Taylor series, Convergence

Question

I need to find the radius of convergence (R) of the Taylor series expansion of the following function $f(x) =\frac{x^3 -2x+1}{x+7}$, $x_0=5$

The Taylor series of $f(x) = \frac{29}{3}+\frac{\frac{95}{18}}{1!}\left(x-5\right)+\frac{\frac{175}{108}}{2!}\left(x-5\right)^2+\frac{\frac{41}{432}}{3!}\left(x-5\right)^3+\frac{-\frac{41}{1296}}{4!}\left(x-5\right)^4+...$

But, in order to find the R, I need TS collect in the form of $\sum_{n=0}^\infty \frac{f^{(n)}(a)}{n!}x^n$. From this, I can extract $a_n$ term and use it to get $R = \frac{1}{\lim_{n\to \infty} \lvert \frac{a_{n+1}}{a_n}\rvert}$

Does anyone has an idea how to write TS in this $\sum_{n=0}^\infty \frac{f^{(n)}(a)}{n!}x^n$ form?

Or are there any other possible ways to solve it?

Thanks

Answer 1

For simplicity, I let $x=y+5$ to make $$\frac{x^3 -2x+1}{x+7}=\frac{y^3+15 y^2+73 y+116}{y+12}$$ and using the long division $$\frac{y^3+15 y^2+73 y+116}{y+12}=\frac{29}{3}+\frac{95 y}{18}+\frac{175 y^2}{216}+\frac{41 y^3}{216 (y+12)}$$

Focusing on the last term and using the classical expansion, we then have $$\frac{ y^3}{ (y+12)}=-\sum_{n=3}^\infty (-1)^n 12^{2-n} y^n$$

If you want to make life easier, , let, for the time being, $y=12 t$ to make $$\frac{41 y^3}{216 (y+12)}=\frac{82 }{3 } \frac{ t^3}{ (t+1)}$$

Answer 2

By long division, the expression is reworked as the sum of a polynomial and a remainder term

$$\frac c{x+7}.$$

The polynomial plays no role, and the derivatives of $(x+7)^{-1}$ at $x=5$ are $(-1)^k\,k!\,12^{-k-1}$, hence for $k>2$, the terms of the Taylor development are $$c\frac{(-1)^k}{12}\left(\frac{x-5}{12}\right)^k.$$