I have recently begun to study John Kingsman's "Poisson processes", and in the first chapter, it defines what a random variable that is Poisson distributed with parameter $\mu$ looks like, i.e.
$\mathbb{P}$({$X$ = $n$}) = $\frac{e^{-\mu}}{n!}$$\mu^n$
However, it says, that if $\mu$ = $\infty$, so if X~Poisson($\infty$), then $\mathbb{P}$({$X$ = +$\infty$}) = 1
I am having trouble understanding why the latter is true, since this would mean that $e^{-\mu}$ tends to zero. I am grateful for any help or piece or advice.
First off, considering the Poisson distribution with $\mu=\infty$ is nonstandard, in my experience, but what the author has done makes sense.
The fact that it is defined to be a “point mass at infinity” should be viewed as a definition and not something you can reason by plugging into the pmf. (Although on the other hand, you’ve conspicuously forgotten to “plug in” $n=\infty$ In fact, your argument seems to say $P(X=n) =0$ for any finite $n,$ which is true when $\mu=\infty$ according to the definition).
For an argument for why the distribution chosen is sensible, compute the limit as $\mu\to\infty$ of $P(X>n)$ for any fixed $n$.