So I have a taylor series $f(x)=\ln(x)$ at $a=8$ is given by: the sum from $0$ to infinity of $cn(x-8)$ . (see photo)
I have all the coefficients but, how do I rewrite a taylor polynomial in a ratio of polynomials so that I can use power series and ratio test to find the interval of convergence ?
The ratio of polynomials I have from the coefficients are
$((-1)^n1*?) / (8^n*n!)$
This is for freshman calculus
$$f(x) = \ln(x)$$
$$f'(x)=\frac1x, c_1 = \frac1{8\cdot 1!}$$
$$f''(x)=-\frac1{x^2}, c_2 = -\frac1{8^2\cdot 2!}$$ $$f^{(3)}(x)=\frac2{x^3}, c_3 = \frac{2}{8^3\cdot 3!}=\frac{1}{8^3\cdot 2}$$
$$f^{(4)}(x) = -\frac{3!}{x^4}, c_4 = -\frac{3!}{8^4\cdot 4!}=-\frac1{8^4\cdot 4}$$
For $n\ge 1$, we have $c_n=\frac{(-1)^{n-1}}{8^n \cdot n}$
Hence $$f(x) = \ln (8) + \sum_{n=0}^\infty \frac{(-1)^{n-1}}{8^n \cdot n}\cdot (x-8)^n$$
and it converges if $$\lim_{n\to \infty}\left|\frac{\frac{(-1)^n}{8^{n+1}(n+1)}(x-8)^{n+1}}{\frac{(-1)^{n-1}}{8^nn}(x-8)^n} \right|<1$$
$$\frac{|x-8|}8<1$$
Also, remember to check the boundary of the interval.