I am still struggling to build my intuition as far as reasoning with ratios of gamma functions.
Reasoning with factorials is significantly clearer.
Consider this example. I would appreciate if anyone could help me to understand how to complete the following with regard to gamma functions.
Let $n > 1$ be any integer.
Clearly:
$$\frac{(2n + 2)!}{(2n)!} = (2n+2)(2n+1) > (n+1)^2 = n^2+2n+1$$
So, changing this to a ratio of Gamma functions, the equivalent is:
$$\frac{\Gamma(2n + 3)}{\Gamma(2n+1)} = (2n+2)(2n+1) > (n+1)^2 = n^2+2n+1$$
So far, so good.
My problem comes down to evaluating when a fraction less than 1 gets applied.
For example, consider the value of $\frac{1.25506}{\ln n}$ which is less than $1$ for $n > e^{1.25506}$
While it is easy to figure out any given value and it is straight forward to generate a graph, how do I show that this value is true for $n > 800$ for example. How would I determine the derivative and show that is increasing (which I suspect it is)?
$$\frac{\Gamma(2n+ 3 - \frac{1.25506}{\ln n})}{\Gamma(2n+1)} > n^2+2n+1$$
In other words, as I leave the safety of factorials, I am at a loss for how to prove or disprove the inequality for all $n > k$ where $k > 800$ for example.
Edit: I think that the inequality may not be true for $\dfrac{5n}{3}$.
I am switching from $\dfrac{5n}{3}$ to $2n$. I believe that this inequality might be true for a reasonably sized $n$.
I believe that the inequality is true for $n=800$