Asymptotic behavior of $\frac{\Gamma \left(2-\frac{1}{p}\right) \Gamma (t+1)}{\Gamma \left(t-\frac{1}{p}+2\right)}$ I'm looking to find $t$ that satisfies the following equation, and treat it as $g(p)$, a function of $p>1$
$$
\frac{\Gamma \left(2-\frac{1}{p}\right) \Gamma (t+1)}{\Gamma
   \left(t-\frac{1}{p}+2\right)}=0.01
$$
Empirically, it seems like the following expression is an upper bound on $\log_{10} g$
$$5(p-1)^\frac{3}{4}$$

notebook

*

*Is this correct/tight upper bound?

*What is the asymptotic behavior of $g(p)$ as $p\to 1$?

*What is asymptotic behavior of $g(p)$ as $p\to\infty$?

 A: We can make use of a few asymptotic expressions for the gamma function as $p \to \infty$:
$$
\Gamma\left(2-\frac1p\right) \sim 1-\frac{1-\gamma}{p}
$$
where $\gamma$ is the Euler-Mascheroni constant, and
$$
\frac{\Gamma\left(t+2-\frac1p\right)}{\Gamma(t+1)} \sim (t+1)^{1-\frac1p}
$$
Then we have
$$
\frac{1-\frac{1-\gamma}{p}}{(t+1)^{1-\frac1p}} \approx \frac{1}{100}
$$
$$
(t+1)^{1-\frac1p} \approx 100\left(1-\frac{1-\gamma}{p}\right)
$$
\begin{align}
t+1 & \approx 100^\frac{p}{p-1} \left(1-\frac{1-\gamma}{p-1}\right) \\
    & \approx 100 \left(1+\frac{\ln 100}{p-1}\right)
                \left(1-\frac{1-\gamma}{p-1}\right) \\
    & \approx 100 \left(1+\frac{\ln 100-1+\gamma}{p-1}\right) \\
    & \approx 100 \left(1+\frac{\ln 100-1+\gamma}{p}\right)
\end{align}
since $\frac{1}{p-1} \sim \frac{1}{p}$, and finally we get
$$
t \approx 100 \left(1+\frac{\ln 100-1+\gamma}{p}\right) - 1
$$

Similarly, as $p \to 1$, we let $N = \frac{p}{p-1}$, and then as $N \to \infty$, we have
$$
\Gamma\left(1+\frac1N\right) \sim 1-\frac{\gamma}{N}
$$
and
$$
\frac{\Gamma\left(t+1+\frac1N\right)}{\Gamma(t+1)} \sim \sqrt[N]{t}
$$
Then we can have
$$
\frac{1-\frac{\gamma}{N}}{\sqrt[N]{t}} \approx \frac{1}{100}
$$
$$
\sqrt[N]{t} \approx 100\left(1-\frac{\gamma}{N}\right)
$$
$$
t \approx 100^N e^{-\gamma}
$$

It should be clear how to generalize this for ratios other than $\frac{1}{100}$.
