# Solving an infinite summation involving exponential and factorial

I'm trying to understand an equality that I found in this biology article.

$$\sum_{i=0}^\infty\frac{e^{-x}x^i(1-y)^i}{i!} = e^{-x\cdot y}$$

Can you help me proving this equation holds true?

The factor $e^{-x}$ does not change as $i$ runs through the list $0,1,2,3,\ldots$, so it can be pulled out, getting $$e^{-x}\sum_{i=1}^\infty\cdots\cdots.$$
Then you have $\displaystyle\sum_{i=0}^\infty \frac{a^i}{i!} = e^a$, where $a=x(1-y)$.
• Why does $\sum \frac{a^i}{i!}=e^a$? Commented Feb 7, 2014 at 0:52
• @Remi.b : Maybe you should make that a separate question. In comments I'll say that it's a standard identity. Try showing that if $f(x)=\sum_{k=0}^\infty \frac{c_k x^k}{k!}$ then $f^{(k)}(0)= c_k$, and figure out what $f^{(k)}(0)$ should be if $f(x) = e^x$. ${}\qquad{}$ Commented Feb 7, 2014 at 1:02
There is a well known series expansion for $e^t$ for $t$ a real number.
Now $\sum\limits_{i=0}^{\infty}\dfrac{(x(1-y))^i}{i!}=e^{x(1-y)}$. Since $e^{-x}$ factor doesn't depend on $i$ we can pull it out. So we have that your = $e^{-x}e^{x-xy}$. Now use $e^{a+b}=e^ae^b$ to get your conclusion.