Upper Bound on $\frac{\Gamma \left( \frac{n}{2}+x\right)}{\Gamma \left( \frac{n}{2}\right)}$

I am looking for an upper bound on the following ratio of Gamma functions \begin{align} \frac{\Gamma \left( \frac{n}{2}+x\right)}{\Gamma \left( \frac{n}{2}\right)} \end{align}

where $x \ge 0$ (real) and $n\ge 1$ (integer).

Here is the bound I was able to come up with \begin{align} \frac{\Gamma \left( \frac{n}{2}+x\right)}{\Gamma \left( \frac{n}{2}\right)} \le \frac{\Gamma \left( \left \lceil \frac{n}{2}+x \right \rceil \right) }{\Gamma \left( \left \lfloor \frac{n}{2} \right \rfloor \right)}=\frac{\left( \left \lceil \frac{n}{2}+x \right \rceil -1 \right) !}{\left( \left \lfloor \frac{n}{2} \right \rfloor -1\right)!} \end{align}

• You can't do better for an upper bound since you have equality for n even and x an integer. – abnry Oct 19 '15 at 13:34
• Thanks. This is what I thought too. I wanted to get rid of the floor and ceiling operation what is the best way to do that? – Boby Oct 19 '15 at 13:42
• For the ceiling: note that $\lceil \frac{n}{2} + x \rceil \le \frac{n}{2} + x +1$. – Hetebrij Oct 19 '15 at 14:07
• For the floor, we have $\lfloor \frac{n}{2} \rfloor = \frac{n}{2} +\frac{( -1)^{n} -1}{4}$. – Hetebrij Oct 19 '15 at 14:10
• @Hetebrij Thanks. – Boby Oct 19 '15 at 16:27