Question

express the product 2.5.8.....(3n-1) in terms of gamma function. First step states to take $$\Gamma(n+\frac{2}{3})$$...
why the particular integer 2/3 ?

Answer 1

the factorial you want is:
\[(3n -1) (3(n-1) - 1) (3(n-2) - 1)... = (3n -1)(3n -4)(3n -7)\]
using
\[\Gamma (x +1 ) = x \ \Gamma (x) \]
and x + 1 = 3n; x = 3n-1 will produce:
\[(3n -1) (3n-2) (3n-3)... = \ wrong\]
\[\Gamma (n +\frac{2}{3} ) = (n - \frac{1}{3} ) \ \Gamma (n - \frac{1}{3} ) \\ = (n - \frac{1}{3} ) \ (n - \frac{4}{3} ) \ \Gamma (n - \frac{4}{3} ) \]

look back up at what i think you actually want to calculate and you will now see the pattern.

Answer 2

Thank you!!. multiplying the last and first terms and equating to n gave the value.