Find the radius of convergence of the Taylor series of $f(x) = \frac{x-3}{x+2}\ln(5+x)$ at $x=0$ I need find radius of convergence for Taylor series in $x = 0$ (over $\mathbb{R}$) and find $x$'s at which series converges to $f$
$$f(x) = \frac{x-3}{x+2}\ln(5+x)$$
My solution
$\ln(5+x) = -\ln(5)\ln(1+\frac{x}{5}) = -\ln(5)\sum_{n \geq 1} \frac{(-1)^{n+1}}{5^nn}x^n, - \frac{1}{5} < x \leq \frac{1}{5}$
$\frac{x-3}{x+2} = \frac{x-3}{2} \sum \frac{ (-1)^n}{2^n}x^n, -\frac{1}{2} < x < \frac{1}{2}$
$\implies$ answer for both questions is intersection: $\frac{1}{5} < x \leq \frac{1}{5}$?

There are several mistakes in your computation. First of all
$$
\log(5+x)=\log5+\log\Bigl(1+\frac x5\Bigr).
$$
The Taylos series of $\log(1+x/5)$ converges for $|x/5|<1$, that is, $-5<x<5$. Similarly the Taylor series of $1/(x+2)$ converges for $|x|<2$.
