Comparing a ratio of gamma functions to a simple polynomial 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 $n=800$

Suppose that we consider the function
$$f(n)=\log \left(\Gamma \left(2n+3-\frac{a}{\log (n)}\right)\right)-\log
   (\Gamma (2 n+1))-2 \log (n+1)$$
Using Stirling approxiamtion followed by Taylor series, we have
$$f(n)=-\left(\frac{a \log (2)}{\log (n)}+a-2\log (2)\right)+\frac{a^2-5 a \log (n)-2 \log
   ^2(n)}{4 n \log ^2(n)}+\cdots$$
Ignoring the second term leads to a lower bound
$$n_{\text{low}}=2^{-\frac{a}{a-2 \log (2)}}$$ which, for the value of $a$, gives $n_{\text{low}}= 756.660$. Including the second term, Newton method converges immediatly at $n=792.720$.
Using Newton method for $f(n)=0$ with $n_0=n_{\text{low}}$, the iterates are 
$$\left(
\begin{array}{cc}
k & n_k \\
 0 & 756.6600 \\
 1 & 791.6120 \\
 2 & 792.7187 \\
 3 & 792.7197
\end{array}
\right)$$
