# Monotonicity of the incomplete gamma ratio

Is it possible to prove that a ratio of an upper incomplete gamma functions is monotonically decreasing?

Specifically, let $\Gamma(s+1,u) = \int_u^\infty \tau^{s} \mathrm{e}^{-\tau}\mathrm{d}\tau$ be the upper incomplete gamma function. Then, I would like to show that $$\frac{\Gamma(s,u+z)}{\Gamma(s,u)}$$ is monotonically decreasing as $u$ grows, assuming $u$ and $z\in \mathbb{R}_+$ .

That is, for $0 < u_1 \leq u_2 \leq... \leq u_n < \infty$

$$\frac{\Gamma(s,u_1+z)}{\Gamma(s,u_1)}\geq \frac{\Gamma(s,u_2+z)}{\Gamma(s,u_2)}\geq ...\geq \frac{\Gamma(s,u_n+z)}{\Gamma(s,u_n)}$$

I tried the first derivative test, without any success.

I think it may be solved by using L'Hôpital's rule, but I'm not sure how.

Any help and ideas are much appreciated.

L'Hopital rule it is. It is enough to show that $$\frac{\partial }{\partial u}\frac{\Gamma(s,u+z)}{\Gamma(s,u)} = \frac{\frac{\partial }{\partial u}\Gamma(s,u+z)\cdot\Gamma(s,u)-\frac{\partial }{\partial u}\Gamma(s,u)\cdot\Gamma(s,u+z)}{\Gamma(s,u)^2}\leq 0$$ or $$(u+z)^s e^{-(u+z)}\int_{u}^{+\infty}t^s e^{-t}\,dt-u^s e^{-u}\int_{u+z}^{+\infty} t^s e^{-t}\,dt\geq 0$$ or $$\int_{0}^{+\infty}\left[(u+t)^s (u+z)^s-u^s(u+t+z)^s\right] e^{-t}\,dt\geq 0$$ which is just a consequence of $$(u+t)(u+z)\geq u(u+t+z)\quad\Longleftrightarrow\quad tz\geq 0.$$