Let's solve the problem step by step: --- **Step 1: Simplify the Function** Given: \( f(x) = \ln \left(\frac{1+x}{1-x}\right) \) First, recall the logarithm property: \[ \ln\left(\frac{a}{b}\right) = \ln(a) - \ln(b) \] So, \[ f(x) = \ln(1+x) - \ln(1-x) \] **Explanation:** We rewrote the original function as the difference of two logarithms, which will help us use known power series for \(\ln(1+x)\). --- **Step 2: Recall Power Series for \(\ln(1+x)\) and \(\ln(1-x)\)** The power series expansion for \(\ln(1+x)\) centered at \(x=0\) is: \[ \ln(1+x) = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} x^n \] For \(\ln(1-x)\), substitute \(-x\) into the above series: \[ \ln(1-x) = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} (-x)^n = -\sum_{n=1}^{\infty} \frac{x^n}{n} \] **Explanation:** We expressed both \(\ln(1+x)\) and \(\ln(1-x)\) as power series, preparing for their subtraction. --- **Step 3: Combine the Series** Now, subtract the series: \[ f(x) = \ln(1+x) - \ln(1-x) = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} x^n - \left(-\sum_{n=1}^{\infty} \frac{x^n}{n}\right) = \sum_{n=1}^{\infty} \left[\frac{(-1)^{n+1}}{n} + \frac{1}{n}\right] x^n \] Simplify the coefficient: \[ \frac{(-1)^{n+1}+1}{n} \] This expression is 2/n when n is odd (since \((-1)^{n+1}+1 = 2\)), and 0 when n is even. So, \[ f(x) = 2\sum_{k=0}^{\infty} \frac{x^{2k+1}}{2k+1} \] **Explanation:** By combining and simplifying the series, we found that only odd powers of \(x\) remain, giving us a compact power series representation. --- ### **Final Answer:** \[ \boxed{ f(x) = \ln\left(\frac{1+x}{1-x}\right) = 2\sum_{n=0}^{\infty} \frac{x^{2n+1}}{2n+1} } \] This is the power series representation for \(f(x)\) centered at \(x=0\). --- If you have questions about the steps or want it written in the same sum notation as the image, let me know!