Answer
$$ \frac{4}{5}$$
Work Step by Step
Given $$1-\frac{1}{4}+\frac{1}{4^{2}}-\frac{1}{4^{3}}+\cdots$$
Since
$$1-\frac{1}{4}+\frac{1}{4^{2}}-\frac{1}{4^{3}}+\cdots=\sum_{n=0}^{\infty} \left(\frac{-1}{4}\right)^n $$
This is a geometric series with $|r|= \frac{1}{4}<1$, so the series converges and has the sum
\begin{align*}
S&=\frac{a}{1-r}\\
&=\frac{1}{1+\frac{1}{4}}\\
&= \frac{4}{5}
\end{align*}
