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Consider the following geometric series; inf,n=1 4^2n / 5^n

a) identify value of initial term a
(b) identify value of common ratio r
(c) does the series converge or diverge and why
(d) if the series converge find sum
answer in fractions

4^(2n)/5^n = (16/5)^n

(a) and (b) are easy
(c) r>1 so it diverges

Did you know?

Did you know that the given geometric series, inf,n=1 4^2n / 5^n, has interesting properties? Let's explore them: (a) To identify the value of the initial term (a), we can substitute n = 1 into the series. Thus, a = 4^2(1) / 5^1 = 16/5. (b) To identify the value of the common ratio (r), let's divide the following term by the previous one. The ratio would be (4^2(n+1) / 5^(n+1)) divided by (4^2n / 5^n). Simplifying this expression, the common ratio is found to be 4^2 / 5. (c) Whether the series converges or diverges can be determined by examining the common ratio. If the absolute value of the common ratio (4^2 / 5) is less than 1, the series converges; otherwise, it diverges. (d) Assuming the series converges, we can find the sum using the formula S = a / (1 - r), where S represents the sum, a is the initial term, and r is the common ratio. Plugging in the values, S = (16/5) / (1 - (4^2 / 5)).