# Prove the asymptote $\Gamma \left(x+\frac{1}{\Gamma(x-y)} \right)/ \Gamma \left(\frac{1}{\Gamma(x-y)} \right) \asymp (x-2)^y$ as $x \to \infty$

I noticed it first for Pochhammer symbols for integers:

$$\left( \frac{1}{n!} \right)_n \asymp \frac{1}{n} \quad \text{as} \quad n \to \infty$$

$$\left( \frac{1}{n!} \right)_{n+1} \asymp 1 \quad \text{as} \quad n \to \infty$$

And so on.

Writing explicitly, for example, the last expression:

$$\left( \frac{1}{n!} \right)_{n+1}=\frac{1}{(n!)^{n+1}} (n!+1)(2n!+1) \cdots (n n!+1)$$

I don't see how to prove the limit.

In general, for the integer case, we seem to have:

$$\left( \frac{1}{(n-k)!} \right)_{n+1} \asymp (n-1)^k \quad \text{as} \quad n \to \infty$$

Generalizing with the Gamma function, it turns to:

$$\frac{\Gamma \left(x+\frac{1}{\Gamma(x-y)} \right) }{ \Gamma \left(\frac{1}{\Gamma(x-y)} \right)} \asymp (x-2)^y \quad \text{as} \quad x \to \infty \tag{1}$$

We should also assume $x >> y$.

Which seems to work numerically (see the plots below): How can we prove $(1)$ for the real case, or at least for the integer case?

• Perhaps Stirling's approximation? – Simply Beautiful Art Aug 30 '16 at 20:54
• @SimpleArt, maybe, I forgot about it. But we have nested Gammas here... – Yuriy S Aug 30 '16 at 20:59
• $$\frac{\Gamma\left(x+\frac1{\Gamma(x-y)}\right)}{\Gamma\left(\frac1{\Gamma(x-y)}\right)}\sim\sqrt{x\left(\frac{x-y}e\right)^{x-y}\sqrt{2\pi (x-y)}}\left(x+\frac{e^{x-y}}{(x-y)^{x-y}\sqrt{2\pi (x-y)}}\right)^xe^{-x}\left(x(\frac{x-y}e)^{x-y}\sqrt{2\pi (x-y)}+1\right)^{\frac{e^{x-y}}{(x-y)^{x-y}\sqrt{2\pi (x-y)}}}$$ Hope you can use this? – Simply Beautiful Art Aug 30 '16 at 21:12
• Regarding the first Pochhammer symbol in your post, note that $$\left( \frac{1}{n!} \right)_n=\prod_{k=0}^{n-1}\left( \frac{1}{n!}+k \right) =\frac{1}{n!}\prod_{k=1}^{n-1}\left[k\cdot\left(1+ \frac{1}{n!\cdot k} \right)\right]=\frac{1}{n}\prod_{k=1}^{n-1}\left(1+ \frac{1}{n!\cdot k} \right)$$ hence $$1\leqslant n\cdot\left( \frac{1}{n!} \right)_n\leqslant\left(1+ \frac{1}{n!} \right)^{n-1}\leqslant e^{n/n!}\to1$$ which proves that the $\asymp$ is actually a full equivalent, and the same holds for ... – Did Aug 30 '16 at 21:32
• $$\left( \frac{1}{n!} \right)_{n+1}=\left( \frac{1}{n!} \right)_n\left( \frac{1}{n!}+n \right)=\left( \frac{1}{n!} \right)_n\cdot n\cdot\left( 1+\frac{1}{n\cdot n!} \right)$$ The same strategy should lead you to a proof for every other quantity you are interested in in your post. – Did Aug 30 '16 at 21:32