The prompt is to determine Maclaurin series of the function $f(x) = -x^5ln(1-2x^3)$ using the Maclaurin series for $g(x) = ln(x+1)$ also to find for what value of x the series f(x) converges, and hence evaluate $f^{100}(0)$ and $f^{101}(0)$.
Here's what I tried, $$g(x) = ln(1+x) = \sum_{n = 1}^{\infty} \frac{(-1)^{n + 1}}{n}x^n$$ $$f(x) =-x^5ln(1 + (-2x^3)) = \sum_{n = 1}^{\infty} \frac{(-x^5)(-1)^{n + 1}}{n}(-2x)^{3n}$$
Using ratio test on f(x) $$a_n = \frac{(-x^5)(-1)^{n + 1}}{n}(-2x)^{3n}$$ $$a_{n+1} = \frac{(-x^5)(-1)^{n + 1}(-1)}{n + 1}(-2x)^{3n}(-2x^3)$$
Applying limits and cancelling terms on numerator and denominator,
$$|2x^3|\lim\limits_{n \to \infty} \frac{n}{n+1}$$
$$|2x^3| < 1$$ $$\frac{-1}{2} < x < \frac{1}{2}$$
To check the border values to find the interval of convergence, $$f(x) =-x^5ln(1 + (-2x^3)) = \sum_{n = 1}^{\infty} \frac{(-x^5)(-1)^{n + 1}}{n}(-2x)^{3n}$$
$$f(\frac{-1}{2}) = \frac{(\frac{-1}{2})^5(-1)^{n + 1}(-2)(\frac{-1}{2})^{3n}}{n} = \frac{-2}{n}$$
$$\lim\limits_{n \to \infty} \frac{-1}{n} = 0$$
This means f(x) converges at $\frac{-1}{2}$ Testing $f(\frac{1}{2})$ now,
$$f(\frac{1}{2}) = \frac{(\frac{1}{2})^5(-1^{n + 1})(-2 \frac{-1}{2})^{3n}}{n} = \frac{(-2)(-1)^{3n}}{n}$$
$$\lim\limits_{n \to \infty} \frac{(-2)}{n} = 0$$ Hence, series converges at $x = \frac{1}{2}$ as well, Interval of convergence is $[\frac{-1}{2}, \frac{1}{2}]$
Finding the n-th derivatives using the series, $$ f^{100}(0) = \frac{-(0)^5(-1)^{n+1}}{100}.(-2)(0)^{300} = 0$$ $$ f^{101}(0) = \frac{-(0)^5(-1)^{n+1}}{101}.(-2)(0)^{303} = 0$$