\(1;\ \frac{1}{4};\ \frac{1}{16};\ \frac{1}{64};...\\\\a_1=\frac{1}{4^{1-1}}=\frac{1}{4^0}=\frac{1}{1}=1\\\\a_2=\frac{1}{4^{2-1}}=\frac{1}{4^1}=\frac{1}{4}\\\\a_3=\frac{1}{4^{3-1}}=\frac{1}{4^2}=\frac{1}{16}\\\\a_4=\frac{1}{4^{4-1}}=\frac{1}{4^3}=\frac{1}{64}\\\vdots\\a_n=\frac{1}{4^{n-1}}\)
\(\lim\limits_{n\to\infty}\frac{1}{4^{n-1}}=\lim\limits_{n\to\infty}\frac{1}{4^n\cdot4^{-1}}=\lim\limits_{n\to\infty}\frac{1}{4^n\cdot\frac{1}{4}}=\lim\limits_{n\to\infty}\frac{4}{4^n}=0\\\\\\Answer:A\)
