First eight terms are:

\(\frac{1}{4^2},\frac{1}{4^3},\frac{1}{4^4},\frac{1}{4^5},\frac{1}{4^6},\frac{1}{4^7},\frac{1}{4^8},\frac{1}{4^9}\)

\(=\frac{1}{16},\frac{1}{64},\frac{1}{256},\frac{1}{1024},\frac{1}{4096},\frac{1}{16384},\frac{1}{65536},\frac{1}{262144}\)

So we see that the denominator is increasing, this means the terms are approaching 0

So the series will converge

That is a geometric series with common ratio \(=\frac14\)

And first term \(=\frac{1}{16}\)

Then we use the infinite geometric series sum formula

\(\frac{a}{1-r}=\frac{\frac{1}{16}}{1-\frac14}=\frac{\frac{1}{16}}{\frac34}=\frac{1}{16}\times\frac43=\frac{1}{12}\)

Answer: \(\frac{1}{12}\)

\(\frac{1}{4^2},\frac{1}{4^3},\frac{1}{4^4},\frac{1}{4^5},\frac{1}{4^6},\frac{1}{4^7},\frac{1}{4^8},\frac{1}{4^9}\)

\(=\frac{1}{16},\frac{1}{64},\frac{1}{256},\frac{1}{1024},\frac{1}{4096},\frac{1}{16384},\frac{1}{65536},\frac{1}{262144}\)

So we see that the denominator is increasing, this means the terms are approaching 0

So the series will converge

That is a geometric series with common ratio \(=\frac14\)

And first term \(=\frac{1}{16}\)

Then we use the infinite geometric series sum formula

\(\frac{a}{1-r}=\frac{\frac{1}{16}}{1-\frac14}=\frac{\frac{1}{16}}{\frac34}=\frac{1}{16}\times\frac43=\frac{1}{12}\)

Answer: \(\frac{1}{12}\)