Answer
The radius of convergent: $1$
The interval of convergence: $(2,4]$
Work Step by Step
Let $a_n=(-1)^n\frac{(x-3)^n}{2n+1}$.
It can be checked $\frac{a_{n+1}}{a_n}=\frac{-(2n+1)(x-3)}{2n+3}$.
We know that $\lim \frac{-(2n+1)}{2n+3}=-1$.
Then, $\lim|\frac{a_{n+1}}{a_n}|=|x-3|$.
Find the interval of convergence:
$|x-3|<1$
$(2,4]$
