# Bound on the tail of a Poisson branching process

I'm trying to understand this argument from "The Probabilsitic Method" book:

Let $$T_c$$ be the time of extinction for a Poisson branching process with parameter $$c$$. The authors prove that $$P[T_c=k] = \frac{e^{-ck}(ck)^{k-1} } {k!} .$$

From this they argue that, by Stirling's approximation, $$P[T_c = k] \sim \frac{1}{2 \pi}k^{-3/2}c^{-1}(ce^{1-c})^k.$$

If we assume $$c<1$$, then $$ce^{1-c}<1$$ and $$P[T_c=k]$$ approaches $$0$$ at exponential speed.

This is where I get lost: This gives a bound on the tail distribution: $$P[T_c \ge u] < e^{-u(\alpha +o(1))},$$ where $$\alpha = c-1-\ln c > 0$$.

Where does this come from? Is it some application of the Chernoff bound (if so, which version?), or is it more elementary than that?

$$P(T_c \ge u) = \sum_{k=u}^\infty P(T_c=k) \le \frac{1}{2\pi} u^{-3/2} c^{-1} \sum_{k=u}^\infty (c e^{1-c})^k = \frac{1}{2\pi} u^{-3/2} c^{-1} \frac{1}{1-ce^{1-c}} \cdot(c e^{1-c})^u.$$ The last term is $$(ce^{1-c})^u = e^{-u\alpha}$$ with $$\alpha := c-1-\ln c$$. The logarithm of the other terms is $$\ln \left(\frac{1}{2\pi} u^{-3/2} c^{-1} \frac{1}{1-ce^{1-c}}\right) = -\frac{3}{2} \ln u + C_c = o(u)$$ where $$C_c = \ln\left(\frac{1}{2\pi} c^{-1} \frac{1}{1-ce^{1-c}}\right)$$. Exponentiating both sides yields $$\frac{1}{2\pi} u^{-3/2} c^{-1} \frac{1}{1-ce^{1-c}} = e^{-u o(1)}.$$