Sum of powered fractions to infinity!

Question

I'm trying to find out the result of $$\lim_{n\rightarrow\infty}\left(\frac 14+\frac 1{4^2}+\frac 1{4^3}+\ldots+\frac 1{4^{n-1}}\right)\tag{a}$$ I know the result probably is $\frac{1}{3}$, but what are the steps for the calculation?

Answer 1

$$\sum_{k=1}^n x^k = \frac{x(x^n-1)}{x-1}$$

since this is just a geometric series. As such,

$$\sum_{k=1}^\infty \frac{1}{4^k}=\lim_{n\to\infty}\frac{\frac{1}{4}(\frac{1}{4^n}-1)}{\frac{1}{4}-1}=\frac{\frac{1}{4}}{1-\frac{1}{4}}=\frac{1}{3}$$