What is the error calculating the sum of this series? I need to determine if $$\sum_{n=1}^\infty \frac{2^n-1}{4^n}$$ converges, in which case I must also find its sum, or diverges. This is my approach:
$i$) $$\frac{2^n-1}{4^n}=\frac{2^n}{4^n}-\frac{1}{4^n}=\frac{2^n}{2^{2n}}-\frac{1}{4^n}=\frac{1}{2^n}-\frac{1}{4^n}$$
Then $$\sum_{n=1}^\infty \frac{2^n-1}{4^n}=\sum_{n=1}^\infty \left(\frac{1}{2^n}-\frac{1}{4^n}\right)=\sum_{n=1}^\infty \frac{1}{2^n}-\sum_{n=1}^\infty \frac{1}{4^n}$$
$$=\sum_{n=1}^\infty \frac{1}{4} \left(\frac{1}{2}\right)^{n-1}-\sum_{n=1}^\infty \frac{1}{16}\left(\frac{1}{4}\right)^{n-1}$$
$ii$) We found our series can be written as the difference of two geometric series with $|r|<1$, and the solution is
$$=\sum_{n=1}^\infty \frac{1}{4} \left(\frac{1}{2}\right)^{n-1}-\sum_{n=1}^\infty \frac{1}{16}\left(\frac{1}{4}\right)^{n-1}=\frac{\frac{1}{4}}{1-\frac{1}{2}}-\frac{\frac{1}{16}}{1-\frac{1}{4}}=\frac{5}{12}$$
I have given this series to an online series calculator, and it claims the result is not $\frac{5}{12}$ but $\frac{2}{3}$, and yet I can not find the mistake in my procedure. Can anyone point it out to me?
EDIT: Please remember, I'm asking more than "what is the right way to solve this problem"; I'm asking precisely what was the mistake I made. Correlated, but not the same.

$\sum_\limits{n=1}^{\infty} k^n = \frac {k}{1-k}\\
\sum_\limits{n=1}^{\infty} (\frac {1}{2})^n = \frac {\frac 12}{1-\frac 12} = 1\\
\sum_\limits{n=1}^{\infty} (\frac {1}{4})^n = \frac {\frac 14}{1-\frac 14} = \frac {1}{3}\\
\sum_\limits{n=1}^{\infty} (\frac {1}{2})^n - \sum_\limits{n=1}^{\infty} (\frac {1}{4})^n = \frac 23$
