Q:

Find the partial sums of the following series and then check their convergence:

$$\sum_{n=0}^{\infty}(\frac{2^n-1}{4^n})
$$
so i figured out that this is equal to:
$$\sum_{n=0}^{\infty}(\frac{1}{2^n})-(\frac{1}{2^{2n}})
$$
and i don't know what to do next.

A:

it is a geometric series $$\sum _{ n=0 }^{ \infty  } \left( \frac { 1 }{ 2^{ n } }  \right) -\sum _{ n=0 }^{ \infty  } \left( \frac { 1 }{ 4^{ n } }  \right) =\frac { 1 }{ 1-\frac { 1 }{ 2 }  } -\frac { 1 }{ 1-\frac { 1 }{ 4 }  } =2-\frac { 4 }{ 3 } =\frac { 2 }{ 3 } $$

