why is

$\displaystyle \sum^{\infty}_{x=1} e^{tx} \left( \frac{1}{3} \right)^{x} = \sum^{\infty}_{x=0} \left( \frac{e^t}{3} \right)^{x} -1$

I get that X which was originally equal to is now equal to 0 but wouldn;t this imply:

$\displaystyle \sum^{\infty}_{x=1} e^{tx} \left( \frac{1}{3} \right)^{x} = \sum^{\infty}_{x=0} e^{t(x+1)}\left( \frac{1}{3} \right)^{x+1}$

how do you get rid of the $\displaystyle +1$ to get -1?