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I found this characteristic function:

$$ E[e^{-\lambda X}] = \sum_{k=0}^{\infty} e^{-\lambda k} P(X=k) $$

How can I find P(X=0)?

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    That looks like a moment-generating function. You want $E[e^{-it X}]$ for a characteristic function.2012-03-06
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    Assume $X \geq 0$. By dominated convergence theorem, we have $$ \lim_{\lambda \to \infty} \mathbb{E} [ e^{-\lambda X} ] = \mathbb{E} [ \lim_{\lambda \to \infty} e^{-\lambda X} ] = \mathbb{E} [ \chi_{\{X = 0\}} ] = P(X = 0). $$2012-03-07

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As Henry says, it's the moment generating function of $X$, and its value is taken at $-\lambda$. We can write $$E[e^{-\lambda X}]=P(X=0)+\sum_{k=1}^{+\infty}e^{-\lambda k}P(X=k),$$ and for $\lambda >0$ $$0\leq \sum_{k=1}^{+\infty}e^{-\lambda k}P(X=k)\leq \sum_{k=1}^{+\infty}e^{-\lambda k}=\frac{e^{-\lambda}}{1-e^{-\lambda}}$$ so $\lim_{\lambda\to+\infty}E[e^{-\lambda X}]=P(X=0)$ and if you know $E[e^{-\lambda X}]$ and its limit when $\lambda\to +\infty$ is not too hard to compute you can deduce $P(X=0)$.