MHB Series involving gamma functions

AI Thread Summary
The discussion revolves around the series involving gamma functions, specifically the sum of the form $\sum_{n=0}^{\infty} \frac{\Gamma(a+n)\Gamma(b+n)}{n! \Gamma(c+n)}$ under the condition $c-a-b > 0$. It is noted that this series can be interpreted through Gauss' hypergeometric theorem, which provides a connection to hypergeometric functions. The transformation of the series into a form involving beta functions and integrals is discussed, leading to a simplified expression for the sum. The final result shows that the series equals $\frac{\Gamma(a) \Gamma(b) \Gamma(c-a-b)}{\Gamma(c-a) \Gamma(c-b)}$. This highlights the relationship between gamma functions and hypergeometric series in mathematical analysis.
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$\displaystyle \sum_{n={\bf 0}}^{\infty} \frac{\Gamma(a+n)\Gamma(b+n)}{n! \Gamma(c+n)} \ \ c-a-b > 0$
 
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I don't like this problem. It's really just Gauss' hypergeometric theorem in disguise.$ \displaystyle F(a,b;c;z) = \sum_{n=0}^{\infty} \frac{(a)_{n} (b)_{n}}{(c)_{n}} \frac{z_{n}}{n!} = \sum_{n=0}^{\infty} \frac{\Gamma(a+n) \Gamma(b+n) \Gamma(c)}{\Gamma(a) \Gamma(b) \Gamma(c+n)} \frac{z^{n}}{n!} $

$\displaystyle = \frac{1}{B(b,c-b)} \int_{0}^{1} x^{b-1} (1-x)^{c-b-1} (1-zx) ^{-a} \ dx \ \ |z|<1 \ \text{or} \ z=1$ let $z=1$$\displaystyle \sum_{n=0}^{\infty} \frac{\Gamma(a+n) \Gamma(b+n) \Gamma(c)}{\Gamma(a) \Gamma(b) \Gamma(c+n)} \frac{1}{n!}
= \frac{B(b,c-a-b)}{B(b,c-b)} = \frac{\Gamma(b) \Gamma(c-a-b) \Gamma(c)}{\Gamma(b) \Gamma(c-b) \Gamma(c-a)} $so $\displaystyle \sum_{n=0}^{\infty} \frac{\Gamma(a+n) \Gamma(b+n)}{n!\Gamma(c+n)} = \frac{\Gamma(a) \Gamma(b) \Gamma(c-a-b)}{\Gamma(c-a) \Gamma(c-b)}$
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
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