Answer
The series converges for all $x$ in the interval $(-2,4)$.
Work Step by Step
We apply the ratio test
$$
\rho=\lim _{n \rightarrow \infty}\left|\frac{a_{n+1}}{a_{n}}\right|=\lim _{n \rightarrow \infty} |\frac{ (n+1)(x-3)^{n+1} }{ n(x-3)^{n}}|\\
=|x-3|\lim _{n \rightarrow \infty} \frac{n+1}{n}=|x-3|.
$$
Hence, the series $\Sigma_{n=1}^{\infty} n(x-3)^{n}$ converges if and only if $|x-3|\lt1$. That is, the interval of convergence is $(-2,4)$.
Now, we check the end points:
At $x=-2$, then
$\Sigma_{n=1}^{\infty} n(x-3)^{n}=\Sigma_{n=2}^{\infty}(-1)^{n}5 n$, diverges (Alternating Series Test).
At $x=4$, then
$\Sigma_{n=1}^{\infty} n(x-3)^{n}=\Sigma_{n=1}^{\infty} n$, diverges (by the limit test ).
Hence, the series converges for all $x$ in the interval $(-2,4)$.
