Let [tex]S_k[/tex] be the k-th partial sum of the infinite series,
[tex]\displaystyle S_k = \sum_{n=1}^k \frac23 \left(\frac14\right)^{n-1} = \frac23 \left(1 + \frac14 + \frac1{4^2} + \cdots + \frac1{4^{k-1}}\right)[/tex]
Multiply both sides by 1/4 :
[tex]\displaystyle \frac14 S_k = \frac23 \left(\frac14 + \frac1{4^2} + \frac1{4^3} + \cdots + \frac1{4^k}\right)[/tex]
Subtract this from [tex]S_k[/tex] and solve for [tex]S_k[/tex] :
[tex]\displaystyle S_k - \frac14 S_k = \frac23 \left(1 - \frac1{4^k}\right)[/tex]
[tex]\displaystyle \frac34 S_k = \frac23 \left(1 - \frac1{4^k}\right)[/tex]
[tex]\displaystyle S_k = \frac89 \left(1 - \frac1{4^k}\right)[/tex]
Then as k goes to infinity, the exponential term will converge to zero, and the sum will converge to
[tex]\displaystyle S = \lim_{k\to\infty} S_k = \boxed{\frac89}[/tex]
Generalizing this result, we have for |r| < 1,
[tex]\displaystyle \sum_{n=1}^\infty a r^{n-1} = \frac a{1-r}[/tex]
