Assume that a random experiment consists in centering a telescopic sight on a random star. Let $A_{n}$ ($n \in \mathbb{Z}^{+}$) denote the event that the telescopic sight spots exactly $n$ stars. Suppose that $P(A_{n})=\frac{6}{\pi^{2}n^{2}}$. Assume further that star shootings occurs independently for various stars and that the probability of a given star to experience this over a one hour period is $10^{-6}$.
What is the probability that one will observe some star shooting within an hour of study?
I am lead to the disastrous conclusion that I should write $P(S)=\sum_{n=1}^{\infty}P(A_{n})*10^{-6}$, which clearly does not converge...
Bit worried about your astronomy but never mind. lets call $q=10^{-6}$
Given that you see n stars probability that you don't see one shooting is $(1-q)^n$ Thus probability that don't see a star shoot in a random experiment is $\sum_n A_n (1-q)^n = \frac{6}{\pi^2} L_2(1-q)$. Where $L_2$ is the special function the polylogarithm.
So probability you do see a shooting star is 1 minus this quantity, which for small q is $\approx \frac{6}{\pi^2} \left(q +\frac{q^2}{4} \right)$.