find the series interval of convergence and, within this interval, the sum of the series as a function of…

Explanation:

Step1: Recognize the series as a geometric series

The given series (\sum_{n = 0}^{\infty}16^{n}x^{n}=\sum_{n=0}^{\infty}(16x)^{n}) is a geometric series of the form (\sum_{n = 0}^{\infty}r^{n}), where (r = 16x).

Step2: Use the formula for the radius of convergence of a geometric series

For a geometric series (\sum_{n=0}^{\infty}r^{n}), the series converges when (|r|\lt1). So, we set (|16x|\lt1). Solve the inequality (|16x|\lt1):

• First, rewrite it as (- 1\lt16x\lt1).
• Then divide each part by (16): (-\frac{1}{16}\lt x\lt\frac{1}{16}).

Step3: Check the endpoints

• When (x =-\frac{1}{16}), the series becomes (\sum_{n = 0}^{\infty}(16\times(-\frac{1}{16}))^{n}=\sum_{n=0}^{\infty}(- 1)^{n}). This series diverges (it oscillates between (1) and (0) as we add more terms).
• When (x=\frac{1}{16}), the series becomes (\sum_{n = 0}^{\infty}(16\times\frac{1}{16})^{n}=\sum_{n=0}^{\infty}1^{n}). This series also diverges (the (n) - th term does not approach (0) as (n\to\infty)).

Step4: Find the sum of the series within the interval of convergence

For a geometric series (\sum_{n = 0}^{\infty}r^{n}) with (|r|\lt1), the sum (S=\frac{1}{1 - r}). Here (r = 16x), so the sum (S(x)=\frac{1}{1-16x}) for (x\in(-\frac{1}{16},\frac{1}{16})).

Answer:

The interval of convergence is (\left(-\frac{1}{16},\frac{1}{16}\right)) and the sum of the series as a function of (x) is (S(x)=\frac{1}{1 - 16x}) for (x\in\left(-\frac{1}{16},\frac{1}{16}\right)).