Evaluating $\sum_{n=1}^\infty\frac{1}{2^{2n-1}}$

What is the summation of this geometric series?$$\sum_{n=1}^\infty\frac{1}{2^{2n-1}}$$

My main confusion is the difference between starting at n=1 versus starting at $n=0$. When you start at $n=1$ I assume that the way that you can apply the geometric sum rule is if the exponent is $n-1(x^{n-1})$. But when you start at n=0 if the exponent is just $n (x^n)$ you can apply it there as well. This summation starts at n=1 so how can I turn it into $x^{n-1}$ so that I can use the geometric sum rule.

We have that
$\sum_{n=1}^{\infty}\frac{1}{2^{2n-1}}=\sum_{n=1}^{\infty}\frac{2}{2^{2n}}=2\sum_{n=1}^{\infty}\frac{1}{4^n}=2\sum_{n=1}^{\infty}(\frac{1}{4})^n$
We want our series to begin from $n=0$ so we add and substract $(\frac{1}{4})^0=1$, hence
$2\sum_{n=1}^{\infty}(\frac{1}{4})^n=2[(\sum_{n=0}^{\infty}(\frac{1}{4})^n)-1]=2(\frac{1}{1-\frac{1}{4}}-1)=2(\frac{4}{3}-1)=\frac{2}{3}.$
We have that $$\sum_{n=0}^{\infty}(\frac{1}{4})^n=\frac{1}{1-\frac{1}{4}}$$ because this is a geometric series.