Evaluating Summation of $5^{-n}$ from $n=4$ to infinity The answer is $\frac1{500}$ but I don't understand why that is so. 
I am given the fact that the summation of $x^{n}$ from $n=0$ to infinity is $\frac1{1-x}$. So if that's the case then I have that $x=\frac15$ and plugging in the values I have $\frac1{1-(\frac15)}= \frac54$.

$$
\sum_{n=4}^\infty\frac{1}{5^n}=\frac{1}{5^4}\sum_{n=4}^\infty\frac{1}{5^{n-4}}=\frac{1}{5^4}\sum_{m=0}^\infty\frac{1}{5^m}=\frac{1}{5^4}\frac{1}{1-1/5}=\frac{1}{500}.
$$
