# $\lim_{y \to \infty} e^{-y}y^s=0$ and $\Gamma(s)$

In order to prove the functional equation of the gamma function,

$$\Gamma(s+1)=s\Gamma(s)$$

You have to show that

$$\lim_{y \to \infty} \frac{y^s}{e^{y}}=0$$

Taking exp and log in the numerator and after some calculations, you can apply L'Hôpital to prove the limit.

Is there a better/quicker proof of this limit?

• Directly check $e^x>x^{\lceil s \rceil}/\lceil s\rceil!$ maybe? (This is for real s, adjust as needed for complex with positive real part.) – Ian Sep 4 '16 at 11:37

Since $e^{y}=\sum_{k\geq0}\frac{y^{k}}{k!}$ we have $$\left|\frac{y^{s}}{e^{y}}\right|\leq\frac{y^{\left\lfloor \textrm{Re}\left(s\right)\right\rfloor +1}}{\sum_{0\leq k\leq\left\lfloor \textrm{Re}\left(s\right)\right\rfloor +2}y^{k}/k!}\leq\left(\left\lfloor \textrm{Re}\left(s\right)\right\rfloor +2\right)!\frac{y^{\left\lfloor \textrm{Re}\left(s\right)\right\rfloor +1}}{y^{\left\lfloor \textrm{Re}\left(s\right)\right\rfloor +2}}=\frac{\left(\left\lfloor \textrm{Re}\left(s\right)\right\rfloor +2\right)!}{y}\rightarrow0.$$