For x&gt;0 and y&gt;x+1, how do we prove that ∑n=1∞x(x+1)…(x+n−1)y(y+1)…(y+n−1)=xy−x−1

So: Γ(x+n)Γ(y−x)Γ(y+n)=∫01tx+n−1(1−t)y−x−1dt summing over n we get, Γ(y−x)∑n≥1Γ(x+n)Γ(y+n)=∫01∑n≥1tx+n−1(1−t)y−x−1dt=∫01tx(1−t)y−x−2dt or, ∑n≥1Γ(x+n)Γ(y+n)=1Γ(y−x)∫01tx+1−1(1−t)y−x−1−1dt =Γ(x+1)Γ(y−x−1)Γ(y−x)Γ(y) =Γ(x+1)(y−x−1)Γ(y) and hence, Γ(y)Γ(x)∑n≥1Γ(x+n)Γ(y+n)=Γ(x+1)(y−x−1)Γ(x)=xy−x−1

"For x>0 and y>x+1, how do we prove..." was answered by our experts

Miguel Davenport

Answered question

2022-01-23

For

x > 0

and

y > x + 1

, how do we prove that

\sum_{n = 1}^{\infty} \frac{x (x + 1) \dots (x + n - 1)}{y (y + 1) \dots (y + n - 1)} = \frac{x}{y - x - 1}

Answer & Explanation

becky4208fj

Beginner2022-01-24Added 10 answers

So:

\frac{Γ (x + n) Γ (y - x)}{Γ (y + n)} = \int_{0}^{1} t^{x + n - 1} {(1 - t)}^{y - x - 1} d t

summing over n we get,

Γ (y - x) \sum_{n \geq 1} \frac{Γ (x + n)}{Γ (y + n)} = \int_{0}^{1} \sum_{n \geq 1} t^{x + n - 1} {(1 - t)}^{y - x - 1} d t = \int_{0}^{1} t^{x} {(1 - t)}^{y - x - 2} d t

or,

\sum_{n \geq 1} \frac{Γ (x + n)}{Γ (y + n)} = \frac{1}{Γ (y - x)} \int_{0}^{1} t^{x + 1 - 1} {(1 - t)}^{y - x - 1 - 1} d t

= \frac{Γ (x + 1) Γ (y - x - 1)}{Γ (y - x) Γ (y)}

= \frac{Γ (x + 1)}{(y - x - 1) Γ (y)}

and hence,

\frac{Γ (y)}{Γ (x)} \sum_{n \geq 1} \frac{Γ (x + n)}{Γ (y + n)} = \frac{Γ (x + 1)}{(y - x - 1) Γ (x)} = \frac{x}{y - x - 1}

Do you have a similar question?

Recalculate according to your conditions!

New Questions in Calculus 2

Didn't find what you were looking for?