how can I prove this statement using the o definition?

Question

If having $$\Omega_i =\frac{\Gamma(i+v+1)\Gamma(i+s+1)}{i!\Gamma(i+v+s+2)},$$ how can I prove that $\lvert \Omega_i \rvert=o(\frac{1}{i}) $. I am really struggling to understand and use the definition of "o"

Answer 1

Take logarithms and use Stirling approximation four times and continue with Taylor expansion.

This should give

$$\log(\Omega_i)=\log \left(\frac{\Gamma (i+s+1) \Gamma (i+v+1)}{i!\, \Gamma (i+s+v+2)}\right)=-\log(i)-\frac{sv+s+v+1 }{i}+O\left(\frac{1}{i^2}\right)$$ Continuing with Taylor, $$\Omega_i=e^{\log(\Omega_i)}=\frac{1}{i}-\frac{sv+s+v+1}{i^2}+O\left(\frac{1}{i^2}\right)$$

Answer 2

we can use this theorem $\Gamma(x+n)$ acts as $x^{n-1}x!$ and get the following: $\Omega_i$ acts as $\frac{i!i^{v}i!i^{s}}{i!i!i^{v+s+1}}=\frac{1}{i}$ Therefore, the prove is done (honestly, that's my professor answer)