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A power series is an infinite series of the form \( \sum_{n=0}^{\infty} a_n(x-c)^n \), where \( a_n \) are coefficients, \( x \) is a variable, and \( c \) is the center of the series. In the given exercise, the power series is \( f(x)=\sum_{n=0}^{\infty}\left(\frac{x}{2}\right)^{n} \). This type of series converges within a particular interval around its center. The size of this interval is determined by the radius of convergence, which tells us where the series converges.Understanding the interval where a power series converges is crucial because it indicates where a function represented by the series actually behaves like the sum of its terms. Outside this interval, the series may diverge, making it no longer represent the function adequately within those values.

The ratio test is a handy method to determine the convergence of a series. For a power series, if you have terms \( a_n \), this test involves computing the limit:\[\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|\]If this limit is less than 1, the series converges. This test is particularly useful for series that have factorials or exponential terms.In our power series \( f(x)=\sum_{n=0}^{\infty}\left(\frac{x}{2}\right)^{n} \), applying the test gives:\[R = \lim_{n \to \infty} \left| \frac{\left(\frac{x}{2}\right)^{n+1}}{\left(\frac{x}{2}\right)^n} \right| = \left| \frac{x}{2} \right| < 1\]Solving the inequality \( \left| \frac{x}{2} \right| < 1 \), we find that \( -2 < x < 2 \), giving a radius of convergence of 2.

When we differentiate a power series, each term is differentiated separately, much like differentiating polynomial terms:\[ f'(x) = \sum_{n=1}^{\infty} n a_n (x-c)^{n-1} \]Differentiating a power series doesn't change its radius of convergence. So, both \( f'(x) \) and \( f''(x) \) have the same radius of convergence as the original series \( f(x) \), which is determined by the ratio test in the solution.This is advantageous because it means we can use differentiation to find new functions with the same interval of convergence as the original one. This property simplifies dealing with power series when it involves calculus operations like differentiation.

Integration of a power series involves integrating each term of the series:\[ \int f(x) dx = \sum_{n=0}^{\infty} \frac{a_n(x-c)^{n+1}}{n+1} + C \]Where \( C \) is the constant of integration. Similar to differentiation, integrating a power series retains the original series' radius of convergence.For the series \( f(x)=\sum_{n=0}^{\infty}\left(\frac{x}{2}\right)^{n} \), the integral of this series would also converge within the interval determined by \( -2 < x < 2 \). This allows for the integration of functions represented by the series to be explored within this everyday range.Understanding that both differentiation and integration preserve the radius of convergence makes it easier to perform these operations without worrying about changing the interval over which the series is valid.