Poisson power series We have a Poisson power series of
$$Y=\sum\limits_{k=0}^{\infty}e^{-\pi\lambda v^2}\frac{(\pi\lambda v^2)^k}{k!}(A)^k $$
If we have a disk with radius $v$
where A is defined as the density of a distance of some node from the origin  placed randomly inside the disk, $A=\frac{2x}{v^2}$.
If i try to plug in k = 0 first, then we have
$$Y=e^{-\pi\lambda v^2}\frac{(\pi\lambda v^2)^0}{0!}(A)^0$$
$$Y=e^{-\pi\lambda v^2}$$
next plug in k=1 and so on, then
$$Y=e^{-\pi\lambda v^2}+e^{-\pi\lambda v^2}\frac{(\pi\lambda v^2)^1}{1!}(A)^1+....+e^{-\pi\lambda v^2}\frac{(\pi\lambda v^2)^{\infty}}{\infty!}(A)^{\infty}$$
however something power to the infinity is undefined, how can we simplified the series so that we can have the result such as
$$\exp \{-\pi\lambda v^2+\pi\lambda v^2 (A)\}$$ 
I try to answer this, rewrite
$Y=e^{-\pi\lambda v^2} \sum\limits_{k=0}^\infty \frac{(\pi\lambda v^2)^k}{k!}A^k$
$Y=e^{-\pi\lambda v^2} \sum\limits_{k=0}^\infty \frac{(\pi\lambda v^2A)^k}{k!}$
based on series formula $e^x = \sum\limits_{k=0}^\infty \frac{x^k}{k!}$ we have
$Y = e^{-\pi\lambda v^2}e^{\pi\lambda v^2A}$ , hence
$Y=\exp\bigg\{-\pi\lambda v^2 + \pi\lambda v^2A\bigg\}$
although the end result is the same, however i'm still not sure about the processes. Is it correct ? 
 A: You need to learn a little about convergence of infinite series
before you tackle this.
First step: Consider the geometric series:
$$A = 1/2 + 1/4 + 1/8 + \cdots  = \sum_{k=1}^\infty 1/2^k.$$
It can be evaluated as follows:
$$(1/2)A = 1/4 + 1/8 + \cdots.$$
So $A - (1/2)A = 1/2$ and $A = 1.$
You can get very close to the correct answer by summing the
first 20 terms, which can be done in Matlab or R. In R,
 k = 1:20; sum(1/2^k)
 ## 0.999999

Next step: Look at a calculus book or math handbook or online
to find the famous infinite series that converges to 
$e = 2.718282\dots.$ This is called a Taylor or Maclauren series.
A more general series is
$$e^a = \frac{a^0}{0!} + \frac{a^1}{1!} + \frac{a^2}{2!} + \frac{a^3}{3!} + \cdots = \sum_{k=0}^\infty \frac{a^k}{k!}.$$
This one is directly related to your series. It converges very
quickly also, meaning that summing just a few terms gets you
very close to the correct value. Again in R:
 a = 1;  k=0:20;  sum(a^k/factorial(k))
 ## 2.718282
 a = 1.5;  k=0:20;  sum(a^k/factorial(k))
 ## 4.481689
 exp(1.5)
 ## 4.481689  # e^1.5

Something like this is probably programmed into your calculator
for the $e^x$ key.
