for example there is a Binomial RV $S_n \sim \mathrm{Binomial}(n,\frac{1}{n})$. I'd like to use Markov inequality to find the tail probability:
$ P(S_n \geq \varepsilon) \leq \frac{\mathbf{E}S_n}{\varepsilon} $
PGF of the Binomial RV is
$ \mathbf{E}z^X=\sum_{j=0}^{n}\binom{n}{k}z^k \frac{1}{n^k}(1-\frac{1}{n})^{n-k}=\Bigg(1+\frac{z-1}{n}\Bigg)^n $
Therefore Markov inequality becomes $ P(S_n \geq \varepsilon)=P(z^{S_n} \geq z^{\varepsilon}) \leq \frac{\mathbf{E}z^{S_n}}{z^{\varepsilon}}=\frac{\Bigg(1+\frac{z-1}{n}\Bigg)^n}{z^{\varepsilon}} $
My question is, how to derive $z$? I know of course that $\mathbf{E}S_n=\frac{d}{dz}\mathbf{E}z^X|_{z=1}=1 $ (in this case), but I wonder if this tail probability can be derived in a different way.
EDIT: as long as I know, this has something to do with 'dominant singularities', but MSE/MO don't seem to have any relevant questions.