91Ó°ÊÓ

First, we must rewrite the function f(x) with base e in order to differentiate it: \(f(x)=\frac{\ln(x+1)}{\ln(3)}\). Then, find the first four derivatives at x=0: \(f(x) = \frac{\ln(x+1)}{\ln(3)}\) \(f'(x) = \frac{1}{(x+1)\ln(3)}\) \(f''(x) = -\frac{1}{(x+1)^2\ln(3)}\) \(f'''(x) = \frac{2}{(x+1)^3\ln(3)}\) \(f^{(4)}(x) = -\frac{6}{(x+1)^4\ln(3)}\)

Now we'll evaluate the first four derivatives at x=0: \(f(0) = \frac{\ln(1)}{\ln(3)} = 0\) \(f'(0) = \frac{1}{(0+1)\ln(3)} = \frac{1}{\ln(3)}\) \(f''(0) = -\frac{1}{(0+1)^2\ln(3)} = -\frac{1}{\ln(3)}\) \(f'''(0) = \frac{2}{(0+1)^3\ln(3)} = \frac{2}{\ln(3)}\) \(f^{(4)}(0) = -\frac{6}{(0+1)^4\ln(3)} = -\frac{6}{\ln(3)}\)

Use the derivatives evaluated at x=0 to write the first four nonzero terms of the Maclaurin series: \(f(x) \approx \frac{1}{\ln(3)}x -\frac{1}{2\ln(3)}x^2 +\frac{2}{3! \ln(3)}x^3 -\frac{6}{4!\ln(3)}x^4\)

Express the function in terms of summation notation starting from n=1 since the first term is 0: \(f(x) = \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n\ln(3)} x^n\)

To determine the interval of convergence, we'll use the Ratio Test: \(\lim_{n\to\infty} \frac{\frac{1}{(n+1)\ln(3)}x^{n+1}}{\frac{1}{n\ln(3)}x^n} = \lim_{n\to\infty} \frac{n}{n+1} |x|\) As \(n \to \infty\), the expression becomes \(|x|\). Since the Ratio Test requires the limit to be less than 1, we have: \(|x| < 1 \Rightarrow -1<x<1\) Now, we need to verify the convergence at the endpoints x=-1 and x=1. At x = -1, the series becomes: \(\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n\ln(3)} (-1)^n =\sum_{n=1}^\infty \frac{(-1)^{2n-1}}{n\ln(3)}\) This series converges since it's an alternating series with decreasing terms, and the limit of the terms as n approaches infinity is 0: At x = 1, the series becomes: \(\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n\ln(3)}\) This is an alternating harmonic series, which also converges. Thus, the interval of convergence is \(-1\leq x\leq1\).