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Frobenius & Special Functions · Easy problem

The Gamma function, which appears in the Frobenius series for JνJ_\nu, satisfies Γ(s+1)=s Γ(s)\Gamma(s+1) = s\,\Gamma(s). Compute Γ(7/2)Γ(3/2)\dfrac{\Gamma(7/2)}{\Gamma(3/2)}. Express your answer as a common fraction.