geometric distribution and expected value How I can resolve this problem ?
Set $X \sim \operatorname{Geo}(p)$, $0 \leqslant p \leqslant 1$.
Evaluate $\operatorname{E}(e^{-X})$.
I can't use $e^{-X}$ in this distribution.
 A: With $f_X(x)=p(1-p)^{x-1}$ we have:
$$\begin{align}
\mathsf{E}(e^{-X})&=\sum_{x=1}^{\infty} e^{-x}p(1-p)^{x-1}\\
&=pe^{-1}\sum_{x=1}^{\infty} e^{-(x-1)}(1-p)^{x-1}\\
&=pe^{-1}\sum_{x=1}^{\infty} (e^{-1}(1-p))^{x-1}\\
&=\frac{pe^{-1}}{1-e^{-1}(1-p)}
\end{align}$$ 
where the last term is calculated using the Geometric sum formula.
A: Sometimes $\operatorname{Geo}(p)$ means a certain distribution of random variables taking values in the set $\{1,2,3,\ldots\}$, and sometimes it means the set $\{0,1,2,3,\ldots\}$ instead. In the former case the expected value is just $1$ unit bigger than in the latter.
I'll assume it's $\{0,1,2,3,\ldots\}$.  You have $\Pr(X=x) = pq^x$.
\begin{align}
\operatorname{E}(e^{-X}) & = \sum_{x=0}^\infty e^{-x} \Pr(X=x) = \sum_{x=0}^\infty e^{-x} pq^x = p \sum_{x=0}^\infty (qe^{-1})^x \\[10pt]
& = p \sum_{x=0}^\infty r^x =p\times \left( \begin{array}{l} \text{the sum of a geometric series with first} \\ \text{term 1 and common ratio }r=qe^{-1}  \end{array} \right).
\end{align}
If you know how to find the sum of a geometric series then you can finish this.
