The question asks for the value of $\Gamma(n+\frac{1}{2})$ given $\Gamma (n+1) = n\Gamma(n)$ and $\Gamma (1/2) = \sqrt{\pi}$. We need to express $\Gamma(n+\frac{1}{2})$ in terms of standard factorials.

Deriving the Gamma Function Formula

We use the property $\Gamma(z+1) = z\Gamma(z)$ repeatedly to expand $\Gamma(n+\frac{1}{2})$.

• Start with the expression: $\Gamma(n+\frac{1}{2})$
• Apply the property $\Gamma(z+1) = z\Gamma(z)$ with $z = n - \frac{1}{2}$: $\Gamma(n+\frac{1}{2}) = (n - \frac{1}{2}) \Gamma(n - \frac{1}{2})$
• Apply the property again with $z = n - \frac{3}{2}$: $= (n - \frac{1}{2}) (n - \frac{3}{2}) \Gamma(n - \frac{3}{2})$
• Continue this process until we reach $\Gamma(\frac{1}{2})$: $= (n - \frac{1}{2}) (n - \frac{3}{2}) \cdots (\frac{1}{2}) \Gamma(\frac{1}{2})$

Expressing the Product

Let's rewrite the terms and the product:

• The expression becomes: $ \Gamma(n+\frac{1}{2}) = \left(\frac{2n-1}{2}\right) \left(\frac{2n-3}{2}\right) \cdots \left(\frac{1}{2}\right) \Gamma(\frac{1}{2}) $
• Combine the fractions: $ \Gamma(n+\frac{1}{2}) = \frac{(2n-1)(2n-3)\cdots(1)}{2^n} \Gamma(\frac{1}{2}) $ The numerator is the product of the first $n$ odd integers, also known as the double factorial $(2n-1)!!$.

Relating to Standard Factorials

We can express the double factorial $(2n-1)!!$ using standard factorials.

• Consider the factorial $(2n)!$: $ (2n)! = (2n)(2n-1)(2n-2)(2n-3)\cdots(2)(1) $
• Separate the even and odd terms: $ (2n)! = [(2n)(2n-2)\cdots(2)] \times [(2n-1)(2n-3)\cdots(1)] $
• Factor out $2$ from the even terms: $ (2n)! = [2^n (n)(n-1)\cdots(1)] \times (2n-1)!! $ $ (2n)! = 2^n n! \times (2n-1)!! $
• Solve for the double factorial: $ (2n-1)!! = \frac{(2n)!}{2^n n!} $

Final Calculation

Substitute the expression for $(2n-1)!!$ back into the equation for $\Gamma(n+\frac{1}{2})$:

• Substitute $(2n-1)!!$: $ \Gamma(n+\frac{1}{2}) = \frac{1}{2^n} \left( \frac{(2n)!}{2^n n!} \right) \Gamma(\frac{1}{2}) $
• Simplify the expression: $ \Gamma(n+\frac{1}{2}) = \frac{(2n)!}{2^{2n} n!} \Gamma(\frac{1}{2}) $
• Use the given value $\Gamma(\frac{1}{2}) = \sqrt{\pi}$: $ \Gamma(n+\frac{1}{2}) = \frac{(2n)!}{2^{2n} n!} \sqrt{\pi} $
• This matches the form $\frac{2n!}{ n! 2^{2n}} \sqrt{\pi}$.