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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?

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$$P(T_c \ge u) = \sum_{k=u}^\infty P(T_c=k) \le \frac{1}{2\pi} u^{-3/2} c^{-1} \sum_{k=u}^\infty (c e^{1-c})^k = \frac{1}{2\pi} u^{-3/2} c^{-1} \frac{1}{1-ce^{1-c}} \cdot(c e^{1-c})^u.$$ 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 $$\frac{1}{2\pi} u^{-3/2} c^{-1} \frac{1}{1-ce^{1-c}} = e^{-u o(1)}.$$

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