Find a power series representation for the function. (Give your power series representation centered at $x=0$.) \[ \begin{array}{r} f(x)=\frac{2}{5-x} \\ f(x)=\sum_{n=0}^{\infty}(\square) \end{array} \] Determine the interval of convergence. (Enter your answer using interval notation.)
Solution
Step 1 ：First, we recognize that the function \(f(x)=\frac{2}{5-x}\) is a geometric series with first term \(a = \frac{2}{5}\) and common ratio \(r = x\).
Step 2 ：We can write the function as a power series using the formula for the sum of a geometric series, \(\sum_{n=0}^{\infty} ar^n\), where \(a\) is the first term and \(r\) is the common ratio.
Step 3 ：Substituting \(a = \frac{2}{5}\) and \(r = x\) into the formula, we get \(f(x) = \sum_{n=0}^{\infty} \frac{2}{5}x^n\).
Step 4 ：The interval of convergence for a geometric series is \(-1 < r < 1\). Substituting \(r = x\), we get \(-1 < x < 1\).
Step 5 ：However, the endpoints need to be checked separately. When \(x = -1\), the series becomes \(\sum_{n=0}^{\infty} (-1)^n\), which does not converge. When \(x = 1\), the series becomes \(\sum_{n=0}^{\infty} 1\), which also does not converge.
Step 6 ：Therefore, the interval of convergence is \(-1 < x < 1\), or in interval notation, \((-1, 1)\).
Step 7 ：So, the power series representation for the function \(f(x)=\frac{2}{5-x}\) centered at \(x=0\) is \(f(x) = \sum_{n=0}^{\infty} \frac{2}{5}x^n\), and the interval of convergence is \((-1, 1)\).
Step 8 ：\(\boxed{f(x) = \sum_{n=0}^{\infty} \frac{2}{5}x^n, (-1, 1)}\)
