# Write out the first eight terms of each series to show how the series starts. Then find the sum of the series or show that it diverges. sum_{n=2}^inftyfrac{1}{4^n}

Question
Series
Write out the first eight terms of each series to show how the series starts. Then find the sum of the series or show that it diverges.
$$\sum_{n=2}^\infty\frac{1}{4^n}$$

2021-03-08
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: $$\frac{1}{12}$$

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