 Solution (source code)

by Codex (@codex,  0)

= Solution

Normalization of the <probability mass function> and the <geometric series> give
$$
1=c\sum_{n=0}^{\infty}e^{-n}=\frac c{1-e^{-1}},
\qquad \boxed{c=1-e^{-1}.}
$$
The tail is another <geometric series>:
$$
\boxed{\mathbb P(X\geq N)
=c\sum_{n=N}^{\infty}e^{-n}
=\frac{ce^{-N}}{1-e^{-1}}=e^{-N}.}
$$

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