - a - 2) Γ (b - 2)}$ $= \frac{(b - 1) (b - 2)}{(b - a - 1) (b - a - 2)}$ ${2 F}_{1} (a, 1, b, 1) = \frac{Γ (b) Γ (b - a - 1)}{Γ (b - a) Γ (b - 1)} = \frac{(b - 1) Γ (b - 1) Γ (b - a - 1)}{(b - a - 1) Γ (b - a - 1) Γ (b - 1)}$ $= \frac{b - 1}{b - a - 1}$ $S = {2 F}_{1} (a, 2, b, 1) - {2 F}_{1} (a, 1, b, 1) =$ $S = \frac{(b - 1) (b - 2)}{(b - a - 1) (b - a - 2)} - \frac{b - 1}{(b - a - 1)} = \frac{b - 1}{b - a - 1} . (\frac{b - 2}{b - a - 2} - 1)$ $= \frac{(b - 1)}{(b - a - 1)} (\frac{a}{b - a - 2}) = \frac{a (b - 1)}{(b - a - 1) (b - a - 2)}$ $what i used {2 F}_{1} (a, b, c, z) = Σ \frac{a_{n} . b_{n}}{c_{n}} . \frac{z^{n}}{n!}$ $(n + 1)! = 1.2 . . . . . . (n + 2 - 1)$ $n = 1.2 . . . . . (n + 1 - 1)$ $β (x, y) = \int_{0}^{1} t^{x - 1} . {(1 - t)}^{y - 1} dt = \frac{Γ (x) . Γ (y)}{Γ (x + y)}$
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