# Infinite sum with factorial

I am learning about Poisson Random Variable and came across a problem with this infinite sum:

$\sum^{∞}_{n=2}\frac{e^{-2}(2)^i}{i!}$

The first step I do is move the constant $e^{-2}$ out.

$e^{-2} * (\sum^{∞}_{n=2}\frac{(2)^i}{i!})$

But I don't know what is the next step after this, I am trying to use this formula: $\frac{a}{1-r}$

I looked through my notes from class, but I can only find basic example that cover $\sum^{∞}_{n=1} 2x$. In the this case, what would be the ratio?

• Hint: What is the Taylor series for the exponential function? – Robert Israel Mar 10 '16 at 23:50
• HINT: $$\sum_{i=0}^{\infty} \frac{x^i}{i!} = e^x$$ – Crostul Mar 10 '16 at 23:51
• Shouldn't the $n$ in the index of the sum be an $i$? – Ryan Goulden Mar 30 '17 at 6:18

Using the series expansion for the exponential function, $$e^{x} = \sum_{r=0}^{\infty} \frac{x^{r}}{r!},$$ then by subtracting the first two terms from both sides: $$\sum_{r=2}^{\infty} \frac{x^{r}}{r!} = e^{x} - 1 - x.$$
It will certainly help to know that $$e^x=\sum_{i=0}^\infty \frac{x^i}{i!}$$
Just fill in $x=2$ and work from there!
The equality I use is called the taylor series at $0$ or the maclaurin series of $e^x$. It's not hard to derive.